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Ch 6Equilibriumसाम्यावस्थाEquilibrium

Physical and chemical equilibria, the equilibrium constant (Kc, Kp), using K and Q, Gibbs energy, Le Chatelier’s principle, acids and bases, pH, ionisation constants, salt hydrolysis, buffers and solubility product.भौतिक और रासायनिक साम्य, साम्य स्थिरांक (Kc, Kp), K और Q का उपयोग, गिब्ज़ ऊर्जा, ला-शातैलिए सिद्धांत, अम्ल और क्षारक, pH, आयनन स्थिरांक, लवण जल-अपघटन, बफ़र और विलेयता गुणनफल।Physical aur chemical equilibrium, equilibrium constant (Kc, Kp), K aur Q ka use, Gibbs energy, Le Chatelier principle, acids aur bases, pH, ionisation constants, salt hydrolysis, buffers aur solubility product.

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👆 Words with a open our interactive diagram (tap any label for its name + function). 3D opens the 3D model.👆 वाले शब्द पर टैप करें — इंटरैक्टिव डायग्राम खुलेगा (हर लेबल पर टैप = नाम + कार्य)। 3D = 3D मॉडल।👆 wale word par tap kijiye — interactive diagram khulega (har label par tap = naam + kaam). 3D = 3D model.
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Diagrams in this chapterइस अध्याय के डायग्रामIs chapter ke diagrams

Attainment of chemical equilibriumरासायनिक साम्य की प्राप्तिAttainment of chemical equilibriumFig 6.4Reaction quotient Q vs equilibrium constant Kअभिक्रिया भागफल Q बनाम साम्य स्थिरांक KReaction quotient Q vs equilibrium constant KFig 6.7Le Chatelier’s principle: effect of changesला-शातैलिए सिद्धांत: परिवर्तनों का प्रभावLe Chatelier’s principle: effect of changesSec 6.8Conjugate acid–base pairsसंयुग्मी अम्ल–क्षार युग्मConjugate acid–base pairsSec 6.10.2The pH scalepH स्केलThe pH scaleTable 6.5
In this chapter (NCERT sections)इस अध्याय में (NCERT खंड)Is chapter mein (NCERT sections)
  1. 6.1Equilibrium in physical processesभौतिक प्रक्रमों में साम्यPhysical processes mein equilibrium
  2. 6.2Equilibrium in chemical processes — dynamic equilibriumरासायनिक प्रक्रमों में साम्य — गतिक साम्यChemical processes mein equilibrium — dynamic equilibrium
  3. 6.3–6.5Law of equilibrium and the equilibrium constantसाम्य नियम और साम्य स्थिरांकEquilibrium ka law aur equilibrium constant
  4. 6.6Applications of equilibrium constantsसाम्य स्थिरांकों के अनुप्रयोगEquilibrium constants ke applications
  5. 6.7Relation between K, Q and Gibbs energyK, Q और गिब्ज़ ऊर्जा का संबंधK, Q aur Gibbs energy ka sambandh
  6. 6.8Factors affecting equilibria — Le Chatelier’s principleसाम्य को प्रभावित करने वाले कारक — ला-शातैलिए सिद्धांतEquilibrium par asar daalne wale factors — Le Chatelier principle
  7. 6.9–6.10Ionic equilibrium; acids, bases and saltsआयनिक साम्य; अम्ल, क्षारक और लवणIonic equilibrium; acids, bases aur salts
  8. 6.11Ionisation of acids and bases; pHअम्लों और क्षारकों का आयनन; pHAcids aur bases ka ionisation; pH
  9. 6.12Buffer solutionsबफ़र विलयनBuffer solutions
  10. 6.13Solubility equilibria of sparingly soluble saltsअल्प विलेय लवणों का विलेयता साम्यKam ghulne wale salts ka solubility equilibrium
  11. ★Yaad rakho — quick revisionयाद रखो — झटपट दोहरावYaad rakho — quick revision
  12. ⚠Common mistakesआम गलतियाँCommon mistakes
  13. ✎NCERT questions — short answersNCERT प्रश्न — संक्षिप्त उत्तरNCERT questions — short answers
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6.1Equilibrium in physical processesभौतिक प्रक्रमों में साम्यPhysical processes mein equilibrium

In a closed system a process can run both ways at once. When the two rates become equal, the measurable properties stop changing — this is equilibrium, and it is dynamic. Examples: ice ⇌ water at 273 K and 1 atm (the normal melting point); water ⇌ vapour in a closed box (the pressure stops rising at the equilibrium vapour pressure; at 1.013 bar water boils at 100 °C); solid iodine ⇌ iodine vapour (sublimation); sugar(solid) ⇌ sugar(solution) in a saturated solution; CO₂(g) ⇌ CO₂(aq) in a soda bottle (Henry’s law: gas solubility ∝ pressure).बंद निकाय में कोई प्रक्रम एक साथ दोनों दिशाओं में चल सकता है। जब दोनों दरें बराबर हो जाती हैं, तो मापने योग्य गुण बदलना बंद कर देते हैं — यही साम्य है, और यह गतिक होता है। उदाहरण: 273 K व 1 atm पर बर्फ़ ⇌ जल (सामान्य गलनांक); बंद डिब्बे में जल ⇌ वाष्प (दाब साम्य वाष्प दाब पर रुक जाता है; 1.013 bar पर जल 100 °C पर उबलता है); ठोस आयोडीन ⇌ आयोडीन वाष्प (ऊर्ध्वपातन); संतृप्त विलयन में चीनी(ठोस) ⇌ चीनी(विलयन); सोडा बोतल में CO₂(g) ⇌ CO₂(aq) (हेनरी नियम: गैस की विलेयता ∝ दाब)।Closed system mein ek process ek saath dono taraf chal sakta hai. Jab dono rates barabar ho jaate hain, to naapne layak properties badalna band kar deti hain — yahi equilibrium hai, aur ye dynamic hota hai. Examples: 273 K aur 1 atm par barf ⇌ paani (normal melting point); band dibbe mein paani ⇌ vapour (pressure equilibrium vapour pressure par ruk jaata hai; 1.013 bar par paani 100 °C par ubalta hai); solid iodine ⇌ iodine vapour (sublimation); saturated solution mein cheeni(solid) ⇌ cheeni(solution); soda bottle mein CO₂(g) ⇌ CO₂(aq) (Henry law: gas ki solubility ∝ pressure).

6.2Equilibrium in chemical processes — dynamic equilibriumरासायनिक प्रक्रमों में साम्य — गतिक साम्यChemical processes mein equilibrium — dynamic equilibrium

Attainment of chemical equilibriumरासायनिक साम्य की प्राप्तिAttainment of chemical equilibriumNCERT Fig 6.4 · 4 tappable partsNCERT चित्र 6.4 · 4 टैप करने लायक हिस्सेNCERT Fig 6.4 · 4 tappable partsOpen

For a reversible reaction A + B ⇌ C + D, the forward rate falls as reactants are used up and the backward rate rises as products form. At chemical equilibrium the two rates are equal and all concentrations stay constant. Haber’s experiment with deuterium (N₂ + 3D₂ ⇌ 2ND₃ added to an H₂/NH₃ equilibrium mixture) showed that D spreads into NH₃ and H₂ — proof that the reactions keep going (dynamic). The same equilibrium is reached from either side (H₂ + I₂ or pure HI).उत्क्रमणीय अभिक्रिया A + B ⇌ C + D में अभिकारक घटने से अग्र दर कम होती है और उत्पाद बनने से पश्च दर बढ़ती है। रासायनिक साम्य पर दोनों दरें बराबर होती हैं और सभी सांद्रताएँ स्थिर रहती हैं। ड्यूटीरियम वाले हाबर के प्रयोग (H₂/NH₃ साम्य मिश्रण में N₂ + 3D₂ ⇌ 2ND₃ मिलाना) से D, NH₃ और H₂ में फैल गया — यह प्रमाण है कि अभिक्रियाएँ चलती रहती हैं (गतिक)। वही साम्य किसी भी ओर से प्राप्त होता है (H₂ + I₂ या शुद्ध HI)।Reversible reaction A + B ⇌ C + D mein reactants ghatne se forward rate kam hota hai aur products banne se backward rate badhta hai. Chemical equilibrium par dono rates barabar hote hain aur saari concentrations constant rehti hain. Deuterium wale Haber experiment (H₂/NH₃ equilibrium mixture mein N₂ + 3D₂ ⇌ 2ND₃ milana) se D, NH₃ aur H₂ mein phail gaya — saboot ki reactions chalti rehti hain (dynamic). Wahi equilibrium kisi bhi taraf se milta hai (H₂ + I₂ ya pure HI).

6.3–6.5Law of equilibrium and the equilibrium constantसाम्य नियम और साम्य स्थिरांकEquilibrium ka law aur equilibrium constant

aA + bB ⇌ cC + dD: Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
law of chemical equilibrium (Guldberg and Waage); concentrations at equilibriumरासायनिक साम्य का नियम (गुल्डबर्ग और वागे); साम्य पर सांद्रताएँchemical equilibrium ka law (Guldberg aur Waage); equilibrium par concentrations
Kp = Kc (RT)<sup>Δn</sup>
Δn = moles of gaseous products − gaseous reactants; R = 0.0831 bar L mol⁻¹ K⁻¹Δn = गैसीय उत्पादों के मोल − गैसीय अभिकारकों के मोल; R = 0.0831 bar L mol⁻¹ K⁻¹Δn = gaseous products ke moles − gaseous reactants ke moles; R = 0.0831 bar L mol⁻¹ K⁻¹
NCERT Problems 6.1–6.66.1 N₂ + 3H₂ ⇌ 2NH₃ at 500 K, [N₂] = 1.5 × 10⁻², [H₂] = 3.0 × 10⁻², [NH₃] = 1.2 × 10⁻² M: Kc = (1.2 × 10⁻²)² / [(1.5 × 10⁻²)(3.0 × 10⁻²)³] = 3.55 × 10².
6.2 N₂ + O₂ ⇌ 2NO, [N₂] = 3.0 × 10⁻³, [O₂] = 4.2 × 10⁻³, [NO] = 2.8 × 10⁻³ M: Kc = (2.8 × 10⁻³)² / (3.0 × 10⁻³ × 4.2 × 10⁻³) = 0.622.
6.3 PCl₅ ⇌ PCl₃ + Cl₂ at 500 K, 1.59 M PCl₃, 1.59 M Cl₂, 1.41 M PCl₅: Kc = 1.59²/1.41 = 1.79.
6.4 2NOCl ⇌ 2NO + Cl₂, Kc = 3.75 × 10⁻⁶ at 1069 K, Δn = 1: Kp = 3.75 × 10⁻⁶ × (0.0831 × 1069) = 3.33 × 10⁻⁴.
6.5 CO + H₂O ⇌ CO₂ + H₂, Kc = 4.24 at 800 K, 0.10 M CO and H₂O at start: x²/(0.1 − x)² = 4.24 → x = 0.067 M. [CO₂] = [H₂] = 0.067 M, [CO] = [H₂O] = 0.033 M.
6.6 CO₂(g) + C(s) ⇌ 2CO(g), Kp = 3.0 at 1000 K, p(CO₂) = 0.48 bar at start: (2x)²/(0.48 − x) = 3 → 4x² + 3x − 1.44 = 0 → x = 0.33. p(CO) = 0.66 bar, p(CO₂) = 0.15 bar.NCERT प्रश्न 6.1–6.66.1 500 K पर N₂ + 3H₂ ⇌ 2NH₃, [N₂] = 1.5 × 10⁻², [H₂] = 3.0 × 10⁻², [NH₃] = 1.2 × 10⁻² M: Kc = (1.2 × 10⁻²)² / [(1.5 × 10⁻²)(3.0 × 10⁻²)³] = 3.55 × 10²।
6.2 N₂ + O₂ ⇌ 2NO: Kc = (2.8 × 10⁻³)² / (3.0 × 10⁻³ × 4.2 × 10⁻³) = 0.622।
6.3 500 K पर PCl₅ ⇌ PCl₃ + Cl₂, 1.59 M PCl₃, 1.59 M Cl₂, 1.41 M PCl₅: Kc = 1.59²/1.41 = 1.79।
6.4 2NOCl ⇌ 2NO + Cl₂, 1069 K पर Kc = 3.75 × 10⁻⁶, Δn = 1: Kp = 3.75 × 10⁻⁶ × (0.0831 × 1069) = 3.33 × 10⁻⁴।
6.5 CO + H₂O ⇌ CO₂ + H₂, 800 K पर Kc = 4.24, आरंभ में 0.10 M CO व H₂O: x²/(0.1 − x)² = 4.24 → x = 0.067 M। [CO₂] = [H₂] = 0.067 M, [CO] = [H₂O] = 0.033 M।
6.6 CO₂(g) + C(s) ⇌ 2CO(g), 1000 K पर Kp = 3.0, आरंभ में p(CO₂) = 0.48 bar: (2x)²/(0.48 − x) = 3 → x = 0.33। p(CO) = 0.66 bar, p(CO₂) = 0.15 bar।NCERT Problems 6.1–6.66.1 500 K par N₂ + 3H₂ ⇌ 2NH₃, [N₂] = 1.5 × 10⁻², [H₂] = 3.0 × 10⁻², [NH₃] = 1.2 × 10⁻² M: Kc = (1.2 × 10⁻²)² / [(1.5 × 10⁻²)(3.0 × 10⁻²)³] = 3.55 × 10².
6.2 N₂ + O₂ ⇌ 2NO: Kc = (2.8 × 10⁻³)² / (3.0 × 10⁻³ × 4.2 × 10⁻³) = 0.622.
6.3 500 K par PCl₅ ⇌ PCl₃ + Cl₂, 1.59 M PCl₃, 1.59 M Cl₂, 1.41 M PCl₅: Kc = 1.59²/1.41 = 1.79.
6.4 2NOCl ⇌ 2NO + Cl₂, 1069 K par Kc = 3.75 × 10⁻⁶, Δn = 1: Kp = 3.75 × 10⁻⁶ × (0.0831 × 1069) = 3.33 × 10⁻⁴.
6.5 CO + H₂O ⇌ CO₂ + H₂, 800 K par Kc = 4.24, shuru mein 0.10 M CO aur H₂O: x²/(0.1 − x)² = 4.24 → x = 0.067 M. [CO₂] = [H₂] = 0.067 M, [CO] = [H₂O] = 0.033 M.
6.6 CO₂(g) + C(s) ⇌ 2CO(g), 1000 K par Kp = 3.0, shuru mein p(CO₂) = 0.48 bar: (2x)²/(0.48 − x) = 3 → x = 0.33. p(CO) = 0.66 bar, p(CO₂) = 0.15 bar.

6.6Applications of equilibrium constantsसाम्य स्थिरांकों के अनुप्रयोगEquilibrium constants ke applications

Reaction quotient Q vs equilibrium constant Kअभिक्रिया भागफल Q बनाम साम्य स्थिरांक KReaction quotient Q vs equilibrium constant KNCERT Fig 6.7 · 3 tappable partsNCERT चित्र 6.7 · 3 टैप करने लायक हिस्सेNCERT Fig 6.7 · 3 tappable partsOpen
NCERT Problems 6.7–6.96.7 2A ⇌ B + C, Kc = 2 × 10⁻³; [A] = [B] = [C] = 3 × 10⁻⁴ M: Qc = (3 × 10⁻⁴)²/(3 × 10⁻⁴)² = 1 > Kc → reaction goes backward.
6.8 13.8 g N₂O₄ (0.15 mol) in 1 L at 400 K; total pressure at equilibrium 9.15 bar. Initial p = nRT/V = 0.15 × 0.083 × 400 = 4.98 bar. N₂O₄ ⇌ 2NO₂: (4.98 − x) + 2x = 9.15 → x = 4.17. p(N₂O₄) = 0.81 bar, p(NO₂) = 8.34 bar; Kp = 8.34²/0.81 = 85.9; Kc = Kp/RT = 85.9/33.2 = 2.6.
6.9 3.00 mol PCl₅ in 1 L at 380 K, Kc = 1.80: x²/(3 − x) = 1.8 → x² + 1.8x − 5.4 = 0 → x = 1.59. [PCl₅] = 1.41 M, [PCl₃] = [Cl₂] = 1.59 M.NCERT प्रश्न 6.7–6.96.7 2A ⇌ B + C, Kc = 2 × 10⁻³; [A] = [B] = [C] = 3 × 10⁻⁴ M: Qc = 1 > Kc → अभिक्रिया पश्च दिशा में।
6.8 400 K पर 1 L में 13.8 g N₂O₄ (0.15 mol); साम्य पर कुल दाब 9.15 bar। आरंभिक p = nRT/V = 0.15 × 0.083 × 400 = 4.98 bar। N₂O₄ ⇌ 2NO₂: (4.98 − x) + 2x = 9.15 → x = 4.17। p(N₂O₄) = 0.81 bar, p(NO₂) = 8.34 bar; Kp = 8.34²/0.81 = 85.9; Kc = Kp/RT = 85.9/33.2 = 2.6।
6.9 380 K पर 1 L में 3.00 mol PCl₅, Kc = 1.80: x²/(3 − x) = 1.8 → x = 1.59। [PCl₅] = 1.41 M, [PCl₃] = [Cl₂] = 1.59 M।NCERT Problems 6.7–6.96.7 2A ⇌ B + C, Kc = 2 × 10⁻³; [A] = [B] = [C] = 3 × 10⁻⁴ M: Qc = 1 > Kc → reaction backward.
6.8 400 K par 1 L mein 13.8 g N₂O₄ (0.15 mol); equilibrium par total pressure 9.15 bar. Initial p = nRT/V = 0.15 × 0.083 × 400 = 4.98 bar. N₂O₄ ⇌ 2NO₂: (4.98 − x) + 2x = 9.15 → x = 4.17. p(N₂O₄) = 0.81 bar, p(NO₂) = 8.34 bar; Kp = 8.34²/0.81 = 85.9; Kc = Kp/RT = 85.9/33.2 = 2.6.
6.9 380 K par 1 L mein 3.00 mol PCl₅, Kc = 1.80: x²/(3 − x) = 1.8 → x = 1.59. [PCl₅] = 1.41 M, [PCl₃] = [Cl₂] = 1.59 M.

6.7Relation between K, Q and Gibbs energyK, Q और गिब्ज़ ऊर्जा का संबंधK, Q aur Gibbs energy ka sambandh

ΔG = ΔG⊖ + RT ln Q
at equilibrium ΔG = 0 and Q = Kसाम्य पर ΔG = 0 और Q = Kequilibrium par ΔG = 0 aur Q = K
ΔG⊖ = −RT ln K, K = e<sup>−ΔG⊖/RT</sup>
ΔG⊖ < 0 → K > 1 (products favoured); ΔG⊖ > 0 → K < 1ΔG⊖ < 0 → K > 1 (उत्पाद अनुकूल); ΔG⊖ > 0 → K < 1ΔG⊖ < 0 → K > 1 (products favoured); ΔG⊖ > 0 → K < 1
NCERT Problems 6.10 and 6.116.10 Phosphorylation of glucose, ΔG⊖ = 13.8 kJ mol⁻¹: ln Kc = −13800/(8.314 × 298) = −5.57 → Kc = 3.8 × 10⁻³.
6.11 Hydrolysis of sucrose, Kc = 2 × 10¹³ at 300 K: ΔG⊖ = −RT ln K = −8.314 × 300 × ln(2 × 10¹³) = −7.64 × 10⁴ J mol⁻¹ (−76.4 kJ mol⁻¹).NCERT प्रश्न 6.10 और 6.116.10 ग्लूकोस का फ़ॉस्फ़ोरिलीकरण, ΔG⊖ = 13.8 kJ mol⁻¹: ln Kc = −13800/(8.314 × 298) = −5.57 → Kc = 3.8 × 10⁻³।
6.11 सुक्रोस का जल-अपघटन, 300 K पर Kc = 2 × 10¹³: ΔG⊖ = −RT ln K = −8.314 × 300 × ln(2 × 10¹³) = −7.64 × 10⁴ J mol⁻¹ (−76.4 kJ mol⁻¹)।NCERT Problems 6.10 aur 6.116.10 Glucose ka phosphorylation, ΔG⊖ = 13.8 kJ mol⁻¹: ln Kc = −13800/(8.314 × 298) = −5.57 → Kc = 3.8 × 10⁻³.
6.11 Sucrose ka hydrolysis, 300 K par Kc = 2 × 10¹³: ΔG⊖ = −RT ln K = −8.314 × 300 × ln(2 × 10¹³) = −7.64 × 10⁴ J mol⁻¹ (−76.4 kJ mol⁻¹).

6.8Factors affecting equilibria — Le Chatelier’s principleसाम्य को प्रभावित करने वाले कारक — ला-शातैलिए सिद्धांतEquilibrium par asar daalne wale factors — Le Chatelier principle

Le Chatelier’s principle: if a system at equilibrium is disturbed by a change in concentration, pressure or temperature, it shifts in the direction that reduces the effect of the change. This guides industry — e.g. Haber process for NH₃ (high pressure ≈ 200 atm, moderate temperature ≈ 700 K, iron catalyst, NH₃ removed).ला-शातैलिए सिद्धांत: यदि साम्यावस्था में निकाय को सांद्रता, दाब या ताप बदलकर विचलित किया जाए, तो वह उस दिशा में खिसकता है जो परिवर्तन के प्रभाव को कम करे। उद्योग में इसका उपयोग होता है — जैसे NH₃ का हाबर प्रक्रम (उच्च दाब ≈ 200 atm, मध्यम ताप ≈ 700 K, आयरन उत्प्रेरक, NH₃ हटाते रहना)।Le Chatelier principle: agar equilibrium par system ko concentration, pressure ya temperature badal kar disturb kiya jaaye, to wo us direction mein khisakta hai jo change ka asar kam kare. Industry mein iska use — jaise NH₃ ka Haber process (high pressure ≈ 200 atm, moderate temperature ≈ 700 K, iron catalyst, NH₃ hatate rehna).

Le Chatelier’s principle: effect of changesला-शातैलिए सिद्धांत: परिवर्तनों का प्रभावLe Chatelier’s principle: effect of changesNCERT Sec 6.8 · 5 tappable partsNCERT अनुभाग 6.8 · 5 टैप करने लायक हिस्सेNCERT Sec 6.8 · 5 tappable partsOpen

6.9–6.10Ionic equilibrium; acids, bases and saltsआयनिक साम्य; अम्ल, क्षारक और लवणIonic equilibrium; acids, bases aur salts

Strong electrolytes (NaCl, HCl, NaOH) ionise almost completely in water; weak electrolytes (CH₃COOH, NH₃) ionise partly, and an equilibrium is set up between ions and un-ionised molecules — ionic equilibrium.प्रबल विद्युत-अपघट्य (NaCl, HCl, NaOH) जल में लगभग पूर्णतः आयनित होते हैं; दुर्बल विद्युत-अपघट्य (CH₃COOH, NH₃) आंशिक रूप से आयनित होते हैं, और आयनों व अनायनित अणुओं के बीच साम्य बनता है — आयनिक साम्य।Strong electrolytes (NaCl, HCl, NaOH) paani mein lagbhag poore ionise hote hain; weak electrolytes (CH₃COOH, NH₃) thode ionise hote hain, aur ions aur un-ionised molecules ke beech equilibrium banta hai — ionic equilibrium.

Conjugate acid–base pairsसंयुग्मी अम्ल–क्षार युग्मConjugate acid–base pairsNCERT Sec 6.10.2 · 6 tappable partsNCERT अनुभाग 6.10.2 · 6 टैप करने लायक हिस्सेNCERT Sec 6.10.2 · 6 tappable partsOpen

In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, NH₃ is the base and H₂O the acid; NH₄⁺ and OH⁻ are their conjugates. In HCl + H₂O → H₃O⁺ + Cl⁻ water acts as a base.NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ में NH₃ क्षारक और H₂O अम्ल है; NH₄⁺ और OH⁻ इनके संयुग्मी हैं। HCl + H₂O → H₃O⁺ + Cl⁻ में जल क्षारक है।NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ mein NH₃ base aur H₂O acid hai; NH₄⁺ aur OH⁻ inke conjugates hain. HCl + H₂O → H₃O⁺ + Cl⁻ mein paani base hai.

NCERT Problems 6.12–6.156.12 Conjugate bases: HF → F⁻, H₂SO₄ → HSO₄⁻, HCO₃⁻ → CO₃²⁻.
6.13 Conjugate acids: NH₂⁻ → NH₃, NH₃ → NH₄⁺, HCOO⁻ → HCOOH.
6.14 H₂O: conj. acid H₃O⁺, conj. base OH⁻; HCO₃⁻: H₂CO₃ / CO₃²⁻; HSO₄⁻: H₂SO₄ / SO₄²⁻; NH₃: NH₄⁺ / NH₂⁻.
6.15 OH⁻ and F⁻ are Lewis bases (donate a lone pair); H⁺ and BCl₃ are Lewis acids (accept a pair; B has only 6 electrons).NCERT प्रश्न 6.12–6.156.12 संयुग्मी क्षारक: HF → F⁻, H₂SO₄ → HSO₄⁻, HCO₃⁻ → CO₃²⁻।
6.13 संयुग्मी अम्ल: NH₂⁻ → NH₃, NH₃ → NH₄⁺, HCOO⁻ → HCOOH।
6.14 H₂O: संयुग्मी अम्ल H₃O⁺, संयुग्मी क्षारक OH⁻; HCO₃⁻: H₂CO₃ / CO₃²⁻; HSO₄⁻: H₂SO₄ / SO₄²⁻; NH₃: NH₄⁺ / NH₂⁻।
6.15 OH⁻ और F⁻ लूइस क्षारक (एकाकी युग्म देते हैं); H⁺ और BCl₃ लूइस अम्ल (युग्म लेते हैं; B पर केवल 6 इलेक्ट्रॉन)।NCERT Problems 6.12–6.156.12 Conjugate bases: HF → F⁻, H₂SO₄ → HSO₄⁻, HCO₃⁻ → CO₃²⁻.
6.13 Conjugate acids: NH₂⁻ → NH₃, NH₃ → NH₄⁺, HCOO⁻ → HCOOH.
6.14 H₂O: conj. acid H₃O⁺, conj. base OH⁻; HCO₃⁻: H₂CO₃ / CO₃²⁻; HSO₄⁻: H₂SO₄ / SO₄²⁻; NH₃: NH₄⁺ / NH₂⁻.
6.15 OH⁻ aur F⁻ Lewis bases (lone pair dete hain); H⁺ aur BCl₃ Lewis acids (pair lete hain; B par sirf 6 electrons).

6.11Ionisation of acids and bases; pHअम्लों और क्षारकों का आयनन; pHAcids aur bases ka ionisation; pH

K<sub>w</sub> = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ (298 K)
ionic product of water; increases with temperatureजल का आयनिक गुणनफल; ताप बढ़ने पर बढ़ता हैpaani ka ionic product; temperature badhne par badhta hai
pH = −log[H₃O⁺], pH + pOH = pK<sub>w</sub> = 14
neutral pH 7, acidic < 7, basic > 7 (at 298 K)उदासीन pH 7, अम्लीय < 7, क्षारीय > 7 (298 K पर)neutral pH 7, acidic < 7, basic > 7 (298 K par)
K<sub>a</sub> = cα²/(1 − α) ≈ cα², [H⁺] = cα = √(K<sub>a</sub>c)
weak acid HA of concentration c, degree of ionisation α; similarly Kb for a weak basec सांद्रता वाला दुर्बल अम्ल HA, आयनन की मात्रा α; इसी प्रकार दुर्बल क्षारक के लिए Kbc concentration wala weak acid HA, degree of ionisation α; isi tarah weak base ke liye Kb
K<sub>a</sub> × K<sub>b</sub> = K<sub>w</sub>, pK<sub>a</sub> + pK<sub>b</sub> = 14
for a conjugate acid–base pairसंयुग्मी अम्ल–क्षारक युग्म के लिएconjugate acid–base pair ke liye
The pH scalepH स्केलThe pH scaleNCERT Table 6.5 · 6 tappable partsNCERT सारणी 6.5 · 6 टैप करने लायक हिस्सेNCERT Table 6.5 · 6 tappable partsOpen
NCERT Problems 6.16–6.25 (pH and ionisation)6.16 Soft drink [H⁺] = 3.8 × 10⁻³ M: pH = 2.42.
6.17 1.0 × 10⁻⁸ M HCl: x(10⁻⁸ + x) = 10⁻¹⁴ → x = 9.5 × 10⁻⁸; [H⁺] = 1.05 × 10⁻⁷, pH = 6.98.
6.18 0.02 M HF, Ka = 3.2 × 10⁻⁴: α = 0.12; [F⁻] = [H₃O⁺] = 2.4 × 10⁻³ M, [HF] = 1.76 × 10⁻² M, pH = 2.62.
6.19 0.1 M monobasic acid, pH 4.50: [H⁺] = 3.16 × 10⁻⁵; Ka = (3.16 × 10⁻⁵)²/0.1 = 1.0 × 10⁻⁸, pKa = 8.
6.20 0.08 M HOCl, Ka = 2.5 × 10⁻⁵: [H⁺] = √(2.5 × 10⁻⁵ × 0.08) = 1.41 × 10⁻³, pH = 2.85.
6.21 0.004 M hydrazine, pH 9.7: pOH 4.3, [OH⁻] = 5.0 × 10⁻⁵; Kb = (5.0 × 10⁻⁵)²/0.004 = 6.25 × 10⁻⁷, pKb = 6.20.
6.22 0.2 M NH₄Cl + 0.1 M NH₃, Kb = 1.77 × 10⁻⁵: [OH⁻] = 1.77 × 10⁻⁵ × 0.1/0.2 = 8.85 × 10⁻⁶, pOH 5.05, pH = 8.95.
6.23 0.05 M NH₃: α = √(1.77 × 10⁻⁵/0.05) = 0.0188; [OH⁻] = 9.4 × 10⁻⁴, pH = 10.97.
6.24 0.10 M NH₃: [OH⁻] = 1.33 × 10⁻³, pH = 11.12. After 50.0 mL + 25.0 mL of 0.10 M HCl: 2.5 mmol NH₃ and 2.5 mmol NH₄⁺ remain → [OH⁻] = Kb = 1.77 × 10⁻⁵, pH = 9.25.
6.25 Ammonium acetate, pKa 4.76, pKb 4.75: pH = 7 + ½(4.76 − 4.75) = 7.005.NCERT प्रश्न 6.16–6.25 (pH और आयनन)6.16 शीतल पेय [H⁺] = 3.8 × 10⁻³ M: pH = 2.42।
6.17 1.0 × 10⁻⁸ M HCl: x(10⁻⁸ + x) = 10⁻¹⁴ → x = 9.5 × 10⁻⁸; [H⁺] = 1.05 × 10⁻⁷, pH = 6.98।
6.18 0.02 M HF, Ka = 3.2 × 10⁻⁴: α = 0.12; [F⁻] = [H₃O⁺] = 2.4 × 10⁻³ M, [HF] = 1.76 × 10⁻² M, pH = 2.62।
6.19 0.1 M एकक्षारकी अम्ल, pH 4.50: Ka = (3.16 × 10⁻⁵)²/0.1 = 1.0 × 10⁻⁸, pKa = 8।
6.20 0.08 M HOCl: [H⁺] = 1.41 × 10⁻³, pH = 2.85।
6.21 0.004 M हाइड्रैज़ीन, pH 9.7: [OH⁻] = 5.0 × 10⁻⁵; Kb = 6.25 × 10⁻⁷, pKb = 6.20।
6.22 0.2 M NH₄Cl + 0.1 M NH₃: [OH⁻] = 8.85 × 10⁻⁶, pOH 5.05, pH = 8.95।
6.23 0.05 M NH₃: α = 0.0188; [OH⁻] = 9.4 × 10⁻⁴, pH = 10.97।
6.24 0.10 M NH₃: pH = 11.12। 50.0 mL में 25.0 mL 0.10 M HCl मिलाने पर 2.5 mmol NH₃ व 2.5 mmol NH₄⁺ → [OH⁻] = Kb, pH = 9.25।
6.25 अमोनियम ऐसीटेट: pH = 7 + ½(4.76 − 4.75) = 7.005।NCERT Problems 6.16–6.25 (pH aur ionisation)6.16 Soft drink [H⁺] = 3.8 × 10⁻³ M: pH = 2.42.
6.17 1.0 × 10⁻⁸ M HCl: x(10⁻⁸ + x) = 10⁻¹⁴ → x = 9.5 × 10⁻⁸; [H⁺] = 1.05 × 10⁻⁷, pH = 6.98.
6.18 0.02 M HF, Ka = 3.2 × 10⁻⁴: α = 0.12; [F⁻] = [H₃O⁺] = 2.4 × 10⁻³ M, [HF] = 1.76 × 10⁻² M, pH = 2.62.
6.19 0.1 M monobasic acid, pH 4.50: Ka = (3.16 × 10⁻⁵)²/0.1 = 1.0 × 10⁻⁸, pKa = 8.
6.20 0.08 M HOCl: [H⁺] = 1.41 × 10⁻³, pH = 2.85.
6.21 0.004 M hydrazine, pH 9.7: [OH⁻] = 5.0 × 10⁻⁵; Kb = 6.25 × 10⁻⁷, pKb = 6.20.
6.22 0.2 M NH₄Cl + 0.1 M NH₃: [OH⁻] = 8.85 × 10⁻⁶, pOH 5.05, pH = 8.95.
6.23 0.05 M NH₃: α = 0.0188; [OH⁻] = 9.4 × 10⁻⁴, pH = 10.97.
6.24 0.10 M NH₃: pH = 11.12. 50.0 mL mein 25.0 mL 0.10 M HCl milane par 2.5 mmol NH₃ aur 2.5 mmol NH₄⁺ → [OH⁻] = Kb, pH = 9.25.
6.25 Ammonium acetate: pH = 7 + ½(4.76 − 4.75) = 7.005.

6.12Buffer solutionsबफ़र विलयनBuffer solutions

A buffer resists change in pH when a little acid or base is added or the solution is diluted. Acidic buffer: weak acid + its salt with a strong base (CH₃COOH + CH₃COONa, pH ≈ 4.75); basic buffer: weak base + its salt with a strong acid (NH₄OH + NH₄Cl, pH ≈ 9.25). Henderson–Hasselbalch: pH = pKa + log([salt]/[acid]); when [salt] = [acid], pH = pKa. Blood is buffered near pH 7.4.बफ़र थोड़ा अम्ल या क्षारक मिलाने या तनु करने पर pH में परिवर्तन का विरोध करता है। अम्लीय बफ़र: दुर्बल अम्ल + प्रबल क्षारक के साथ उसका लवण (CH₃COOH + CH₃COONa, pH ≈ 4.75); क्षारीय बफ़र: दुर्बल क्षारक + प्रबल अम्ल के साथ उसका लवण (NH₄OH + NH₄Cl, pH ≈ 9.25)। हेंडरसन–हैसलबाख: pH = pKa + log([लवण]/[अम्ल]); [लवण] = [अम्ल] होने पर pH = pKa। रक्त का pH लगभग 7.4 पर बफ़र रहता है।Buffer thoda acid ya base milane ya dilute karne par pH change ka virodh karta hai. Acidic buffer: weak acid + strong base ke saath uska salt (CH₃COOH + CH₃COONa, pH ≈ 4.75); basic buffer: weak base + strong acid ke saath uska salt (NH₄OH + NH₄Cl, pH ≈ 9.25). Henderson–Hasselbalch: pH = pKa + log([salt]/[acid]); [salt] = [acid] hone par pH = pKa. Blood ka pH lagbhag 7.4 par buffered rehta hai.

6.13Solubility equilibria of sparingly soluble saltsअल्प विलेय लवणों का विलेयता साम्यKam ghulne wale salts ka solubility equilibrium

MₓXᵧ(s) ⇌ xMᵖ⁺(aq) + yXᑫ⁻(aq): K<sub>sp</sub> = (xS)ˣ(yS)ʸ
solubility product; S = molar solubilityविलेयता गुणनफल; S = मोलर विलेयताsolubility product; S = molar solubility
Q<sub>sp</sub> (ionic product) > K<sub>sp</sub> ⟶ precipitate forms
Qsp < Ksp → more solid dissolvesQsp < Ksp → और ठोस घुलता हैQsp < Ksp → aur solid ghulta hai

Examples: AgCl, S = √Ksp; Ag₂CrO₄, Ksp = 4S³; A₂X₃, Ksp = 108S⁵. Compare solubilities through S, not Ksp, when the formulas differ. Common ion effect on solubility: a common ion lowers solubility (pure NaCl is precipitated by passing HCl gas into its saturated solution; Ni(OH)₂ is much less soluble in NaOH). Salts of weak acids (phosphates) dissolve more at low pH.उदाहरण: AgCl, S = √Ksp; Ag₂CrO₄, Ksp = 4S³; A₂X₃, Ksp = 108S⁵। सूत्र भिन्न हों तो विलेयता की तुलना S से करें, Ksp से नहीं। विलेयता पर समआयन प्रभाव: समान आयन विलेयता घटाता है (संतृप्त NaCl विलयन में HCl गैस प्रवाहित करने पर शुद्ध NaCl अवक्षेपित; NaOH में Ni(OH)₂ बहुत कम घुलता है)। दुर्बल अम्लों के लवण (फ़ॉस्फ़ेट) कम pH पर अधिक घुलते हैं।Examples: AgCl, S = √Ksp; Ag₂CrO₄, Ksp = 4S³; A₂X₃, Ksp = 108S⁵. Formula alag hon to solubility ki tulna S se kijiye, Ksp se nahi. Solubility par common ion effect: common ion solubility ghatata hai (saturated NaCl solution mein HCl gas bhejne par pure NaCl precipitate; NaOH mein Ni(OH)₂ bahut kam ghulta hai). Weak acids ke salts (phosphates) kam pH par zyada ghulte hain.

NCERT Problems 6.26–6.286.26 A₂X₃, Ksp = 1.1 × 10⁻²³: Ksp = (2S)²(3S)³ = 108S⁵ → S⁵ = 1.0 × 10⁻²⁵ → S = 1.0 × 10⁻⁵ mol L⁻¹.
6.27 AgCN (Ksp 6 × 10⁻¹⁷): S₁ = √(6 × 10⁻¹⁷) = 7.8 × 10⁻⁹ M; Ni(OH)₂ (Ksp 2.0 × 10⁻¹⁵): 4S₂³ = 2 × 10⁻¹⁵ → S₂ = 7.9 × 10⁻⁶ M. So Ni(OH)₂ is more soluble (the book prints S₂ = 0.58 × 10⁻⁴; either way Ni(OH)₂ is more soluble).
6.28 Ni(OH)₂ in 0.10 M NaOH: Ksp = S(0.10 + 2S)² ≈ S(0.10)² → S = 2.0 × 10⁻¹³ M.NCERT प्रश्न 6.26–6.286.26 A₂X₃, Ksp = 1.1 × 10⁻²³: 108S⁵ = 1.1 × 10⁻²³ → S = 1.0 × 10⁻⁵ mol L⁻¹।
6.27 AgCN: S₁ = √(6 × 10⁻¹⁷) = 7.8 × 10⁻⁹ M; Ni(OH)₂: 4S₂³ = 2 × 10⁻¹⁵ → S₂ = 7.9 × 10⁻⁶ M। अतः Ni(OH)₂ अधिक विलेय (पुस्तक में S₂ = 0.58 × 10⁻⁴ छपा है; निष्कर्ष वही)।
6.28 0.10 M NaOH में Ni(OH)₂: S(0.10)² = 2.0 × 10⁻¹⁵ → S = 2.0 × 10⁻¹³ M।NCERT Problems 6.26–6.286.26 A₂X₃, Ksp = 1.1 × 10⁻²³: 108S⁵ = 1.1 × 10⁻²³ → S = 1.0 × 10⁻⁵ mol L⁻¹.
6.27 AgCN: S₁ = √(6 × 10⁻¹⁷) = 7.8 × 10⁻⁹ M; Ni(OH)₂: 4S₂³ = 2 × 10⁻¹⁵ → S₂ = 7.9 × 10⁻⁶ M. Isliye Ni(OH)₂ zyada soluble (book mein S₂ = 0.58 × 10⁻⁴ chhapa hai; conclusion wahi).
6.28 0.10 M NaOH mein Ni(OH)₂: S(0.10)² = 2.0 × 10⁻¹⁵ → S = 2.0 × 10⁻¹³ M.

★ Yaad rakho — quick revision★ याद रखो — झटपट दोहराव★ Yaad rakho — quick revision

  1. Equilibrium is dynamic: forward rate = backward rate; reachable from either side; only in a closed system.साम्य गतिक है: अग्र दर = पश्च दर; किसी भी ओर से प्राप्य; केवल बंद निकाय में।Equilibrium dynamic hai: forward rate = backward rate; dono taraf se possible; sirf closed system mein.
  2. Kc = [products]^coeff / [reactants]^coeff; leave out pure solids, pure liquids and the solvent.Kc = [उत्पाद]^गुणांक / [अभिकारक]^गुणांक; शुद्ध ठोस, शुद्ध द्रव और विलायक नहीं लिखते।Kc = [products]^coeff / [reactants]^coeff; pure solids, pure liquids aur solvent nahi likhte.
  3. Kp = Kc(RT)^Δn (Δn = gas moles of products − reactants); reverse reaction K′ = 1/K.Kp = Kc(RT)^Δn (Δn = उत्पादों − अभिकारकों के गैस मोल); उलटी अभिक्रिया K′ = 1/K।Kp = Kc(RT)^Δn (Δn = products − reactants ke gas moles); ulti reaction K′ = 1/K.
  4. Q < K forward, Q = K equilibrium, Q > K backward.Q < K अग्र, Q = K साम्य, Q > K पश्च।Q < K forward, Q = K equilibrium, Q > K backward.
  5. ΔG⊖ = −RT ln K = −2.303 RT log K.ΔG⊖ = −RT ln K = −2.303 RT log K।ΔG⊖ = −RT ln K = −2.303 RT log K.
  6. Le Chatelier: only temperature changes K; a catalyst changes neither K nor the composition.ला-शातैलिए: केवल ताप K बदलता है; उत्प्रेरक न K बदलता है न संघटन।Le Chatelier: sirf temperature K badalta hai; catalyst na K badalta hai na composition.
  7. Brönsted acid = H⁺ donor, base = H⁺ acceptor; Lewis acid = electron-pair acceptor.ब्रॉन्स्टेड अम्ल = H⁺ दाता, क्षारक = H⁺ ग्राही; लूइस अम्ल = इलेक्ट्रॉन-युग्म ग्राही।Brönsted acid = H⁺ donor, base = H⁺ acceptor; Lewis acid = electron-pair acceptor.
  8. Kw = 1.0 × 10⁻¹⁴ at 298 K; pH + pOH = 14; pH = −log[H⁺].298 K पर Kw = 1.0 × 10⁻¹⁴; pH + pOH = 14; pH = −log[H⁺]।298 K par Kw = 1.0 × 10⁻¹⁴; pH + pOH = 14; pH = −log[H⁺].
  9. Weak acid: [H⁺] = √(Ka c), α = √(Ka/c); Ka × Kb = Kw.दुर्बल अम्ल: [H⁺] = √(Ka c), α = √(Ka/c); Ka × Kb = Kw।Weak acid: [H⁺] = √(Ka c), α = √(Ka/c); Ka × Kb = Kw.
  10. Salt of weak acid + strong base → basic; strong acid + weak base → acidic; weak + weak: pH = 7 + ½(pKa − pKb).दुर्बल अम्ल + प्रबल क्षारक का लवण → क्षारीय; प्रबल अम्ल + दुर्बल क्षारक → अम्लीय; दुर्बल + दुर्बल: pH = 7 + ½(pKa − pKb)।Weak acid + strong base ka salt → basic; strong acid + weak base → acidic; weak + weak: pH = 7 + ½(pKa − pKb).
  11. Buffer: pH = pKa + log([salt]/[acid]); acetate buffer ≈ 4.75, ammonia buffer ≈ 9.25.बफ़र: pH = pKa + log([लवण]/[अम्ल]); ऐसीटेट बफ़र ≈ 4.75, अमोनिया बफ़र ≈ 9.25।Buffer: pH = pKa + log([salt]/[acid]); acetate buffer ≈ 4.75, ammonia buffer ≈ 9.25.
  12. Ksp = (xS)ˣ(yS)ʸ; ionic product > Ksp → precipitate; a common ion lowers solubility.Ksp = (xS)ˣ(yS)ʸ; आयनिक गुणनफल > Ksp → अवक्षेप; समआयन विलेयता घटाता है।Ksp = (xS)ˣ(yS)ʸ; ionic product > Ksp → precipitate; common ion solubility ghatata hai.

⚠ Common mistakes⚠ आम गलतियाँ⚠ Common mistakes

  • Writing solids or water in K: for CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = p(CO₂) only.K में ठोस या जल लिखना: CaCO₃(s) ⇌ CaO(s) + CO₂(g) के लिए केवल Kp = p(CO₂)।K mein solids ya paani likhna: CaCO₃(s) ⇌ CaO(s) + CO₂(g) ke liye sirf Kp = p(CO₂).
  • Δn counts only gas moles, products minus reactants — a wrong sign of Δn turns Kp = Kc × RT into Kc/RT.Δn में केवल गैस मोल, उत्पाद − अभिकारक — Δn का गलत चिह्न Kp = Kc × RT को Kc/RT बना देता है।Δn mein sirf gas moles, products − reactants — Δn ka galat sign Kp = Kc × RT ko Kc/RT bana deta hai.
  • Thinking pressure or a catalyst changes K. Only temperature does.यह मानना कि दाब या उत्प्रेरक K बदलते हैं। केवल ताप बदलता है।Ye maanna ki pressure ya catalyst K badalte hain. Sirf temperature badalta hai.
  • pH of 10⁻⁸ M HCl is not 8 — add water’s own H⁺ (pH ≈ 6.98). An acid can never give a basic pH.10⁻⁸ M HCl का pH 8 नहीं — जल का अपना H⁺ जोड़ें (pH ≈ 6.98)। अम्ल कभी क्षारीय pH नहीं देता।10⁻⁸ M HCl ka pH 8 nahi — paani ka apna H⁺ jodiye (pH ≈ 6.98). Acid kabhi basic pH nahi deta.
  • Counting OH⁻ once for Ca(OH)₂ or H⁺ once for H₂SO₄ — each gives 2 per formula unit.Ca(OH)₂ के लिए OH⁻ या H₂SO₄ के लिए H⁺ को एक बार गिनना — प्रत्येक प्रति सूत्र इकाई 2 देता है।Ca(OH)₂ ke liye OH⁻ ya H₂SO₄ ke liye H⁺ ko ek baar ginna — har ek per formula unit 2 deta hai.
  • Comparing solubility of salts with different formulas (AgCl vs Ag₂CrO₄) through Ksp — compare S instead.भिन्न सूत्र वाले लवणों (AgCl बनाम Ag₂CrO₄) की विलेयता Ksp से तुलना करना — S से तुलना करें।Alag formula wale salts (AgCl vs Ag₂CrO₄) ki solubility Ksp se compare karna — S se compare kijiye.

✎ NCERT questions — short answers✎ NCERT प्रश्न — संक्षिप्त उत्तर✎ NCERT questions — short answers

Solved problems in the chapter (tap to see the solution)पाठ में हल किए गए प्रश्न (हल देखने के लिए टैप करें)Chapter ke solved problems (solution dekhne ke liye tap kijiye)

Problem 6.1: Kc for N₂ + 3H₂ ⇌ 2NH₃ at 500 K ([N₂] 1.5 × 10⁻², [H₂] 3.0 × 10⁻², [NH₃] 1.2 × 10⁻² M).उदाहरण 6.1: N₂ + 3H₂ ⇌ 2NH₃ का Kc (500 K)।Problem 6.1: N₂ + 3H₂ ⇌ 2NH₃ ka Kc (500 K).
Kc = 3.55 × 10²Kc = 3.55 × 10²Kc = 3.55 × 10²
Problem 6.2: Kc for N₂ + O₂ ⇌ 2NO.उदाहरण 6.2: N₂ + O₂ ⇌ 2NO का Kc।Problem 6.2: N₂ + O₂ ⇌ 2NO ka Kc.
Kc = 0.622Kc = 0.622Kc = 0.622
Problem 6.3: Kc for PCl₅ ⇌ PCl₃ + Cl₂ (1.59, 1.59, 1.41 M).उदाहरण 6.3: PCl₅ ⇌ PCl₃ + Cl₂ का Kc।Problem 6.3: PCl₅ ⇌ PCl₃ + Cl₂ ka Kc.
Kc = 1.79Kc = 1.79Kc = 1.79
Problem 6.4: Kp for 2NOCl ⇌ 2NO + Cl₂, Kc = 3.75 × 10⁻⁶ at 1069 K.उदाहरण 6.4: 2NOCl ⇌ 2NO + Cl₂ का Kp।Problem 6.4: 2NOCl ⇌ 2NO + Cl₂ ka Kp.
Kp = Kc(RT)¹ = 3.33 × 10⁻⁴Kp = Kc(RT)¹ = 3.33 × 10⁻⁴Kp = Kc(RT)¹ = 3.33 × 10⁻⁴
Problem 6.5: Equilibrium concentrations for CO + H₂O ⇌ CO₂ + H₂, Kc = 4.24, 0.1 M each at start.उदाहरण 6.5: CO + H₂O ⇌ CO₂ + H₂ की साम्य सांद्रताएँ।Problem 6.5: CO + H₂O ⇌ CO₂ + H₂ ki equilibrium concentrations.
[CO₂] = [H₂] = 0.067 M; [CO] = [H₂O] = 0.033 M[CO₂] = [H₂] = 0.067 M; [CO] = [H₂O] = 0.033 M[CO₂] = [H₂] = 0.067 M; [CO] = [H₂O] = 0.033 M
Problem 6.6: CO₂ + C ⇌ 2CO, Kp = 3.0, p(CO₂) = 0.48 bar at start.उदाहरण 6.6: CO₂ + C ⇌ 2CO, Kp = 3.0।Problem 6.6: CO₂ + C ⇌ 2CO, Kp = 3.0.
p(CO) = 0.66 bar, p(CO₂) = 0.15 barp(CO) = 0.66 bar, p(CO₂) = 0.15 barp(CO) = 0.66 bar, p(CO₂) = 0.15 bar
Problem 6.7: Direction of 2A ⇌ B + C (Kc 2 × 10⁻³) when all are 3 × 10⁻⁴ M.उदाहरण 6.7: 2A ⇌ B + C की दिशा।Problem 6.7: 2A ⇌ B + C ki direction.
Qc = 1 > Kc → backward.Qc = 1 > Kc → पश्च दिशा।Qc = 1 > Kc → backward.
Problem 6.8: N₂O₄ ⇌ 2NO₂ at 400 K, total pressure 9.15 bar.उदाहरण 6.8: N₂O₄ ⇌ 2NO₂, 400 K, कुल दाब 9.15 bar।Problem 6.8: N₂O₄ ⇌ 2NO₂, 400 K, total pressure 9.15 bar.
p(N₂O₄) = 0.81, p(NO₂) = 8.34 bar; Kp = 85.9, Kc = 2.6p(N₂O₄) = 0.81, p(NO₂) = 8.34 bar; Kp = 85.9, Kc = 2.6p(N₂O₄) = 0.81, p(NO₂) = 8.34 bar; Kp = 85.9, Kc = 2.6
Problem 6.9: Composition when 3.00 mol PCl₅ (1 L, 380 K, Kc 1.80) reaches equilibrium.उदाहरण 6.9: 3.00 mol PCl₅ का साम्य संघटन।Problem 6.9: 3.00 mol PCl₅ ka equilibrium composition.
[PCl₅] = 1.41 M, [PCl₃] = [Cl₂] = 1.59 M[PCl₅] = 1.41 M, [PCl₃] = [Cl₂] = 1.59 M[PCl₅] = 1.41 M, [PCl₃] = [Cl₂] = 1.59 M
Problem 6.10: Kc for glucose phosphorylation, ΔG⊖ = 13.8 kJ mol⁻¹ (298 K).उदाहरण 6.10: ग्लूकोस फ़ॉस्फ़ोरिलीकरण का Kc।Problem 6.10: Glucose phosphorylation ka Kc.
Kc = e^(−5.57) = 3.8 × 10⁻³Kc = e^(−5.57) = 3.8 × 10⁻³Kc = e^(−5.57) = 3.8 × 10⁻³
Problem 6.11: ΔG⊖ for sucrose hydrolysis, Kc = 2 × 10¹³ at 300 K.उदाहरण 6.11: सुक्रोस जल-अपघटन का ΔG⊖।Problem 6.11: Sucrose hydrolysis ka ΔG⊖.
ΔG⊖ = −7.64 × 10⁴ J mol⁻¹ΔG⊖ = −7.64 × 10⁴ J mol⁻¹ΔG⊖ = −7.64 × 10⁴ J mol⁻¹
Problem 6.12: Conjugate bases of HF, H₂SO₄, HCO₃⁻.उदाहरण 6.12: HF, H₂SO₄, HCO₃⁻ के संयुग्मी क्षारक।Problem 6.12: HF, H₂SO₄, HCO₃⁻ ke conjugate bases.
F⁻, HSO₄⁻, CO₃²⁻F⁻, HSO₄⁻, CO₃²⁻F⁻, HSO₄⁻, CO₃²⁻
Problem 6.13: Conjugate acids of NH₂⁻, NH₃, HCOO⁻.उदाहरण 6.13: NH₂⁻, NH₃, HCOO⁻ के संयुग्मी अम्ल।Problem 6.13: NH₂⁻, NH₃, HCOO⁻ ke conjugate acids.
NH₃, NH₄⁺, HCOOHNH₃, NH₄⁺, HCOOHNH₃, NH₄⁺, HCOOH
Problem 6.14: Conjugate acid / base of H₂O, HCO₃⁻, HSO₄⁻, NH₃.उदाहरण 6.14: H₂O, HCO₃⁻, HSO₄⁻, NH₃ के संयुग्मी अम्ल / क्षारक।Problem 6.14: H₂O, HCO₃⁻, HSO₄⁻, NH₃ ke conjugate acid / base.
H₃O⁺/OH⁻; H₂CO₃/CO₃²⁻; H₂SO₄/SO₄²⁻; NH₄⁺/NH₂⁻H₃O⁺/OH⁻; H₂CO₃/CO₃²⁻; H₂SO₄/SO₄²⁻; NH₄⁺/NH₂⁻H₃O⁺/OH⁻; H₂CO₃/CO₃²⁻; H₂SO₄/SO₄²⁻; NH₄⁺/NH₂⁻
Problem 6.15: Classify OH⁻, F⁻, H⁺, BCl₃ as Lewis acids or bases.उदाहरण 6.15: OH⁻, F⁻, H⁺, BCl₃ को लूइस अम्ल/क्षारक में बाँटिए।Problem 6.15: OH⁻, F⁻, H⁺, BCl₃ ko Lewis acid/base mein baantiye.
Bases: OH⁻, F⁻. Acids: H⁺, BCl₃.क्षारक: OH⁻, F⁻। अम्ल: H⁺, BCl₃।Bases: OH⁻, F⁻. Acids: H⁺, BCl₃.
Problem 6.16: pH of soft drink with [H⁺] = 3.8 × 10⁻³ M.उदाहरण 6.16: [H⁺] = 3.8 × 10⁻³ M वाले शीतल पेय का pH।Problem 6.16: [H⁺] = 3.8 × 10⁻³ M wale soft drink ka pH.
pH = 2.42pH = 2.42pH = 2.42
Problem 6.17: pH of 1.0 × 10⁻⁸ M HCl.उदाहरण 6.17: 1.0 × 10⁻⁸ M HCl का pH।Problem 6.17: 1.0 × 10⁻⁸ M HCl ka pH.
pH = 6.98pH = 6.98pH = 6.98
Problem 6.18: α and pH of 0.02 M HF (Ka 3.2 × 10⁻⁴).उदाहरण 6.18: 0.02 M HF का α और pH।Problem 6.18: 0.02 M HF ka α aur pH.
α = 0.12; pH = 2.62α = 0.12; pH = 2.62α = 0.12; pH = 2.62
Problem 6.19: Ka of a 0.1 M monobasic acid with pH 4.50.उदाहरण 6.19: pH 4.50 वाले 0.1 M एकक्षारकी अम्ल का Ka।Problem 6.19: pH 4.50 wale 0.1 M monobasic acid ka Ka.
Ka = 1.0 × 10⁻⁸, pKa = 8Ka = 1.0 × 10⁻⁸, pKa = 8Ka = 1.0 × 10⁻⁸, pKa = 8
Problem 6.20: pH of 0.08 M HOCl (Ka 2.5 × 10⁻⁵).उदाहरण 6.20: 0.08 M HOCl का pH।Problem 6.20: 0.08 M HOCl ka pH.
pH = 2.85pH = 2.85pH = 2.85
Problem 6.21: Kb of hydrazine (0.004 M, pH 9.7).उदाहरण 6.21: हाइड्रैज़ीन का Kb।Problem 6.21: Hydrazine ka Kb.
Kb = 6.25 × 10⁻⁷, pKb = 6.20Kb = 6.25 × 10⁻⁷, pKb = 6.20Kb = 6.25 × 10⁻⁷, pKb = 6.20
Problem 6.22: pH of 0.2 M NH₄Cl + 0.1 M NH₃.उदाहरण 6.22: 0.2 M NH₄Cl + 0.1 M NH₃ का pH।Problem 6.22: 0.2 M NH₄Cl + 0.1 M NH₃ ka pH.
pH = 8.95pH = 8.95pH = 8.95
Problem 6.23: α and pH of 0.05 M NH₃.उदाहरण 6.23: 0.05 M NH₃ का α और pH।Problem 6.23: 0.05 M NH₃ ka α aur pH.
α = 0.0188; pH = 10.97α = 0.0188; pH = 10.97α = 0.0188; pH = 10.97
Problem 6.24: pH of 0.10 M NH₃, and after adding 25 mL 0.10 M HCl to 50 mL.उदाहरण 6.24: 0.10 M NH₃ का pH, और 50 mL में 25 mL 0.10 M HCl मिलाने के बाद।Problem 6.24: 0.10 M NH₃ ka pH, aur 50 mL mein 25 mL 0.10 M HCl ke baad.
pH 11.12; then 9.25pH 11.12; then 9.25pH 11.12; then 9.25
Problem 6.25: pH of ammonium acetate (pKa 4.76, pKb 4.75).उदाहरण 6.25: अमोनियम ऐसीटेट का pH।Problem 6.25: Ammonium acetate ka pH.
pH = 7.005pH = 7.005pH = 7.005
Problem 6.26: Solubility of A₂X₃, Ksp = 1.1 × 10⁻²³.उदाहरण 6.26: A₂X₃ की विलेयता।Problem 6.26: A₂X₃ ki solubility.
S = 1.0 × 10⁻⁵ mol L⁻¹S = 1.0 × 10⁻⁵ mol L⁻¹S = 1.0 × 10⁻⁵ mol L⁻¹
Problem 6.27: Which is more soluble: Ni(OH)₂ (Ksp 2.0 × 10⁻¹⁵) or AgCN (6 × 10⁻¹⁷)?उदाहरण 6.27: कौन अधिक विलेय: Ni(OH)₂ या AgCN?Problem 6.27: Kaun zyada soluble: Ni(OH)₂ ya AgCN?
Ni(OH)₂ (S ≈ 7.9 × 10⁻⁶ M) > AgCN (7.8 × 10⁻⁹ M).Ni(OH)₂ (S ≈ 7.9 × 10⁻⁶ M) > AgCN (7.8 × 10⁻⁹ M)।Ni(OH)₂ (S ≈ 7.9 × 10⁻⁶ M) > AgCN (7.8 × 10⁻⁹ M).
Problem 6.28: Molar solubility of Ni(OH)₂ in 0.10 M NaOH.उदाहरण 6.28: 0.10 M NaOH में Ni(OH)₂ की मोलर विलेयता।Problem 6.28: 0.10 M NaOH mein Ni(OH)₂ ki molar solubility.
S = 2.0 × 10⁻¹³ MS = 2.0 × 10⁻¹³ MS = 2.0 × 10⁻¹³ M

Exercisesअभ्यासExercises

6.1 A liquid is in equilibrium with its vapour in a sealed container at fixed temperature; the volume is suddenly increased. (a) initial effect on vapour pressure? (b) how do rates of evaporation and condensation change initially? (c) what happens when equilibrium is restored, and the final vapour pressure?6.1 बंद पात्र में द्रव अपनी वाष्प के साथ साम्य में है; आयतन अचानक बढ़ाया गया। (a) वाष्प दाब पर आरंभिक प्रभाव? (b) वाष्पन व संघनन की दरें आरंभ में कैसे बदलेंगी? (c) साम्य पुनः स्थापित होने पर क्या होगा, अंतिम वाष्प दाब?6.1 Band container mein liquid apni vapour ke saath equilibrium mein hai; volume achanak badhaya. (a) vapour pressure par shuru ka asar? (b) evaporation aur condensation rates shuru mein kaise badlenge? (c) equilibrium wapas aane par kya hoga, final vapour pressure?
(a) Vapour pressure falls at first (same vapour in a larger volume). (b) Evaporation rate stays the same (same T and surface); condensation rate drops (fewer molecules per unit volume hit the surface). (c) More liquid evaporates until the rates are equal again; the final vapour pressure is the same as before — it depends only on temperature.(a) आरंभ में वाष्प दाब घटता है (वही वाष्प बड़े आयतन में)। (b) वाष्पन दर वही रहती है (वही ताप व सतह); संघनन दर घटती है (प्रति इकाई आयतन कम अणु)। (c) और द्रव वाष्पित होता है जब तक दरें फिर बराबर न हों; अंतिम वाष्प दाब पहले जितना — यह केवल ताप पर निर्भर है।(a) Shuru mein vapour pressure ghatta hai (wahi vapour bade volume mein). (b) Evaporation rate same (same T aur surface); condensation rate ghatta hai (per unit volume kam molecules). (c) Aur liquid evaporate hota hai jab tak rates phir barabar na hon; final vapour pressure pehle jitna — sirf temperature par depend.
6.2 Kc for 2SO₂ + O₂ ⇌ 2SO₃ with [SO₂] = 0.60, [O₂] = 0.82, [SO₃] = 1.90 M.6.2 2SO₂ + O₂ ⇌ 2SO₃ का Kc, [SO₂] = 0.60, [O₂] = 0.82, [SO₃] = 1.90 M।6.2 2SO₂ + O₂ ⇌ 2SO₃ ka Kc, [SO₂] = 0.60, [O₂] = 0.82, [SO₃] = 1.90 M.
Kc = [SO₃]²/([SO₂]²[O₂]) = 1.90²/(0.60² × 0.82) = 3.61/0.2952 = 12.2 M⁻¹.Kc = [SO₃]²/([SO₂]²[O₂]) = 1.90²/(0.60² × 0.82) = 3.61/0.2952 = 12.2 M⁻¹.Kc = [SO₃]²/([SO₂]²[O₂]) = 1.90²/(0.60² × 0.82) = 3.61/0.2952 = 12.2 M⁻¹.
6.3 At 10⁵ Pa total pressure, iodine vapour contains 40 % by volume of I atoms (I₂ ⇌ 2I). Calculate Kp.6.3 10⁵ Pa कुल दाब पर आयोडीन वाष्प में आयतन से 40 % I परमाणु हैं (I₂ ⇌ 2I)। Kp ज्ञात कीजिए।6.3 10⁵ Pa total pressure par iodine vapour mein volume se 40 % I atoms hain (I₂ ⇌ 2I). Kp nikaaliye.
p(I) = 0.40 × 10⁵ Pa, p(I₂) = 0.60 × 10⁵ Pa. Kp = p(I)²/p(I₂) = (0.4 × 10⁵)²/(0.6 × 10⁵) = 2.67 × 10⁴ Pa.p(I) = 0.40 × 10⁵ Pa, p(I₂) = 0.60 × 10⁵ Pa। Kp = (0.4 × 10⁵)²/(0.6 × 10⁵) = 2.67 × 10⁴ Pa।p(I) = 0.40 × 10⁵ Pa, p(I₂) = 0.60 × 10⁵ Pa. Kp = (0.4 × 10⁵)²/(0.6 × 10⁵) = 2.67 × 10⁴ Pa.
6.4 Write Kc for: (i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) (ii) 2Cu(NO₃)₂(s) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g) (iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq) (iv) Fe³⁺(aq) + 3OH⁻(aq) ⇌ Fe(OH)₃(s) (v) I₂(s) + 5F₂ ⇌ 2IF₅6.4 Kc लिखिए: (i) 2NOCl ⇌ 2NO + Cl₂ (ii) 2Cu(NO₃)₂(s) ⇌ 2CuO(s) + 4NO₂ + O₂ (iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH + C₂H₅OH (iv) Fe³⁺ + 3OH⁻ ⇌ Fe(OH)₃(s) (v) I₂(s) + 5F₂ ⇌ 2IF₅6.4 Kc likhiye: (i) 2NOCl ⇌ 2NO + Cl₂ (ii) 2Cu(NO₃)₂(s) ⇌ 2CuO(s) + 4NO₂ + O₂ (iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH + C₂H₅OH (iv) Fe³⁺ + 3OH⁻ ⇌ Fe(OH)₃(s) (v) I₂(s) + 5F₂ ⇌ 2IF₅
(i) Kc = [NO]²[Cl₂]/[NOCl]² (ii) Kc = [NO₂]⁴[O₂] (solids omitted) (iii) Kc = [CH₃COOH][C₂H₅OH]/[CH₃COOC₂H₅] (water is the solvent, omitted) (iv) Kc = 1/([Fe³⁺][OH⁻]³) (v) Kc = [IF₅]²/[F₂]⁵(i) Kc = [NO]²[Cl₂]/[NOCl]² (ii) Kc = [NO₂]⁴[O₂] (ठोस नहीं लिखे) (iii) Kc = [CH₃COOH][C₂H₅OH]/[CH₃COOC₂H₅] (जल विलायक है) (iv) Kc = 1/([Fe³⁺][OH⁻]³) (v) Kc = [IF₅]²/[F₂]⁵(i) Kc = [NO]²[Cl₂]/[NOCl]² (ii) Kc = [NO₂]⁴[O₂] (solids nahi likhe) (iii) Kc = [CH₃COOH][C₂H₅OH]/[CH₃COOC₂H₅] (paani solvent hai) (iv) Kc = 1/([Fe³⁺][OH⁻]³) (v) Kc = [IF₅]²/[F₂]⁵
6.5 Find Kc from Kp: (i) 2NOCl ⇌ 2NO + Cl₂, Kp = 1.8 × 10⁻² at 500 K (ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = 167 at 1073 K6.5 Kp से Kc: (i) 2NOCl ⇌ 2NO + Cl₂, 500 K पर Kp = 1.8 × 10⁻² (ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g), 1073 K पर Kp = 1676.5 Kp se Kc: (i) 2NOCl ⇌ 2NO + Cl₂, 500 K par Kp = 1.8 × 10⁻² (ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g), 1073 K par Kp = 167
Kc = Kp/(RT)Δn, Δn = 1 in both. (i) Kc = 1.8 × 10⁻²/(0.0831 × 500) = 4.33 × 10⁻⁴ (ii) Kc = 167/(0.0831 × 1073) = 1.87Kc = Kp/(RT)Δn, Δn = 1 in both. (i) Kc = 1.8 × 10⁻²/(0.0831 × 500) = 4.33 × 10⁻⁴ (ii) Kc = 167/(0.0831 × 1073) = 1.87Kc = Kp/(RT)Δn, Δn = 1 in both. (i) Kc = 1.8 × 10⁻²/(0.0831 × 500) = 4.33 × 10⁻⁴ (ii) Kc = 167/(0.0831 × 1073) = 1.87
6.6 NO + O₃ ⇌ NO₂ + O₂, Kc = 6.3 × 10¹⁴ at 1000 K. Kc for the reverse reaction?6.6 NO + O₃ ⇌ NO₂ + O₂, 1000 K पर Kc = 6.3 × 10¹⁴। उलटी अभिक्रिया का Kc?6.6 NO + O₃ ⇌ NO₂ + O₂, 1000 K par Kc = 6.3 × 10¹⁴. Ulti reaction ka Kc?
Kc′ = 1/Kc = 1/(6.3 × 10¹⁴) = 1.59 × 10⁻¹⁵Kc′ = 1/Kc = 1/(6.3 × 10¹⁴) = 1.59 × 10⁻¹⁵Kc′ = 1/Kc = 1/(6.3 × 10¹⁴) = 1.59 × 10⁻¹⁵
6.7 Why can pure liquids and solids be ignored while writing the equilibrium constant expression?6.7 साम्य स्थिरांक लिखते समय शुद्ध द्रव और ठोस क्यों छोड़ दिए जाते हैं?6.7 Equilibrium constant likhte waqt pure liquids aur solids kyun chhod diye jaate hain?
The molar concentration of a pure solid or liquid = density/molar mass, which is constant at a given temperature however much is present. These constants are absorbed into K (their activity is taken as 1).शुद्ध ठोस या द्रव की मोलर सांद्रता = घनत्व/मोलर द्रव्यमान, जो दिए गए ताप पर स्थिर रहती है चाहे कितनी भी मात्रा हो। ये स्थिरांक K में मिल जाते हैं (इनकी सक्रियता 1 मानी जाती है)।Pure solid ya liquid ki molar concentration = density/molar mass, jo diye gaye temperature par constant rehti hai chahe kitni bhi matra ho. Ye constants K mein mil jaate hain (inki activity 1 maani jaati hai).
6.8 2N₂ + O₂ ⇌ 2N₂O; 0.482 mol N₂ and 0.933 mol O₂ in a 10 L vessel, Kc = 2.0 × 10⁻³⁷. Composition at equilibrium?6.8 2N₂ + O₂ ⇌ 2N₂O; 10 L पात्र में 0.482 mol N₂ और 0.933 mol O₂, Kc = 2.0 × 10⁻³⁷। साम्य संघटन?6.8 2N₂ + O₂ ⇌ 2N₂O; 10 L vessel mein 0.482 mol N₂ aur 0.933 mol O₂, Kc = 2.0 × 10⁻³⁷. Equilibrium composition?
Kc is tiny, so almost nothing reacts: [N₂] ≈ 0.0482 M, [O₂] ≈ 0.0933 M. [N₂O]² = Kc[N₂]²[O₂] = 2.0 × 10⁻³⁷ × 0.0482² × 0.0933 → [N₂O] = 6.6 × 10⁻²¹ M.Kc बहुत छोटा है, लगभग कुछ नहीं बदलता: [N₂] ≈ 0.0482 M, [O₂] ≈ 0.0933 M। [N₂O]² = Kc[N₂]²[O₂] → [N₂O] = 6.6 × 10⁻²¹ M।Kc bahut chhota hai, lagbhag kuch nahi badalta: [N₂] ≈ 0.0482 M, [O₂] ≈ 0.0933 M. [N₂O]² = Kc[N₂]²[O₂] → [N₂O] = 6.6 × 10⁻²¹ M.
6.9 2NO + Br₂ ⇌ 2NOBr; 0.087 mol NO and 0.0437 mol Br₂ give 0.0518 mol NOBr at equilibrium. Amounts of NO and Br₂?6.9 2NO + Br₂ ⇌ 2NOBr; 0.087 mol NO और 0.0437 mol Br₂ से साम्य पर 0.0518 mol NOBr। NO और Br₂ की मात्राएँ?6.9 2NO + Br₂ ⇌ 2NOBr; 0.087 mol NO aur 0.0437 mol Br₂ se equilibrium par 0.0518 mol NOBr. NO aur Br₂ ki matra?
NO used = 0.0518 mol, Br₂ used = 0.0259 mol. NO left = 0.087 − 0.0518 = 0.0352 mol; Br₂ left = 0.0437 − 0.0259 = 0.0178 mol.खर्च NO = 0.0518 mol, Br₂ = 0.0259 mol। शेष NO = 0.0352 mol; शेष Br₂ = 0.0178 mol।Kharch NO = 0.0518 mol, Br₂ = 0.0259 mol. Bacha NO = 0.0352 mol; bacha Br₂ = 0.0178 mol.
6.10 2SO₂ + O₂ ⇌ 2SO₃, Kp = 2.0 × 10¹⁰ bar⁻¹ at 450 K. Kc?6.10 2SO₂ + O₂ ⇌ 2SO₃, 450 K पर Kp = 2.0 × 10¹⁰ bar⁻¹। Kc?6.10 2SO₂ + O₂ ⇌ 2SO₃, 450 K par Kp = 2.0 × 10¹⁰ bar⁻¹. Kc?
Δn = 2 − 3 = −1, so Kc = Kp × RT = 2.0 × 10¹⁰ × 0.0831 × 450 = 7.48 × 10¹¹ M⁻¹.Δn = 2 − 3 = −1, so Kc = Kp × RT = 2.0 × 10¹⁰ × 0.0831 × 450 = 7.48 × 10¹¹ M⁻¹.Δn = 2 − 3 = −1, so Kc = Kp × RT = 2.0 × 10¹⁰ × 0.0831 × 450 = 7.48 × 10¹¹ M⁻¹.
6.11 HI at 0.2 atm in a flask; at equilibrium p(HI) = 0.04 atm. Kp for 2HI ⇌ H₂ + I₂?6.11 फ़्लास्क में 0.2 atm HI; साम्य पर p(HI) = 0.04 atm। 2HI ⇌ H₂ + I₂ का Kp?6.11 Flask mein 0.2 atm HI; equilibrium par p(HI) = 0.04 atm. 2HI ⇌ H₂ + I₂ ka Kp?
HI decomposed = 0.16 atm → p(H₂) = p(I₂) = 0.08 atm. Kp = 0.08 × 0.08 / 0.04² = 4.0.HI decomposed = 0.16 atm → p(H₂) = p(I₂) = 0.08 atm. Kp = 0.08 × 0.08 / 0.04² = 4.0.HI decomposed = 0.16 atm → p(H₂) = p(I₂) = 0.08 atm. Kp = 0.08 × 0.08 / 0.04² = 4.0.
6.12 1.57 mol N₂, 1.92 mol H₂ and 8.13 mol NH₃ in 20 L at 500 K; Kc = 1.7 × 10². At equilibrium? If not, which direction?6.12 500 K पर 20 L में 1.57 mol N₂, 1.92 mol H₂ और 8.13 mol NH₃; Kc = 1.7 × 10²। साम्य पर? नहीं तो किस दिशा में?6.12 500 K par 20 L mein 1.57 mol N₂, 1.92 mol H₂ aur 8.13 mol NH₃; Kc = 1.7 × 10². Equilibrium par? Nahi to kis direction?
[N₂] = 0.0785, [H₂] = 0.096, [NH₃] = 0.4065 M. Qc = 0.4065²/(0.0785 × 0.096³) = 2.38 × 10³ > Kc, so not at equilibrium; the net reaction goes backward (NH₃ decomposes).[N₂] = 0.0785, [H₂] = 0.096, [NH₃] = 0.4065 M। Qc = 2.38 × 10³ > Kc, अतः साम्य पर नहीं; नेट अभिक्रिया पश्च दिशा में (NH₃ विघटित)।[N₂] = 0.0785, [H₂] = 0.096, [NH₃] = 0.4065 M. Qc = 2.38 × 10³ > Kc, isliye equilibrium par nahi; net reaction backward (NH₃ tootega).
6.13 Kc = [NH₃]⁴[O₂]⁵ / ([NO]⁴[H₂O]⁶). Write the balanced equation.6.13 Kc = [NH₃]⁴[O₂]⁵ / ([NO]⁴[H₂O]⁶)। संतुलित समीकरण लिखिए।6.13 Kc = [NH₃]⁴[O₂]⁵ / ([NO]⁴[H₂O]⁶). Balanced equation likhiye.
4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g) (products in the numerator).4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g) (अंश में उत्पाद)।4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g) (numerator mein products).
6.14 1 mol H₂O and 1 mol CO in 10 L at 725 K; 40 % of the water reacts: H₂O + CO ⇌ H₂ + CO₂. Equilibrium constant?6.14 725 K पर 10 L में 1 mol H₂O और 1 mol CO; 40 % जल अभिक्रिया करता है: H₂O + CO ⇌ H₂ + CO₂। साम्य स्थिरांक?6.14 725 K par 10 L mein 1 mol H₂O aur 1 mol CO; 40 % paani react karta hai: H₂O + CO ⇌ H₂ + CO₂. Equilibrium constant?
At equilibrium: H₂O = CO = 0.6 mol, H₂ = CO₂ = 0.4 mol (in 10 L). Kc = (0.04 × 0.04)/(0.06 × 0.06) = 0.44.साम्य पर: H₂O = CO = 0.6 mol, H₂ = CO₂ = 0.4 mol (10 L में)। Kc = (0.04 × 0.04)/(0.06 × 0.06) = 0.44।Equilibrium par: H₂O = CO = 0.6 mol, H₂ = CO₂ = 0.4 mol (10 L mein). Kc = (0.04 × 0.04)/(0.06 × 0.06) = 0.44.
6.15 H₂ + I₂ ⇌ 2HI, K = 54.8 at 700 K. If [HI] = 0.5 M at equilibrium (starting from HI), find [H₂] and [I₂].6.15 H₂ + I₂ ⇌ 2HI, 700 K पर K = 54.8। साम्य पर [HI] = 0.5 M (HI से आरंभ), [H₂] और [I₂]?6.15 H₂ + I₂ ⇌ 2HI, 700 K par K = 54.8. Equilibrium par [HI] = 0.5 M (HI se shuru), [H₂] aur [I₂]?
For 2HI ⇌ H₂ + I₂, K′ = 1/54.8. [H₂] = [I₂] = x: x²/0.5² = 1/54.8 → x = 0.068 M.2HI ⇌ H₂ + I₂ के लिए K′ = 1/54.8। [H₂] = [I₂] = x: x²/0.5² = 1/54.8 → x = 0.068 M।2HI ⇌ H₂ + I₂ ke liye K′ = 1/54.8. [H₂] = [I₂] = x: x²/0.5² = 1/54.8 → x = 0.068 M.
6.16 Equilibrium concentrations when [ICl] = 0.78 M at start: 2ICl ⇌ I₂ + Cl₂, Kc = 0.14.6.16 आरंभ में [ICl] = 0.78 M: 2ICl ⇌ I₂ + Cl₂, Kc = 0.14। साम्य सांद्रताएँ?6.16 Shuru mein [ICl] = 0.78 M: 2ICl ⇌ I₂ + Cl₂, Kc = 0.14. Equilibrium concentrations?
x²/(0.78 − 2x)² = 0.14 → x/(0.78 − 2x) = 0.374 → x = 0.167. [I₂] = [Cl₂] = 0.167 M, [ICl] = 0.446 M.x²/(0.78 − 2x)² = 0.14 → x/(0.78 − 2x) = 0.374 → x = 0.167. [I₂] = [Cl₂] = 0.167 M, [ICl] = 0.446 M.x²/(0.78 − 2x)² = 0.14 → x/(0.78 − 2x) = 0.374 → x = 0.167. [I₂] = [Cl₂] = 0.167 M, [ICl] = 0.446 M.
6.17 C₂H₆ ⇌ C₂H₄ + H₂, Kp = 0.04 atm at 899 K. C₂H₆ placed at 4.0 atm; equilibrium pressure of C₂H₆?6.17 C₂H₆ ⇌ C₂H₄ + H₂, 899 K पर Kp = 0.04 atm। 4.0 atm पर C₂H₆ रखा; साम्य पर C₂H₆ का दाब?6.17 C₂H₆ ⇌ C₂H₄ + H₂, 899 K par Kp = 0.04 atm. 4.0 atm par C₂H₆ rakha; equilibrium par C₂H₆ ka pressure?
p²/(4.0 − p) = 0.04 → p² + 0.04p − 0.16 = 0 → p = 0.38 atm. p(C₂H₆) = 4.0 − 0.38 = 3.62 atm.p²/(4.0 − p) = 0.04 → p = 0.38 atm। p(C₂H₆) = 3.62 atm।p²/(4.0 − p) = 0.04 → p = 0.38 atm. p(C₂H₆) = 3.62 atm.
6.18 CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l). (i) Write Qc. (ii) From 1.00 mol acid and 0.18 mol ethanol, 0.171 mol ester at equilibrium (293 K) — find K. (iii) From 0.5 mol ethanol and 1.0 mol acid, 0.214 mol ester after some time — equilibrium reached?6.18 CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)। (i) Qc लिखिए। (ii) 1.00 mol अम्ल और 0.18 mol एथेनॉल से साम्य पर 0.171 mol एस्टर — K? (iii) 0.5 mol एथेनॉल और 1.0 mol अम्ल से कुछ समय बाद 0.214 mol एस्टर — साम्य?6.18 CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l). (i) Qc likhiye. (ii) 1.00 mol acid aur 0.18 mol ethanol se equilibrium par 0.171 mol ester — K? (iii) 0.5 mol ethanol aur 1.0 mol acid se kuch der baad 0.214 mol ester — equilibrium?
(i) Qc = [CH₃COOC₂H₅][H₂O]/([CH₃COOH][C₂H₅OH]). (ii) Ester = water = 0.171, acid = 0.829, ethanol = 0.009 mol: K = 0.171²/(0.829 × 0.009) = 3.92. (iii) Ester = water = 0.214, acid = 0.786, ethanol = 0.286: Qc = 0.214²/(0.786 × 0.286) = 0.204 < K, so not at equilibrium; more ester will form.(i) Qc = [CH₃COOC₂H₅][H₂O]/([CH₃COOH][C₂H₅OH])। (ii) एस्टर = जल = 0.171, अम्ल = 0.829, एथेनॉल = 0.009 mol: K = 3.92। (iii) Qc = 0.214²/(0.786 × 0.286) = 0.204 < K, अतः साम्य नहीं; और एस्टर बनेगा।(i) Qc = [CH₃COOC₂H₅][H₂O]/([CH₃COOH][C₂H₅OH]). (ii) Ester = paani = 0.171, acid = 0.829, ethanol = 0.009 mol: K = 3.92. (iii) Qc = 0.214²/(0.786 × 0.286) = 0.204 < K, isliye equilibrium nahi; aur ester banega.
6.19 Pure PCl₅ at 473 K; at equilibrium [PCl₅] = 0.5 × 10⁻¹ M, Kc = 8.3 × 10⁻³. [PCl₃] and [Cl₂]?6.19 473 K पर शुद्ध PCl₅; साम्य पर [PCl₅] = 0.5 × 10⁻¹ M, Kc = 8.3 × 10⁻³। [PCl₃] और [Cl₂]?6.19 473 K par pure PCl₅; equilibrium par [PCl₅] = 0.5 × 10⁻¹ M, Kc = 8.3 × 10⁻³. [PCl₃] aur [Cl₂]?
x² = 8.3 × 10⁻³ × 0.05 = 4.15 × 10⁻⁴ → [PCl₃] = [Cl₂] = 0.020 M.x² = 8.3 × 10⁻³ × 0.05 = 4.15 × 10⁻⁴ → [PCl₃] = [Cl₂] = 0.020 M.x² = 8.3 × 10⁻³ × 0.05 = 4.15 × 10⁻⁴ → [PCl₃] = [Cl₂] = 0.020 M.
6.20 FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g), Kp = 0.265 at 1050 K; initial p(CO) = 1.4 atm, p(CO₂) = 0.80 atm. Equilibrium pressures?6.20 FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g), 1050 K पर Kp = 0.265; आरंभ में p(CO) = 1.4, p(CO₂) = 0.80 atm। साम्य दाब?6.20 FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g), 1050 K par Kp = 0.265; shuru mein p(CO) = 1.4, p(CO₂) = 0.80 atm. Equilibrium pressures?
Qp = 0.80/1.4 = 0.571 > Kp → goes backward. (0.80 − x)/(1.4 + x) = 0.265 → x = 0.339. p(CO) = 1.74 atm, p(CO₂) = 0.46 atm.Qp = 0.571 > Kp → पश्च दिशा। (0.80 − x)/(1.4 + x) = 0.265 → x = 0.339। p(CO) = 1.74 atm, p(CO₂) = 0.46 atm।Qp = 0.571 > Kp → backward. (0.80 − x)/(1.4 + x) = 0.265 → x = 0.339. p(CO) = 1.74 atm, p(CO₂) = 0.46 atm.
6.21 N₂ + 3H₂ ⇌ 2NH₃, Kc = 0.061 at 500 K; mixture has 3.0 M N₂, 2.0 M H₂, 0.5 M NH₃. At equilibrium? Direction?6.21 N₂ + 3H₂ ⇌ 2NH₃, 500 K पर Kc = 0.061; मिश्रण में 3.0 M N₂, 2.0 M H₂, 0.5 M NH₃। साम्य? दिशा?6.21 N₂ + 3H₂ ⇌ 2NH₃, 500 K par Kc = 0.061; mixture mein 3.0 M N₂, 2.0 M H₂, 0.5 M NH₃. Equilibrium? Direction?
Qc = 0.5²/(3.0 × 2.0³) = 0.0104 < 0.061 → not at equilibrium; goes forward (more NH₃).Qc = 0.0104 < 0.061 → साम्य नहीं; अग्र दिशा (और NH₃)।Qc = 0.0104 < 0.061 → equilibrium nahi; forward (aur NH₃).
6.22 2BrCl ⇌ Br₂ + Cl₂, Kc = 32 at 500 K. Initially pure BrCl at 3.3 × 10⁻³ M. [BrCl] at equilibrium?6.22 2BrCl ⇌ Br₂ + Cl₂, 500 K पर Kc = 32। आरंभ में शुद्ध BrCl 3.3 × 10⁻³ M। साम्य पर [BrCl]?6.22 2BrCl ⇌ Br₂ + Cl₂, 500 K par Kc = 32. Shuru mein pure BrCl 3.3 × 10⁻³ M. Equilibrium par [BrCl]?
x/(3.3 × 10⁻³ − 2x) = √32 = 5.66 → x = 1.52 × 10⁻³. [BrCl] = 3.3 × 10⁻³ − 3.03 × 10⁻³ = 2.7 × 10⁻⁴ M.x/(3.3 × 10⁻³ − 2x) = √32 = 5.66 → x = 1.52 × 10⁻³. [BrCl] = 3.3 × 10⁻³ − 3.03 × 10⁻³ = 2.7 × 10⁻⁴ M.x/(3.3 × 10⁻³ − 2x) = √32 = 5.66 → x = 1.52 × 10⁻³. [BrCl] = 3.3 × 10⁻³ − 3.03 × 10⁻³ = 2.7 × 10⁻⁴ M.
6.23 C(s) + CO₂ ⇌ 2CO at 1127 K and 1 atm; the gas is 90.55 % CO by mass. Kc?6.23 1127 K व 1 atm पर C(s) + CO₂ ⇌ 2CO; गैस में द्रव्यमान से 90.55 % CO। Kc?6.23 1127 K aur 1 atm par C(s) + CO₂ ⇌ 2CO; gas mein mass se 90.55 % CO. Kc?
In 100 g: CO = 90.55/28 = 3.23 mol, CO₂ = 9.45/44 = 0.215 mol; x(CO) = 0.938. p(CO) = 0.938, p(CO₂) = 0.062 atm. Kp = 0.938²/0.062 = 14.1; Kc = Kp/RT = 14.1/(0.0821 × 1127) = 0.153.100 g में: CO = 3.23 mol, CO₂ = 0.215 mol; x(CO) = 0.938। Kp = 0.938²/0.062 = 14.1; Kc = Kp/RT = 0.153।100 g mein: CO = 3.23 mol, CO₂ = 0.215 mol; x(CO) = 0.938. Kp = 0.938²/0.062 = 14.1; Kc = Kp/RT = 0.153.
6.24 NO + ½O₂ ⇌ NO₂ at 298 K; ΔfG⊖: NO₂ 52.0, NO 87.0, O₂ 0 kJ mol⁻¹. Find (a) ΔG⊖ (b) K.6.24 NO + ½O₂ ⇌ NO₂, 298 K; ΔfG⊖: NO₂ 52.0, NO 87.0, O₂ 0 kJ mol⁻¹। (a) ΔG⊖ (b) K?6.24 NO + ½O₂ ⇌ NO₂, 298 K; ΔfG⊖: NO₂ 52.0, NO 87.0, O₂ 0 kJ mol⁻¹. (a) ΔG⊖ (b) K?
(a) ΔG⊖ = 52.0 − 87.0 = −35.0 kJ mol⁻¹. (b) log K = 35000/(2.303 × 8.314 × 298) = 6.134 → K = 1.36 × 10⁶.(a) ΔG⊖ = 52.0 − 87.0 = −35.0 kJ mol⁻¹. (b) log K = 35000/(2.303 × 8.314 × 298) = 6.134 → K = 1.36 × 10⁶.(a) ΔG⊖ = 52.0 − 87.0 = −35.0 kJ mol⁻¹. (b) log K = 35000/(2.303 × 8.314 × 298) = 6.134 → K = 1.36 × 10⁶.
6.25 Do moles of products increase, decrease or stay the same when pressure is lowered (volume increased)? (a) PCl₅ ⇌ PCl₃ + Cl₂ (b) CaO(s) + CO₂ ⇌ CaCO₃(s) (c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)6.25 दाब घटाने (आयतन बढ़ाने) पर उत्पादों के मोल बढ़ेंगे, घटेंगे या वही? (a) PCl₅ ⇌ PCl₃ + Cl₂ (b) CaO(s) + CO₂ ⇌ CaCO₃(s) (c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)6.25 Pressure ghatane (volume badhane) par products ke moles badhenge, ghatenge ya same? (a) PCl₅ ⇌ PCl₃ + Cl₂ (b) CaO(s) + CO₂ ⇌ CaCO₃(s) (c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)
Lower pressure favours more gas moles. (a) increase (1 → 2) (b) decrease (CO₂ is released, CaCO₃ falls) (c) same (4 = 4 gas moles).कम दाब अधिक गैस मोल वाली ओर। (a) बढ़ेंगे (1 → 2) (b) घटेंगे (c) वही (4 = 4)।Kam pressure zyada gas moles wali taraf. (a) badhenge (1 → 2) (b) ghatenge (c) same (4 = 4).
6.26 Which are affected by increasing pressure, and in which direction? (i) COCl₂ ⇌ CO + Cl₂ (ii) CH₄ + 2S₂ ⇌ CS₂ + 2H₂S (iii) CO₂ + C(s) ⇌ 2CO (iv) 2H₂ + CO ⇌ CH₃OH (v) CaCO₃(s) ⇌ CaO(s) + CO₂ (vi) 4NH₃ + 5O₂ ⇌ 4NO + 6H₂O(g)6.26 दाब बढ़ाने पर कौन प्रभावित, किस दिशा में? (i) COCl₂ ⇌ CO + Cl₂ (ii) CH₄ + 2S₂ ⇌ CS₂ + 2H₂S (iii) CO₂ + C(s) ⇌ 2CO (iv) 2H₂ + CO ⇌ CH₃OH (v) CaCO₃(s) ⇌ CaO(s) + CO₂ (vi) 4NH₃ + 5O₂ ⇌ 4NO + 6H₂O(g)6.26 Pressure badhane par kaun affect, kis direction? (i) COCl₂ ⇌ CO + Cl₂ (ii) CH₄ + 2S₂ ⇌ CS₂ + 2H₂S (iii) CO₂ + C(s) ⇌ 2CO (iv) 2H₂ + CO ⇌ CH₃OH (v) CaCO₃(s) ⇌ CaO(s) + CO₂ (vi) 4NH₃ + 5O₂ ⇌ 4NO + 6H₂O(g)
(i) backward (ii) not affected (3 = 3) (iii) backward (iv) forward (v) backward (vi) backward (9 → 10 gas moles).(i) पश्च (ii) अप्रभावित (3 = 3) (iii) पश्च (iv) अग्र (v) पश्च (vi) पश्च (9 → 10 गैस मोल)।(i) backward (ii) affect nahi (3 = 3) (iii) backward (iv) forward (v) backward (vi) backward (9 → 10 gas moles).
6.27 H₂ + Br₂ ⇌ 2HBr, K = 1.6 × 10⁵ at 1024 K. 10.0 bar of HBr is put in a sealed container. Equilibrium pressures?6.27 H₂ + Br₂ ⇌ 2HBr, 1024 K पर K = 1.6 × 10⁵। बंद पात्र में 10.0 bar HBr। साम्य दाब?6.27 H₂ + Br₂ ⇌ 2HBr, 1024 K par K = 1.6 × 10⁵. Band container mein 10.0 bar HBr. Equilibrium pressures?
For 2HBr ⇌ H₂ + Br₂, K′ = 1/(1.6 × 10⁵). p²/(10 − 2p)² = K′ → p/(10 − 2p) = 2.5 × 10⁻³ → p = 2.49 × 10⁻² bar. p(H₂) = p(Br₂) = 2.5 × 10⁻² bar, p(HBr) ≈ 9.95 bar.2HBr ⇌ H₂ + Br₂ के लिए K′ = 1/(1.6 × 10⁵)। p/(10 − 2p) = 2.5 × 10⁻³ → p(H₂) = p(Br₂) = 2.5 × 10⁻² bar, p(HBr) ≈ 9.95 bar।2HBr ⇌ H₂ + Br₂ ke liye K′ = 1/(1.6 × 10⁵). p/(10 − 2p) = 2.5 × 10⁻³ → p(H₂) = p(Br₂) = 2.5 × 10⁻² bar, p(HBr) ≈ 9.95 bar.
6.28 CH₄ + H₂O ⇌ CO + 3H₂ (endothermic). (a) Write Kp. (b) Effect on Kp and composition of (i) increasing pressure (ii) increasing temperature (iii) a catalyst.6.28 CH₄ + H₂O ⇌ CO + 3H₂ (ऊष्माशोषी)। (a) Kp लिखिए। (b) Kp और संघटन पर प्रभाव: (i) दाब बढ़ाना (ii) ताप बढ़ाना (iii) उत्प्रेरक।6.28 CH₄ + H₂O ⇌ CO + 3H₂ (endothermic). (a) Kp likhiye. (b) Kp aur composition par asar: (i) pressure badhana (ii) temperature badhana (iii) catalyst.
(a) Kp = p(CO)·p(H₂)³ / (p(CH₄)·p(H₂O)). (b) (i) Kp unchanged; shifts backward (4 → 2 gas moles), less H₂. (ii) Kp increases; shifts forward (endothermic). (iii) No change in Kp or composition; equilibrium reached faster.(a) Kp = p(CO)·p(H₂)³ / (p(CH₄)·p(H₂O))। (b) (i) Kp वही; पश्च दिशा (कम H₂)। (ii) Kp बढ़ता है; अग्र दिशा (ऊष्माशोषी)। (iii) Kp व संघटन नहीं बदलते; साम्य जल्दी।(a) Kp = p(CO)·p(H₂)³ / (p(CH₄)·p(H₂O)). (b) (i) Kp same; backward (kam H₂). (ii) Kp badhta hai; forward (endothermic). (iii) Kp aur composition nahi badalte; equilibrium jaldi.
6.29 Effect on 2H₂ + CO ⇌ CH₃OH of: (a) adding H₂ (b) adding CH₃OH (c) removing CO (d) removing CH₃OH6.29 2H₂ + CO ⇌ CH₃OH पर प्रभाव: (a) H₂ मिलाना (b) CH₃OH मिलाना (c) CO हटाना (d) CH₃OH हटाना6.29 2H₂ + CO ⇌ CH₃OH par asar: (a) H₂ milana (b) CH₃OH milana (c) CO hatana (d) CH₃OH hatana
(a) forward (b) backward (c) backward (d) forward.(a) अग्र (b) पश्च (c) पश्च (d) अग्र।(a) forward (b) backward (c) backward (d) forward.
6.30 PCl₅ ⇌ PCl₃ + Cl₂, Kc = 8.3 × 10⁻³ at 473 K, ΔrH⊖ = 124.0 kJ mol⁻¹. (a) Kc expression (b) Kc of reverse reaction (c) effect on Kc of (i) more PCl₅ (ii) higher pressure (iii) higher temperature6.30 PCl₅ ⇌ PCl₃ + Cl₂, 473 K पर Kc = 8.3 × 10⁻³, ΔrH⊖ = 124.0 kJ mol⁻¹। (a) Kc व्यंजक (b) उलटी अभिक्रिया का Kc (c) Kc पर प्रभाव: (i) और PCl₅ (ii) अधिक दाब (iii) अधिक ताप6.30 PCl₅ ⇌ PCl₃ + Cl₂, 473 K par Kc = 8.3 × 10⁻³, ΔrH⊖ = 124.0 kJ mol⁻¹. (a) Kc expression (b) ulti reaction ka Kc (c) Kc par asar: (i) aur PCl₅ (ii) zyada pressure (iii) zyada temperature
(a) Kc = [PCl₃][Cl₂]/[PCl₅] (b) 1/(8.3 × 10⁻³) = 120.5 (c) (i) no change (ii) no change (iii) Kc increases (endothermic).(a) Kc = [PCl₃][Cl₂]/[PCl₅] (b) 1/(8.3 × 10⁻³) = 120.5 (c) (i) कोई परिवर्तन नहीं (ii) कोई परिवर्तन नहीं (iii) Kc बढ़ता है (ऊष्माशोषी)।(a) Kc = [PCl₃][Cl₂]/[PCl₅] (b) 1/(8.3 × 10⁻³) = 120.5 (c) (i) koi change nahi (ii) koi change nahi (iii) Kc badhta hai (endothermic).
6.31 CO + H₂O ⇌ CO₂ + H₂ at 400 °C, Kp = 10.1; p(CO) = p(H₂O) = 4.0 bar initially. p(H₂) at equilibrium?6.31 CO + H₂O ⇌ CO₂ + H₂, 400 °C पर Kp = 10.1; आरंभ में p(CO) = p(H₂O) = 4.0 bar। साम्य पर p(H₂)?6.31 CO + H₂O ⇌ CO₂ + H₂, 400 °C par Kp = 10.1; shuru mein p(CO) = p(H₂O) = 4.0 bar. Equilibrium par p(H₂)?
x²/(4.0 − x)² = 10.1 → x/(4.0 − x) = 3.18 → p(H₂) = x = 3.04 bar.x/(4.0 − x) = 3.18 → p(H₂) = 3.04 bar।x/(4.0 − x) = 3.18 → p(H₂) = 3.04 bar.
6.32 Which will have appreciable amounts of both reactants and products? (a) Cl₂ ⇌ 2Cl, Kc = 5 × 10⁻³⁹ (b) Cl₂ + 2NO ⇌ 2NOCl, Kc = 3.7 × 10⁸ (c) Cl₂ + 2NO₂ ⇌ 2NO₂Cl, Kc = 1.86.32 किसमें अभिकारक और उत्पाद दोनों पर्याप्त होंगे? (a) Kc = 5 × 10⁻³⁹ (b) Kc = 3.7 × 10⁸ (c) Kc = 1.86.32 Kismein reactants aur products dono kaafi honge? (a) Kc = 5 × 10⁻³⁹ (b) Kc = 3.7 × 10⁸ (c) Kc = 1.8
(c), because Kc = 1.8 lies between 10⁻³ and 10³. (a) mostly Cl₂, (b) mostly NOCl.(c), क्योंकि Kc = 1.8, 10⁻³ और 10³ के बीच है। (a) अधिकतर Cl₂, (b) अधिकतर NOCl।(c), kyunki Kc = 1.8, 10⁻³ aur 10³ ke beech hai. (a) zyadatar Cl₂, (b) zyadatar NOCl.
6.33 3O₂ ⇌ 2O₃, Kc = 2.0 × 10⁻⁵⁰ at 25 °C; [O₂] = 1.6 × 10⁻² M. [O₃]?6.33 3O₂ ⇌ 2O₃, 25 °C पर Kc = 2.0 × 10⁻⁵⁰; [O₂] = 1.6 × 10⁻² M। [O₃]?6.33 3O₂ ⇌ 2O₃, 25 °C par Kc = 2.0 × 10⁻⁵⁰; [O₂] = 1.6 × 10⁻² M. [O₃]?
[O₃]² = Kc[O₂]³ = 2.0 × 10⁻⁵⁰ × (1.6 × 10⁻²)³ = 8.2 × 10⁻⁵⁶ → [O₃] = 2.86 × 10⁻²⁸ M.[O₃]² = Kc[O₂]³ = 2.0 × 10⁻⁵⁰ × (1.6 × 10⁻²)³ = 8.2 × 10⁻⁵⁶ → [O₃] = 2.86 × 10⁻²⁸ M.[O₃]² = Kc[O₂]³ = 2.0 × 10⁻⁵⁰ × (1.6 × 10⁻²)³ = 8.2 × 10⁻⁵⁶ → [O₃] = 2.86 × 10⁻²⁸ M.
6.34 CO + 3H₂ ⇌ CH₄ + H₂O at 1300 K in 1 L: 0.30 mol CO, 0.10 mol H₂, 0.02 mol H₂O, Kc = 3.90. [CH₄]?6.34 1300 K पर 1 L में CO + 3H₂ ⇌ CH₄ + H₂O: 0.30 mol CO, 0.10 mol H₂, 0.02 mol H₂O, Kc = 3.90। [CH₄]?6.34 1300 K par 1 L mein CO + 3H₂ ⇌ CH₄ + H₂O: 0.30 mol CO, 0.10 mol H₂, 0.02 mol H₂O, Kc = 3.90. [CH₄]?
[CH₄] = Kc[CO][H₂]³/[H₂O] = 3.90 × 0.30 × (0.10)³/0.02 = 5.85 × 10⁻² M.[CH₄] = Kc[CO][H₂]³/[H₂O] = 3.90 × 0.30 × (0.10)³/0.02 = 5.85 × 10⁻² M.[CH₄] = Kc[CO][H₂]³/[H₂O] = 3.90 × 0.30 × (0.10)³/0.02 = 5.85 × 10⁻² M.
6.35 What is a conjugate acid–base pair? Find the conjugate acid/base of HNO₂, CN⁻, HClO₄, F⁻, OH⁻, CO₃²⁻ and S²⁻.6.35 संयुग्मी अम्ल–क्षारक युग्म क्या है? HNO₂, CN⁻, HClO₄, F⁻, OH⁻, CO₃²⁻ और S²⁻ के संयुग्मी अम्ल/क्षारक बताइए।6.35 Conjugate acid–base pair kya hai? HNO₂, CN⁻, HClO₄, F⁻, OH⁻, CO₃²⁻ aur S²⁻ ke conjugate acid/base bataiye.
A pair that differs by one proton (H⁺). HNO₂ → base NO₂⁻; CN⁻ → acid HCN; HClO₄ → base ClO₄⁻; F⁻ → acid HF; OH⁻ → acid H₂O (base O²⁻); CO₃²⁻ → acid HCO₃⁻; S²⁻ → acid HS⁻.एक प्रोटॉन (H⁺) से भिन्न युग्म। HNO₂ → क्षारक NO₂⁻; CN⁻ → अम्ल HCN; HClO₄ → क्षारक ClO₄⁻; F⁻ → अम्ल HF; OH⁻ → अम्ल H₂O (क्षारक O²⁻); CO₃²⁻ → अम्ल HCO₃⁻; S²⁻ → अम्ल HS⁻।Ek proton (H⁺) se alag pair. HNO₂ → base NO₂⁻; CN⁻ → acid HCN; HClO₄ → base ClO₄⁻; F⁻ → acid HF; OH⁻ → acid H₂O (base O²⁻); CO₃²⁻ → acid HCO₃⁻; S²⁻ → acid HS⁻.
6.36 Which are Lewis acids? H₂O, BF₃, H⁺, NH₄⁺6.36 कौन लूइस अम्ल हैं? H₂O, BF₃, H⁺, NH₄⁺6.36 Kaun Lewis acids hain? H₂O, BF₃, H⁺, NH₄⁺
BF₃ and H⁺ (electron-pair acceptors). H₂O is a Lewis base; NH₄⁺ has no vacant orbital or lone pair (it is a Brönsted acid).BF₃ और H⁺ (इलेक्ट्रॉन-युग्म ग्राही)। H₂O लूइस क्षारक है; NH₄⁺ में न रिक्त कक्षक न एकाकी युग्म (यह ब्रॉन्स्टेड अम्ल है)।BF₃ aur H⁺ (electron-pair acceptors). H₂O Lewis base hai; NH₄⁺ mein na khaali orbital na lone pair (ye Brönsted acid hai).
6.37 Conjugate bases of the Brönsted acids HF, H₂SO₄ and HCO₃⁻?6.37 ब्रॉन्स्टेड अम्ल HF, H₂SO₄ और HCO₃⁻ के संयुग्मी क्षारक?6.37 Brönsted acids HF, H₂SO₄ aur HCO₃⁻ ke conjugate bases?
F⁻, HSO₄⁻, CO₃²⁻F⁻, HSO₄⁻, CO₃²⁻F⁻, HSO₄⁻, CO₃²⁻
6.38 Conjugate acids of the Brönsted bases NH₂⁻, NH₃ and HCOO⁻?6.38 ब्रॉन्स्टेड क्षारक NH₂⁻, NH₃ और HCOO⁻ के संयुग्मी अम्ल?6.38 Brönsted bases NH₂⁻, NH₃ aur HCOO⁻ ke conjugate acids?
NH₃, NH₄⁺, HCOOHNH₃, NH₄⁺, HCOOHNH₃, NH₄⁺, HCOOH
6.39 H₂O, HCO₃⁻, HSO₄⁻ and NH₃ can act as both Brönsted acids and bases. Give the conjugate acid and base of each.6.39 H₂O, HCO₃⁻, HSO₄⁻ और NH₃ अम्ल व क्षारक दोनों हैं। प्रत्येक का संयुग्मी अम्ल और क्षारक?6.39 H₂O, HCO₃⁻, HSO₄⁻ aur NH₃ acid aur base dono hain. Har ek ka conjugate acid aur base?
H₂O: acid H₃O⁺ / base OH⁻; HCO₃⁻: H₂CO₃ / CO₃²⁻; HSO₄⁻: H₂SO₄ / SO₄²⁻; NH₃: NH₄⁺ / NH₂⁻.H₂O: अम्ल H₃O⁺ / क्षारक OH⁻; HCO₃⁻: H₂CO₃ / CO₃²⁻; HSO₄⁻: H₂SO₄ / SO₄²⁻; NH₃: NH₄⁺ / NH₂⁻।H₂O: acid H₃O⁺ / base OH⁻; HCO₃⁻: H₂CO₃ / CO₃²⁻; HSO₄⁻: H₂SO₄ / SO₄²⁻; NH₃: NH₄⁺ / NH₂⁻.
6.40 Classify as Lewis acids or bases and show how they act: (a) OH⁻ (b) F⁻ (c) H⁺ (d) BCl₃6.40 लूइस अम्ल या क्षारक में बाँटिए और दिखाइए: (a) OH⁻ (b) F⁻ (c) H⁺ (d) BCl₃6.40 Lewis acid ya base mein baantiye aur dikhaiye: (a) OH⁻ (b) F⁻ (c) H⁺ (d) BCl₃
(a) OH⁻: base — gives a lone pair to H⁺ (OH⁻ + H⁺ → H₂O). (b) F⁻: base — F⁻ + BF₃ → BF₄⁻. (c) H⁺: acid — accepts a pair (H⁺ + NH₃ → NH₄⁺). (d) BCl₃: acid — B has 6 electrons and accepts a pair (BCl₃ + NH₃ → Cl₃B←NH₃).(a) OH⁻: क्षारक — H⁺ को एकाकी युग्म देता है (OH⁻ + H⁺ → H₂O)। (b) F⁻: क्षारक — F⁻ + BF₃ → BF₄⁻। (c) H⁺: अम्ल — युग्म ग्रहण करता है (H⁺ + NH₃ → NH₄⁺)। (d) BCl₃: अम्ल — B पर 6 इलेक्ट्रॉन, युग्म लेता है (BCl₃ + NH₃ → Cl₃B←NH₃)।(a) OH⁻: base — H⁺ ko lone pair deta hai (OH⁻ + H⁺ → H₂O). (b) F⁻: base — F⁻ + BF₃ → BF₄⁻. (c) H⁺: acid — pair leta hai (H⁺ + NH₃ → NH₄⁺). (d) BCl₃: acid — B par 6 electrons, pair leta hai (BCl₃ + NH₃ → Cl₃B←NH₃).
6.41 [H⁺] in a soft drink is 3.8 × 10⁻³ M. pH?6.41 शीतल पेय में [H⁺] = 3.8 × 10⁻³ M। pH?6.41 Soft drink mein [H⁺] = 3.8 × 10⁻³ M. pH?
pH = −log(3.8 × 10⁻³) = 2.42pH = −log(3.8 × 10⁻³) = 2.42pH = −log(3.8 × 10⁻³) = 2.42
6.42 pH of vinegar is 3.76. [H⁺]?6.42 सिरके का pH 3.76 है। [H⁺]?6.42 Vinegar ka pH 3.76 hai. [H⁺]?
[H⁺] = 10⁻³·⁷⁶ = 1.74 × 10⁻⁴ M[H⁺] = 10⁻³·⁷⁶ = 1.74 × 10⁻⁴ M[H⁺] = 10⁻³·⁷⁶ = 1.74 × 10⁻⁴ M
6.43 Ka of HF, HCOOH and HCN at 298 K are 6.8 × 10⁻⁴, 1.8 × 10⁻⁴ and 4.8 × 10⁻⁹. Kb of their conjugate bases?6.43 298 K पर HF, HCOOH और HCN के Ka: 6.8 × 10⁻⁴, 1.8 × 10⁻⁴, 4.8 × 10⁻⁹। संयुग्मी क्षारकों के Kb?6.43 298 K par HF, HCOOH aur HCN ke Ka: 6.8 × 10⁻⁴, 1.8 × 10⁻⁴, 4.8 × 10⁻⁹. Conjugate bases ke Kb?
Kb = Kw/Ka: F⁻ 1.47 × 10⁻¹¹, HCOO⁻ 5.6 × 10⁻¹¹, CN⁻ 2.08 × 10⁻⁶Kb = Kw/Ka: F⁻ 1.47 × 10⁻¹¹, HCOO⁻ 5.6 × 10⁻¹¹, CN⁻ 2.08 × 10⁻⁶Kb = Kw/Ka: F⁻ 1.47 × 10⁻¹¹, HCOO⁻ 5.6 × 10⁻¹¹, CN⁻ 2.08 × 10⁻⁶
6.44 Ka of phenol = 1.0 × 10⁻¹⁰. [phenolate] in 0.05 M phenol? Degree of ionisation if the solution is also 0.01 M in sodium phenolate?6.44 फ़ीनॉल का Ka = 1.0 × 10⁻¹⁰। 0.05 M फ़ीनॉल में [फ़ीनॉक्साइड]? यदि विलयन 0.01 M सोडियम फ़ीनॉक्साइड भी हो तो आयनन की मात्रा?6.44 Phenol ka Ka = 1.0 × 10⁻¹⁰. 0.05 M phenol mein [phenolate]? Agar solution 0.01 M sodium phenolate bhi ho to degree of ionisation?
[C₆H₅O⁻] = √(1.0 × 10⁻¹⁰ × 0.05) = 2.2 × 10⁻⁶ M. With 0.01 M phenolate (common ion): α = Ka/[C₆H₅O⁻] = 10⁻¹⁰/0.01 = 1.0 × 10⁻⁸.[C₆H₅O⁻] = √(1.0 × 10⁻¹⁰ × 0.05) = 2.2 × 10⁻⁶ M। 0.01 M फ़ीनॉक्साइड (समआयन) के साथ: α = 10⁻¹⁰/0.01 = 1.0 × 10⁻⁸।[C₆H₅O⁻] = √(1.0 × 10⁻¹⁰ × 0.05) = 2.2 × 10⁻⁶ M. 0.01 M phenolate (common ion) ke saath: α = 10⁻¹⁰/0.01 = 1.0 × 10⁻⁸.
6.45 Ka1 of H₂S = 9.1 × 10⁻⁸. [HS⁻] in 0.1 M H₂S? Effect if the solution is also 0.1 M in HCl? With Ka2 = 1.2 × 10⁻¹³, [S²⁻] in both cases?6.45 H₂S का Ka1 = 9.1 × 10⁻⁸। 0.1 M H₂S में [HS⁻]? 0.1 M HCl भी हो तो? Ka2 = 1.2 × 10⁻¹³ हो तो दोनों स्थितियों में [S²⁻]?6.45 H₂S ka Ka1 = 9.1 × 10⁻⁸. 0.1 M H₂S mein [HS⁻]? 0.1 M HCl bhi ho to? Ka2 = 1.2 × 10⁻¹³ ho to dono cases mein [S²⁻]?
Water: [HS⁻] = √(9.1 × 10⁻⁸ × 0.1) = 9.54 × 10⁻⁵ M; [S²⁻] ≈ Ka2 = 1.2 × 10⁻¹³ M. In 0.1 M HCl ([H⁺] = 0.1): [HS⁻] = Ka1 × 0.1/0.1 = 9.1 × 10⁻⁸ M (suppressed); [S²⁻] = Ka2[HS⁻]/[H⁺] = 1.2 × 10⁻¹³ × 9.1 × 10⁻⁸/0.1 = 1.09 × 10⁻¹⁹ M.जल में: [HS⁻] = 9.54 × 10⁻⁵ M; [S²⁻] ≈ Ka2 = 1.2 × 10⁻¹³ M। 0.1 M HCl में: [HS⁻] = 9.1 × 10⁻⁸ M (दबा हुआ); [S²⁻] = 1.2 × 10⁻¹³ × 9.1 × 10⁻⁸/0.1 = 1.09 × 10⁻¹⁹ M।Paani mein: [HS⁻] = 9.54 × 10⁻⁵ M; [S²⁻] ≈ Ka2 = 1.2 × 10⁻¹³ M. 0.1 M HCl mein: [HS⁻] = 9.1 × 10⁻⁸ M (daba hua); [S²⁻] = 1.2 × 10⁻¹³ × 9.1 × 10⁻⁸/0.1 = 1.09 × 10⁻¹⁹ M.
6.46 Ka of acetic acid = 1.74 × 10⁻⁵. Degree of dissociation in 0.05 M solution, [acetate] and pH?6.46 ऐसीटिक अम्ल का Ka = 1.74 × 10⁻⁵। 0.05 M में α, [ऐसीटेट] और pH?6.46 Acetic acid ka Ka = 1.74 × 10⁻⁵. 0.05 M mein α, [acetate] aur pH?
α = √(Ka/c) = √(1.74 × 10⁻⁵/0.05) = 0.0187; [CH₃COO⁻] = cα = 9.3 × 10⁻⁴ M; pH = 3.03.α = √(Ka/c) = √(1.74 × 10⁻⁵/0.05) = 0.0187; [CH₃COO⁻] = cα = 9.3 × 10⁻⁴ M; pH = 3.03.α = √(Ka/c) = √(1.74 × 10⁻⁵/0.05) = 0.0187; [CH₃COO⁻] = cα = 9.3 × 10⁻⁴ M; pH = 3.03.
6.47 pH of 0.01 M organic acid is 4.15. [anion], Ka and pKa?6.47 0.01 M कार्बनिक अम्ल का pH 4.15। [ऋणायन], Ka और pKa?6.47 0.01 M organic acid ka pH 4.15. [anion], Ka aur pKa?
[A⁻] = [H⁺] = 10⁻⁴·¹⁵ = 7.08 × 10⁻⁵ M; Ka = (7.08 × 10⁻⁵)²/0.01 = 5.0 × 10⁻⁷; pKa = 6.30.[A⁻] = [H⁺] = 10⁻⁴·¹⁵ = 7.08 × 10⁻⁵ M; Ka = (7.08 × 10⁻⁵)²/0.01 = 5.0 × 10⁻⁷; pKa = 6.30.[A⁻] = [H⁺] = 10⁻⁴·¹⁵ = 7.08 × 10⁻⁵ M; Ka = (7.08 × 10⁻⁵)²/0.01 = 5.0 × 10⁻⁷; pKa = 6.30.
6.48 pH (complete dissociation): (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH6.48 pH (पूर्ण वियोजन): (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH6.48 pH (poora dissociation): (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH
(a) 2.52 (b) pOH 2.30 → 11.70 (c) 2.70 (d) pOH 2.70 → 11.30(a) 2.52 (b) pOH 2.30 → 11.70 (c) 2.70 (d) pOH 2.70 → 11.30(a) 2.52 (b) pOH 2.30 → 11.70 (c) 2.70 (d) pOH 2.70 → 11.30
6.49 pH of: (a) 2 g TlOH in 2 L (b) 0.3 g Ca(OH)₂ in 500 mL (c) 0.3 g NaOH in 200 mL (d) 1 mL of 13.6 M HCl diluted to 1 L6.49 pH: (a) 2 L में 2 g TlOH (b) 500 mL में 0.3 g Ca(OH)₂ (c) 200 mL में 0.3 g NaOH (d) 1 mL 13.6 M HCl को 1 L तक तनु6.49 pH: (a) 2 L mein 2 g TlOH (b) 500 mL mein 0.3 g Ca(OH)₂ (c) 200 mL mein 0.3 g NaOH (d) 1 mL 13.6 M HCl ko 1 L tak dilute
(a) TlOH (221.4 g mol⁻¹): [OH⁻] = 2/221.4/2 = 4.52 × 10⁻³ M → pH 11.65 (b) Ca(OH)₂: 0.3/74 = 4.05 × 10⁻³ mol in 0.5 L → 8.1 × 10⁻³ M, [OH⁻] = 1.62 × 10⁻² → pH 12.21 (c) NaOH: 7.5 × 10⁻³ mol in 0.2 L = 0.0375 M → pH 12.57 (d) [H⁺] = 13.6 × 10⁻³ M → pH 1.87(a) TlOH (221.4 g mol⁻¹): [OH⁻] = 2/221.4/2 = 4.52 × 10⁻³ M → pH 11.65 (b) Ca(OH)₂: 0.3/74 = 4.05 × 10⁻³ mol in 0.5 L → 8.1 × 10⁻³ M, [OH⁻] = 1.62 × 10⁻² → pH 12.21 (c) NaOH: 7.5 × 10⁻³ mol in 0.2 L = 0.0375 M → pH 12.57 (d) [H⁺] = 13.6 × 10⁻³ M → pH 1.87(a) TlOH (221.4 g mol⁻¹): [OH⁻] = 2/221.4/2 = 4.52 × 10⁻³ M → pH 11.65 (b) Ca(OH)₂: 0.3/74 = 4.05 × 10⁻³ mol in 0.5 L → 8.1 × 10⁻³ M, [OH⁻] = 1.62 × 10⁻² → pH 12.21 (c) NaOH: 7.5 × 10⁻³ mol in 0.2 L = 0.0375 M → pH 12.57 (d) [H⁺] = 13.6 × 10⁻³ M → pH 1.87
6.50 Degree of ionisation of 0.1 M bromoacetic acid is 0.132. pH and pKa?6.50 0.1 M ब्रोमोऐसीटिक अम्ल की आयनन मात्रा 0.132। pH और pKa?6.50 0.1 M bromoacetic acid ki degree of ionisation 0.132. pH aur pKa?
[H⁺] = 0.1 × 0.132 = 0.0132 M → pH 1.88. Ka = cα²/(1 − α) = 0.1 × 0.132²/0.868 = 2.0 × 10⁻³ → pKa 2.70.[H⁺] = 0.1 × 0.132 = 0.0132 M → pH 1.88. Ka = cα²/(1 − α) = 0.1 × 0.132²/0.868 = 2.0 × 10⁻³ → pKa 2.70.[H⁺] = 0.1 × 0.132 = 0.0132 M → pH 1.88. Ka = cα²/(1 − α) = 0.1 × 0.132²/0.868 = 2.0 × 10⁻³ → pKa 2.70.
6.51 pH of 0.005 M codeine is 9.95. Kb and pKb?6.51 0.005 M कोडीन का pH 9.95। Kb और pKb?6.51 0.005 M codeine ka pH 9.95. Kb aur pKb?
pOH = 4.05 → [OH⁻] = 8.9 × 10⁻⁵ M. Kb = (8.9 × 10⁻⁵)²/0.005 = 1.6 × 10⁻⁶; pKb = 5.80.pOH = 4.05 → [OH⁻] = 8.9 × 10⁻⁵ M. Kb = (8.9 × 10⁻⁵)²/0.005 = 1.6 × 10⁻⁶; pKb = 5.80.pOH = 4.05 → [OH⁻] = 8.9 × 10⁻⁵ M. Kb = (8.9 × 10⁻⁵)²/0.005 = 1.6 × 10⁻⁶; pKb = 5.80.
6.52 pH of 0.001 M aniline (Kb = 4.27 × 10⁻¹⁰), its degree of ionisation, and Ka of its conjugate acid.6.52 0.001 M ऐनिलीन (Kb = 4.27 × 10⁻¹⁰) का pH, आयनन मात्रा, और संयुग्मी अम्ल का Ka।6.52 0.001 M aniline (Kb = 4.27 × 10⁻¹⁰) ka pH, degree of ionisation, aur conjugate acid ka Ka.
[OH⁻] = √(4.27 × 10⁻¹⁰ × 0.001) = 6.53 × 10⁻⁷ M → pOH 6.19, pH 7.81; α = 6.53 × 10⁻⁷/0.001 = 6.53 × 10⁻⁴; Ka(C₆H₅NH₃⁺) = Kw/Kb = 2.34 × 10⁻⁵.[OH⁻] = √(4.27 × 10⁻¹⁰ × 0.001) = 6.53 × 10⁻⁷ M → pOH 6.19, pH 7.81; α = 6.53 × 10⁻⁷/0.001 = 6.53 × 10⁻⁴; Ka(C₆H₅NH₃⁺) = Kw/Kb = 2.34 × 10⁻⁵.[OH⁻] = √(4.27 × 10⁻¹⁰ × 0.001) = 6.53 × 10⁻⁷ M → pOH 6.19, pH 7.81; α = 6.53 × 10⁻⁷/0.001 = 6.53 × 10⁻⁴; Ka(C₆H₅NH₃⁺) = Kw/Kb = 2.34 × 10⁻⁵.
6.53 Degree of ionisation of 0.05 M acetic acid (pKa 4.74). How does it change if the solution is also (a) 0.01 M (b) 0.1 M in HCl?6.53 0.05 M ऐसीटिक अम्ल (pKa 4.74) की आयनन मात्रा। विलयन (a) 0.01 M (b) 0.1 M HCl भी हो तो?6.53 0.05 M acetic acid (pKa 4.74) ki degree of ionisation. Solution (a) 0.01 M (b) 0.1 M HCl bhi ho to?
Ka = 10⁻⁴·⁷⁴ = 1.82 × 10⁻⁵; α = √(Ka/c) = 0.019. With HCl, [H⁺] ≈ [HCl] (common ion): (a) α = Ka/0.01 = 1.82 × 10⁻³ (b) α = Ka/0.1 = 1.82 × 10⁻⁴ — ionisation is suppressed.Ka = 1.82 × 10⁻⁵; α = 0.019। HCl के साथ [H⁺] ≈ [HCl] (समआयन): (a) α = 1.82 × 10⁻³ (b) α = 1.82 × 10⁻⁴ — आयनन दब जाता है।Ka = 1.82 × 10⁻⁵; α = 0.019. HCl ke saath [H⁺] ≈ [HCl] (common ion): (a) α = 1.82 × 10⁻³ (b) α = 1.82 × 10⁻⁴ — ionisation dab jaata hai.
6.54 Kb of dimethylamine = 5.4 × 10⁻⁴. Degree of ionisation in 0.02 M solution? Percentage ionised if the solution is also 0.1 M in NaOH?6.54 डाइमेथिलऐमीन का Kb = 5.4 × 10⁻⁴। 0.02 M में आयनन मात्रा? 0.1 M NaOH भी हो तो कितना प्रतिशत आयनित?6.54 Dimethylamine ka Kb = 5.4 × 10⁻⁴. 0.02 M mein degree of ionisation? 0.1 M NaOH bhi ho to kitna percent ionised?
α ≈ √(Kb/c) = √(5.4 × 10⁻⁴/0.02) = 0.164 (solving the quadratic gives 0.15). In 0.1 M NaOH: α = Kb/[OH⁻] = 5.4 × 10⁻⁴/0.1 = 5.4 × 10⁻³ → 0.54 %.α ≈ √(5.4 × 10⁻⁴/0.02) = 0.164 (द्विघात हल से 0.15)। 0.1 M NaOH में: α = 5.4 × 10⁻⁴/0.1 = 5.4 × 10⁻³ → 0.54 %।α ≈ √(5.4 × 10⁻⁴/0.02) = 0.164 (quadratic se 0.15). 0.1 M NaOH mein: α = 5.4 × 10⁻⁴/0.1 = 5.4 × 10⁻³ → 0.54 %.
6.55 [H⁺] of: (a) muscle fluid pH 6.83 (b) stomach fluid 1.2 (c) blood 7.38 (d) saliva 6.46.55 [H⁺]: (a) पेशी द्रव pH 6.83 (b) जठर द्रव 1.2 (c) रक्त 7.38 (d) लार 6.46.55 [H⁺]: (a) muscle fluid pH 6.83 (b) stomach fluid 1.2 (c) blood 7.38 (d) saliva 6.4
(a) 1.48 × 10⁻⁷ M (b) 6.3 × 10⁻² M (c) 4.17 × 10⁻⁸ M (d) 3.98 × 10⁻⁷ M(a) 1.48 × 10⁻⁷ M (b) 6.3 × 10⁻² M (c) 4.17 × 10⁻⁸ M (d) 3.98 × 10⁻⁷ M(a) 1.48 × 10⁻⁷ M (b) 6.3 × 10⁻² M (c) 4.17 × 10⁻⁸ M (d) 3.98 × 10⁻⁷ M
6.56 [H⁺] of milk 6.8, black coffee 5.0, tomato juice 4.2, lemon juice 2.2, egg white 7.86.56 दूध 6.8, ब्लैक कॉफ़ी 5.0, टमाटर रस 4.2, नींबू रस 2.2, अंडे की सफ़ेदी 7.8 का [H⁺]6.56 Doodh 6.8, black coffee 5.0, tamatar ras 4.2, nimbu ras 2.2, ande ki safedi 7.8 ka [H⁺]
Milk 1.58 × 10⁻⁷, coffee 1.0 × 10⁻⁵, tomato 6.31 × 10⁻⁵, lemon 6.31 × 10⁻³, egg white 1.58 × 10⁻⁸ MMilk 1.58 × 10⁻⁷, coffee 1.0 × 10⁻⁵, tomato 6.31 × 10⁻⁵, lemon 6.31 × 10⁻³, egg white 1.58 × 10⁻⁸ MMilk 1.58 × 10⁻⁷, coffee 1.0 × 10⁻⁵, tomato 6.31 × 10⁻⁵, lemon 6.31 × 10⁻³, egg white 1.58 × 10⁻⁸ M
6.57 0.561 g KOH in 200 mL at 298 K. [K⁺], [H⁺], [OH⁻] and pH?6.57 298 K पर 200 mL में 0.561 g KOH। [K⁺], [H⁺], [OH⁻] और pH?6.57 298 K par 200 mL mein 0.561 g KOH. [K⁺], [H⁺], [OH⁻] aur pH?
n = 0.561/56.1 = 0.010 mol in 0.2 L → [K⁺] = [OH⁻] = 0.05 M; [H⁺] = 10⁻¹⁴/0.05 = 2 × 10⁻¹³ M; pH = 12.70.n = 0.561/56.1 = 0.010 mol in 0.2 L → [K⁺] = [OH⁻] = 0.05 M; [H⁺] = 10⁻¹⁴/0.05 = 2 × 10⁻¹³ M; pH = 12.70.n = 0.561/56.1 = 0.010 mol in 0.2 L → [K⁺] = [OH⁻] = 0.05 M; [H⁺] = 10⁻¹⁴/0.05 = 2 × 10⁻¹³ M; pH = 12.70.
6.58 Solubility of Sr(OH)₂ at 298 K is 19.23 g L⁻¹. [Sr²⁺], [OH⁻] and pH?6.58 298 K पर Sr(OH)₂ की विलेयता 19.23 g L⁻¹। [Sr²⁺], [OH⁻] और pH?6.58 298 K par Sr(OH)₂ ki solubility 19.23 g L⁻¹. [Sr²⁺], [OH⁻] aur pH?
M = 121.6 g mol⁻¹: [Sr²⁺] = 19.23/121.6 = 0.158 M; [OH⁻] = 0.316 M; pOH 0.50 → pH 13.50.M = 121.6 g mol⁻¹: [Sr²⁺] = 19.23/121.6 = 0.158 M; [OH⁻] = 0.316 M; pOH 0.50 → pH 13.50.M = 121.6 g mol⁻¹: [Sr²⁺] = 19.23/121.6 = 0.158 M; [OH⁻] = 0.316 M; pOH 0.50 → pH 13.50.
6.59 Ka of propanoic acid = 1.32 × 10⁻⁵. α and pH of 0.05 M; α if also 0.01 M in HCl?6.59 प्रोपेनॉइक अम्ल का Ka = 1.32 × 10⁻⁵। 0.05 M का α और pH; 0.01 M HCl भी हो तो α?6.59 Propanoic acid ka Ka = 1.32 × 10⁻⁵. 0.05 M ka α aur pH; 0.01 M HCl bhi ho to α?
α = √(1.32 × 10⁻⁵/0.05) = 0.0162; [H⁺] = 8.1 × 10⁻⁴ → pH 3.09. With 0.01 M HCl: α = Ka/0.01 = 1.32 × 10⁻³.α = √(1.32 × 10⁻⁵/0.05) = 0.0162; [H⁺] = 8.1 × 10⁻⁴ → pH 3.09. With 0.01 M HCl: α = Ka/0.01 = 1.32 × 10⁻³.α = √(1.32 × 10⁻⁵/0.05) = 0.0162; [H⁺] = 8.1 × 10⁻⁴ → pH 3.09. With 0.01 M HCl: α = Ka/0.01 = 1.32 × 10⁻³.
6.60 pH of 0.1 M cyanic acid (HCNO) is 2.34. Ka and degree of ionisation?6.60 0.1 M साइनिक अम्ल (HCNO) का pH 2.34। Ka और आयनन मात्रा?6.60 0.1 M cyanic acid (HCNO) ka pH 2.34. Ka aur degree of ionisation?
[H⁺] = 10⁻²·³⁴ = 4.57 × 10⁻³ M; Ka = (4.57 × 10⁻³)²/0.1 = 2.09 × 10⁻⁴; α = 4.57 × 10⁻³/0.1 = 0.0457.[H⁺] = 10⁻²·³⁴ = 4.57 × 10⁻³ M; Ka = (4.57 × 10⁻³)²/0.1 = 2.09 × 10⁻⁴; α = 4.57 × 10⁻³/0.1 = 0.0457.[H⁺] = 10⁻²·³⁴ = 4.57 × 10⁻³ M; Ka = (4.57 × 10⁻³)²/0.1 = 2.09 × 10⁻⁴; α = 4.57 × 10⁻³/0.1 = 0.0457.
6.61 Ka of nitrous acid = 4.5 × 10⁻⁴. pH of 0.04 M sodium nitrite and its degree of hydrolysis?6.61 नाइट्रस अम्ल का Ka = 4.5 × 10⁻⁴। 0.04 M सोडियम नाइट्राइट का pH और जल-अपघटन मात्रा?6.61 Nitrous acid ka Ka = 4.5 × 10⁻⁴. 0.04 M sodium nitrite ka pH aur degree of hydrolysis?
NO₂⁻ + H₂O ⇌ HNO₂ + OH⁻; Kh = Kw/Ka = 2.22 × 10⁻¹¹. [OH⁻] = √(Kh c) = 9.43 × 10⁻⁷ M → pOH 6.03, pH 7.97; h = [OH⁻]/c = 2.36 × 10⁻⁵.NO₂⁻ + H₂O ⇌ HNO₂ + OH⁻; Kh = Kw/Ka = 2.22 × 10⁻¹¹. [OH⁻] = √(Kh c) = 9.43 × 10⁻⁷ M → pOH 6.03, pH 7.97; h = [OH⁻]/c = 2.36 × 10⁻⁵.NO₂⁻ + H₂O ⇌ HNO₂ + OH⁻; Kh = Kw/Ka = 2.22 × 10⁻¹¹. [OH⁻] = √(Kh c) = 9.43 × 10⁻⁷ M → pOH 6.03, pH 7.97; h = [OH⁻]/c = 2.36 × 10⁻⁵.
6.62 0.02 M pyridinium hydrochloride has pH 3.44. Kb of pyridine?6.62 0.02 M पिरिडीनियम हाइड्रोक्लोराइड का pH 3.44। पिरिडीन का Kb?6.62 0.02 M pyridinium hydrochloride ka pH 3.44. Pyridine ka Kb?
[H⁺] = 3.63 × 10⁻⁴ M; Kh = [H⁺]²/c = 6.6 × 10⁻⁶ = Kw/Kb → Kb = 10⁻¹⁴/6.6 × 10⁻⁶ = 1.5 × 10⁻⁹.[H⁺] = 3.63 × 10⁻⁴ M; Kh = [H⁺]²/c = 6.6 × 10⁻⁶ = Kw/Kb → Kb = 10⁻¹⁴/6.6 × 10⁻⁶ = 1.5 × 10⁻⁹.[H⁺] = 3.63 × 10⁻⁴ M; Kh = [H⁺]²/c = 6.6 × 10⁻⁶ = Kw/Kb → Kb = 10⁻¹⁴/6.6 × 10⁻⁶ = 1.5 × 10⁻⁹.
6.63 Neutral, acidic or basic? NaCl, KBr, NaCN, NH₄NO₃, NaNO₂, KF6.63 उदासीन, अम्लीय या क्षारीय? NaCl, KBr, NaCN, NH₄NO₃, NaNO₂, KF6.63 Neutral, acidic ya basic? NaCl, KBr, NaCN, NH₄NO₃, NaNO₂, KF
NaCl, KBr: neutral (strong acid + strong base). NaCN, NaNO₂, KF: basic (anion of a weak acid hydrolyses). NH₄NO₃: acidic (NH₄⁺ of a weak base hydrolyses).NaCl, KBr: उदासीन (प्रबल अम्ल + प्रबल क्षारक)। NaCN, NaNO₂, KF: क्षारीय (दुर्बल अम्ल का ऋणायन जल-अपघटित)। NH₄NO₃: अम्लीय (दुर्बल क्षारक का NH₄⁺ जल-अपघटित)।NaCl, KBr: neutral (strong acid + strong base). NaCN, NaNO₂, KF: basic (weak acid ka anion hydrolyse). NH₄NO₃: acidic (weak base ka NH₄⁺ hydrolyse).
6.64 Ka of chloroacetic acid = 1.35 × 10⁻³. pH of 0.1 M acid and of its 0.1 M sodium salt?6.64 क्लोरोऐसीटिक अम्ल का Ka = 1.35 × 10⁻³। 0.1 M अम्ल और उसके 0.1 M सोडियम लवण का pH?6.64 Chloroacetic acid ka Ka = 1.35 × 10⁻³. 0.1 M acid aur uske 0.1 M sodium salt ka pH?
Acid: [H⁺] = √(1.35 × 10⁻³ × 0.1) = 1.16 × 10⁻² → pH 1.94 (quadratic: 1.96). Salt: Kh = Kw/Ka = 7.4 × 10⁻¹², [OH⁻] = √(Kh × 0.1) = 8.6 × 10⁻⁷ → pOH 6.07, pH 7.93.Acid: [H⁺] = √(1.35 × 10⁻³ × 0.1) = 1.16 × 10⁻² → pH 1.94 (quadratic: 1.96). Salt: Kh = Kw/Ka = 7.4 × 10⁻¹², [OH⁻] = √(Kh × 0.1) = 8.6 × 10⁻⁷ → pOH 6.07, pH 7.93.Acid: [H⁺] = √(1.35 × 10⁻³ × 0.1) = 1.16 × 10⁻² → pH 1.94 (quadratic: 1.96). Salt: Kh = Kw/Ka = 7.4 × 10⁻¹², [OH⁻] = √(Kh × 0.1) = 8.6 × 10⁻⁷ → pOH 6.07, pH 7.93.
6.65 Kw at 310 K is 2.7 × 10⁻¹⁴. pH of neutral water at this temperature?6.65 310 K पर Kw = 2.7 × 10⁻¹⁴। इस ताप पर उदासीन जल का pH?6.65 310 K par Kw = 2.7 × 10⁻¹⁴. Is temperature par neutral paani ka pH?
[H⁺] = √(2.7 × 10⁻¹⁴) = 1.64 × 10⁻⁷ M → pH = 6.78 (still neutral, since [H⁺] = [OH⁻]).[H⁺] = √(2.7 × 10⁻¹⁴) = 1.64 × 10⁻⁷ M → pH = 6.78 (still neutral, since [H⁺] = [OH⁻]).[H⁺] = √(2.7 × 10⁻¹⁴) = 1.64 × 10⁻⁷ M → pH = 6.78 (still neutral, since [H⁺] = [OH⁻]).
6.66 pH of: (a) 10 mL 0.2 M Ca(OH)₂ + 25 mL 0.1 M HCl (b) 10 mL 0.01 M H₂SO₄ + 10 mL 0.01 M Ca(OH)₂ (c) 10 mL 0.1 M H₂SO₄ + 10 mL 0.1 M KOH6.66 pH: (a) 10 mL 0.2 M Ca(OH)₂ + 25 mL 0.1 M HCl (b) 10 mL 0.01 M H₂SO₄ + 10 mL 0.01 M Ca(OH)₂ (c) 10 mL 0.1 M H₂SO₄ + 10 mL 0.1 M KOH6.66 pH: (a) 10 mL 0.2 M Ca(OH)₂ + 25 mL 0.1 M HCl (b) 10 mL 0.01 M H₂SO₄ + 10 mL 0.01 M Ca(OH)₂ (c) 10 mL 0.1 M H₂SO₄ + 10 mL 0.1 M KOH
(a) OH⁻ = 4 mmol, H⁺ = 2.5 mmol → 1.5 mmol OH⁻ in 35 mL = 0.0429 M → pH 12.63. (b) H⁺ 0.2 mmol = OH⁻ 0.2 mmol → pH 7. (c) H⁺ 2 mmol, OH⁻ 1 mmol → 1 mmol H⁺ in 20 mL = 0.05 M → pH 1.30.(a) OH⁻ = 4 mmol, H⁺ = 2.5 mmol → 35 mL में 1.5 mmol OH⁻ = 0.0429 M → pH 12.63। (b) H⁺ 0.2 mmol = OH⁻ 0.2 mmol → pH 7। (c) H⁺ 2 mmol, OH⁻ 1 mmol → 20 mL में 1 mmol H⁺ = 0.05 M → pH 1.30।(a) OH⁻ = 4 mmol, H⁺ = 2.5 mmol → 35 mL mein 1.5 mmol OH⁻ = 0.0429 M → pH 12.63. (b) H⁺ 0.2 mmol = OH⁻ 0.2 mmol → pH 7. (c) H⁺ 2 mmol, OH⁻ 1 mmol → 20 mL mein 1 mmol H⁺ = 0.05 M → pH 1.30.
6.67 Solubilities and ion molarities at 298 K of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide (Ksp from Table 6.9: 1.1 × 10⁻¹², 1.2 × 10⁻¹⁰, 1.0 × 10⁻³⁸, 1.6 × 10⁻⁵, 4.5 × 10⁻²⁹).6.67 298 K पर सिल्वर क्रोमेट, बेरियम क्रोमेट, फ़ेरिक हाइड्रॉक्साइड, लेड क्लोराइड और मरक्यूरस आयोडाइड की विलेयता व आयन मोलरता (सारणी 6.9 के Ksp)।6.67 298 K par silver chromate, barium chromate, ferric hydroxide, lead chloride aur mercurous iodide ki solubility aur ion molarity (Table 6.9 ke Ksp).
Ag₂CrO₄: 4S³ = 1.1 × 10⁻¹² → S = 6.5 × 10⁻⁵ M, [Ag⁺] = 1.3 × 10⁻⁴, [CrO₄²⁻] = 6.5 × 10⁻⁵ M. BaCrO₄: S = √Ksp = 1.1 × 10⁻⁵ M = [Ba²⁺] = [CrO₄²⁻]. Fe(OH)₃: 27S⁴ = 10⁻³⁸ → S = 1.39 × 10⁻¹⁰ M, [OH⁻] = 4.2 × 10⁻¹⁰ M. PbCl₂: 4S³ = 1.6 × 10⁻⁵ → S = 1.59 × 10⁻² M, [Cl⁻] = 3.2 × 10⁻² M. Hg₂I₂: 4S³ = 4.5 × 10⁻²⁹ → S = 2.24 × 10⁻¹⁰ M = [Hg₂²⁺], [I⁻] = 4.5 × 10⁻¹⁰ M.Ag₂CrO₄: 4S³ = 1.1 × 10⁻¹² → S = 6.5 × 10⁻⁵ M, [Ag⁺] = 1.3 × 10⁻⁴, [CrO₄²⁻] = 6.5 × 10⁻⁵ M. BaCrO₄: S = √Ksp = 1.1 × 10⁻⁵ M = [Ba²⁺] = [CrO₄²⁻]. Fe(OH)₃: 27S⁴ = 10⁻³⁸ → S = 1.39 × 10⁻¹⁰ M, [OH⁻] = 4.2 × 10⁻¹⁰ M. PbCl₂: 4S³ = 1.6 × 10⁻⁵ → S = 1.59 × 10⁻² M, [Cl⁻] = 3.2 × 10⁻² M. Hg₂I₂: 4S³ = 4.5 × 10⁻²⁹ → S = 2.24 × 10⁻¹⁰ M = [Hg₂²⁺], [I⁻] = 4.5 × 10⁻¹⁰ M.Ag₂CrO₄: 4S³ = 1.1 × 10⁻¹² → S = 6.5 × 10⁻⁵ M, [Ag⁺] = 1.3 × 10⁻⁴, [CrO₄²⁻] = 6.5 × 10⁻⁵ M. BaCrO₄: S = √Ksp = 1.1 × 10⁻⁵ M = [Ba²⁺] = [CrO₄²⁻]. Fe(OH)₃: 27S⁴ = 10⁻³⁸ → S = 1.39 × 10⁻¹⁰ M, [OH⁻] = 4.2 × 10⁻¹⁰ M. PbCl₂: 4S³ = 1.6 × 10⁻⁵ → S = 1.59 × 10⁻² M, [Cl⁻] = 3.2 × 10⁻² M. Hg₂I₂: 4S³ = 4.5 × 10⁻²⁹ → S = 2.24 × 10⁻¹⁰ M = [Hg₂²⁺], [I⁻] = 4.5 × 10⁻¹⁰ M.
6.68 Ksp of Ag₂CrO₄ and AgBr are 1.1 × 10⁻¹² and 5.0 × 10⁻¹³. Ratio of molarities of their saturated solutions?6.68 Ag₂CrO₄ और AgBr के Ksp 1.1 × 10⁻¹² और 5.0 × 10⁻¹³। संतृप्त विलयनों की मोलरता का अनुपात?6.68 Ag₂CrO₄ aur AgBr ke Ksp 1.1 × 10⁻¹² aur 5.0 × 10⁻¹³. Saturated solutions ki molarity ka ratio?
S(Ag₂CrO₄) = (1.1 × 10⁻¹²/4)^(1/3) = 6.5 × 10⁻⁵ M; S(AgBr) = √(5.0 × 10⁻¹³) = 7.07 × 10⁻⁷ M. Ratio = ≈ 92.S(Ag₂CrO₄) = (1.1 × 10⁻¹²/4)^(1/3) = 6.5 × 10⁻⁵ M; S(AgBr) = √(5.0 × 10⁻¹³) = 7.07 × 10⁻⁷ M. Ratio = ≈ 92.S(Ag₂CrO₄) = (1.1 × 10⁻¹²/4)^(1/3) = 6.5 × 10⁻⁵ M; S(AgBr) = √(5.0 × 10⁻¹³) = 7.07 × 10⁻⁷ M. Ratio = ≈ 92.
6.69 Equal volumes of 0.002 M sodium iodate and cupric chlorate are mixed. Will copper iodate precipitate? (Ksp = 7.4 × 10⁻⁸)6.69 0.002 M सोडियम आयोडेट और क्यूप्रिक क्लोरेट के समान आयतन मिलाए। क्या कॉपर आयोडेट अवक्षेपित होगा? (Ksp = 7.4 × 10⁻⁸)6.69 0.002 M sodium iodate aur cupric chlorate ke barabar volume milaye. Kya copper iodate precipitate hoga? (Ksp = 7.4 × 10⁻⁸)
After mixing [Cu²⁺] = [IO₃⁻] = 0.001 M. Q = [Cu²⁺][IO₃⁻]² = 0.001 × (0.001)² = 1 × 10⁻⁹ < Ksp → no precipitate.मिलाने पर [Cu²⁺] = [IO₃⁻] = 0.001 M। Q = 1 × 10⁻⁹ < Ksp → अवक्षेप नहीं।Milane par [Cu²⁺] = [IO₃⁻] = 0.001 M. Q = 1 × 10⁻⁹ < Ksp → precipitate nahi.
6.70 Ka of benzoic acid = 6.46 × 10⁻⁵, Ksp of silver benzoate = 2.5 × 10⁻¹³. How many times more soluble is silver benzoate in a buffer of pH 3.19 than in pure water?6.70 बेन्ज़ोइक अम्ल का Ka = 6.46 × 10⁻⁵, सिल्वर बेन्ज़ोएट का Ksp = 2.5 × 10⁻¹³। pH 3.19 बफ़र में यह शुद्ध जल से कितने गुना अधिक विलेय है?6.70 Benzoic acid ka Ka = 6.46 × 10⁻⁵, silver benzoate ka Ksp = 2.5 × 10⁻¹³. pH 3.19 buffer mein ye pure paani se kitne guna zyada soluble hai?
In the buffer [H⁺] = 6.46 × 10⁻⁴, so [H⁺]/Ka = 10: only 1/11 of dissolved benzoate stays as C₆H₅COO⁻. S² = Ksp(1 + [H⁺]/Ka) = 11 Ksp. Ratio = √11 = 3.32 times.बफ़र में [H⁺] = 6.46 × 10⁻⁴, अतः [H⁺]/Ka = 10। S² = Ksp(1 + [H⁺]/Ka) = 11 Ksp। अनुपात = √11 = 3.32 गुना।Buffer mein [H⁺] = 6.46 × 10⁻⁴, isliye [H⁺]/Ka = 10. S² = Ksp(1 + [H⁺]/Ka) = 11 Ksp. Ratio = √11 = 3.32 guna.
6.71 Maximum concentration of equimolar FeSO₄ and Na₂S solutions so that, mixed in equal volumes, FeS does not precipitate (Ksp = 6.3 × 10⁻¹⁸)?6.71 समान मोलर FeSO₄ और Na₂S विलयनों की अधिकतम सांद्रता ताकि समान आयतन मिलाने पर FeS अवक्षेपित न हो (Ksp = 6.3 × 10⁻¹⁸)?6.71 Equimolar FeSO₄ aur Na₂S solutions ki maximum concentration taaki barabar volume milane par FeS precipitate na ho (Ksp = 6.3 × 10⁻¹⁸)?
After mixing each is c/2: (c/2)² = 6.3 × 10⁻¹⁸ → c/2 = 2.51 × 10⁻⁹ → c = 5.0 × 10⁻⁹ M.After mixing each is c/2: (c/2)² = 6.3 × 10⁻¹⁸ → c/2 = 2.51 × 10⁻⁹ → c = 5.0 × 10⁻⁹ M.After mixing each is c/2: (c/2)² = 6.3 × 10⁻¹⁸ → c/2 = 2.51 × 10⁻⁹ → c = 5.0 × 10⁻⁹ M.
6.72 Minimum volume of water to dissolve 1 g of CaSO₄ at 298 K (Ksp = 9.1 × 10⁻⁶)?6.72 298 K पर 1 g CaSO₄ घोलने के लिए न्यूनतम जल (Ksp = 9.1 × 10⁻⁶)?6.72 298 K par 1 g CaSO₄ gholne ke liye minimum paani (Ksp = 9.1 × 10⁻⁶)?
S = √(9.1 × 10⁻⁶) = 3.02 × 10⁻³ mol L⁻¹ × 136 g mol⁻¹ = 0.41 g L⁻¹. Volume = 1/0.41 = 2.43 L.S = √(9.1 × 10⁻⁶) = 3.02 × 10⁻³ mol L⁻¹ × 136 g mol⁻¹ = 0.41 g L⁻¹. Volume = 1/0.41 = 2.43 L.S = √(9.1 × 10⁻⁶) = 3.02 × 10⁻³ mol L⁻¹ × 136 g mol⁻¹ = 0.41 g L⁻¹. Volume = 1/0.41 = 2.43 L.
6.73 [S²⁻] in 0.1 M HCl saturated with H₂S is 1.0 × 10⁻¹⁹ M. 10 mL of it is added to 5 mL of 0.04 M FeSO₄, MnCl₂, ZnCl₂ and CdCl₂. Where will precipitation occur?6.73 H₂S से संतृप्त 0.1 M HCl में [S²⁻] = 1.0 × 10⁻¹⁹ M। इसका 10 mL, 0.04 M FeSO₄, MnCl₂, ZnCl₂ और CdCl₂ के 5 mL में मिलाया। कहाँ अवक्षेप बनेगा?6.73 H₂S se saturated 0.1 M HCl mein [S²⁻] = 1.0 × 10⁻¹⁹ M. Iska 10 mL, 0.04 M FeSO₄, MnCl₂, ZnCl₂ aur CdCl₂ ke 5 mL mein milaya. Kahan precipitate banega?
In 15 mL: [S²⁻] = 6.67 × 10⁻²⁰ M, [M²⁺] = 0.0133 M; ionic product = 8.9 × 10⁻²². Ksp: FeS 6.3 × 10⁻¹⁸, MnS 2.5 × 10⁻¹³ (larger → no ppt); ZnS 1.6 × 10⁻²⁴, CdS 8.0 × 10⁻²⁷ (smaller → ppt). ZnS and CdS precipitate.15 mL में: [S²⁻] = 6.67 × 10⁻²⁰ M, [M²⁺] = 0.0133 M; आयनिक गुणनफल = 8.9 × 10⁻²²। FeS (6.3 × 10⁻¹⁸) व MnS (2.5 × 10⁻¹³) का Ksp बड़ा → अवक्षेप नहीं; ZnS (1.6 × 10⁻²⁴) व CdS (8.0 × 10⁻²⁷) छोटा → ZnS और CdS अवक्षेपित।15 mL mein: [S²⁻] = 6.67 × 10⁻²⁰ M, [M²⁺] = 0.0133 M; ionic product = 8.9 × 10⁻²². FeS (6.3 × 10⁻¹⁸) aur MnS (2.5 × 10⁻¹³) ka Ksp bada → precipitate nahi; ZnS (1.6 × 10⁻²⁴) aur CdS (8.0 × 10⁻²⁷) chhota → ZnS aur CdS precipitate.

🎯 Quiz — 14 questions🎯 क्विज़ — 14 प्रश्न🎯 Quiz — 14 questions

Tap an option. The answer and a short reason appear at once.किसी विकल्प पर टैप करें। उत्तर और छोटा कारण तुरंत दिखेगा।Kisi option par tap kijiye. Answer aur chhota reason turant dikhega.

1. At equilibrium:साम्य पर:Equilibrium par:

2. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp equals:CaCO₃(s) ⇌ CaO(s) + CO₂(g) के लिए Kp:CaCO₃(s) ⇌ CaO(s) + CO₂(g) ke liye Kp:

3. For which reaction is Kp = Kc?किस अभिक्रिया के लिए Kp = Kc?Kis reaction ke liye Kp = Kc?

4. If K for A ⇌ B is 4, K for 2B ⇌ 2A is:A ⇌ B का K = 4 हो, तो 2B ⇌ 2A का K:A ⇌ B ka K = 4 ho, to 2B ⇌ 2A ka K:

5. If Qc > Kc, the reaction:यदि Qc > Kc, तो अभिक्रिया:Agar Qc > Kc, to reaction:

6. Which changes the value of K?K का मान कौन बदलता है?K ki value kaun badalta hai?

7. For N₂ + 3H₂ ⇌ 2NH₃ (ΔH < 0), the yield of NH₃ increases with:N₂ + 3H₂ ⇌ 2NH₃ (ΔH < 0) में NH₃ की लब्धि बढ़ती है:N₂ + 3H₂ ⇌ 2NH₃ (ΔH < 0) mein NH₃ ki yield badhti hai:

8. Which is a Lewis acid but not a Brönsted acid?कौन लूइस अम्ल है पर ब्रॉन्स्टेड अम्ल नहीं?Kaun Lewis acid hai par Brönsted acid nahi?

9. The conjugate base of HCO₃⁻ is:HCO₃⁻ का संयुग्मी क्षारक:HCO₃⁻ ka conjugate base:

10. pH of 0.001 M NaOH at 298 K is:298 K पर 0.001 M NaOH का pH:298 K par 0.001 M NaOH ka pH:

11. An aqueous solution of NH₄Cl is:NH₄Cl का जलीय विलयन:NH₄Cl ka aqueous solution:

12. Which mixture is a buffer?कौन-सा मिश्रण बफ़र है?Kaun sa mixture buffer hai?

13. For Ag₂CrO₄ with molar solubility S, Ksp is:मोलर विलेयता S वाले Ag₂CrO₄ का Ksp:Molar solubility S wale Ag₂CrO₄ ka Ksp:

14. Adding NaCl to a saturated AgCl solution:संतृप्त AgCl विलयन में NaCl मिलाने पर:Saturated AgCl solution mein NaCl milane par:

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