CBSE · NCERT · Class 11 Chemistry · Chapter 6

NCERT Solutions: Class 11 Chemistry Chapter 6 - Equilibrium

68 textbook Q&A68 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Equilibrium, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 68
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1Exercises68 questions
Q.6.1A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. a) What is the initial effect of the change on vapour pressure? b) How do rates of evaporation and condensation change initially? c) What happens when equilibrium is restored finally and what will be the final vapour pressure?v
Solution

On increasing the volume, the same vapour occupies a larger space, so its pressure falls at once. The rate of condensation decreases because vapour molecules strike the liquid surface less frequently, while the rate of evaporation is initially almost unchanged. More liquid evaporates until equilibrium is re-established; at the same temperature the final vapour pressure again equals the original equilibrium vapour pressure.

Answer:

Initially the vapour pressure decreases, evaporation is greater than condensation, and finally the same vapour pressure is restored.

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Q.6.2What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ? 2SO2(g) + O2(g) ⇌ 2SO3(g)v
Solution

For 2SO2(g) + O2(g) ⇌ 2SO3(g), Kc = [SO3]^2/([SO2]^2[O2]). Substituting the concentrations, Kc = (1.90)^2/[(0.60)^2(0.82)] = 12.23, so Kc is approximately 12.2.

Answer:

Kc = 12.2.

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Q.6.3At a certain temperature and total pressure of 10^5 Pa, iodine vapour contains 40% by volume of I atoms I2 (g) ⇌ 2I (g) Calculate Kp for the equilibrium.v
Solution

Volume percent equals mole percent for gases. Thus pI = 0.40 x 10^5 Pa and pI2 = 0.60 x 10^5 Pa. For I2(g) ⇌ 2I(g), Kp = (pI)^2/pI2 = (4.0 x 10^4)^2/(6.0 x 10^4) = 2.67 x 10^4 Pa.

Answer:

Kp = 2.67 x 10^4 Pa.

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Q.6.4Write the expression for the equilibrium constant, Kc for each of the following reactions: (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 ⇌ 2IF5v
Solution

Pure solids and pure liquids are omitted from Kc because their activities are constant. Therefore only gaseous and aqueous species appear in the expressions with powers equal to their stoichiometric coefficients.

Answer:

(i) Kc = [NO]^2[Cl2]/[NOCl]^2; (ii) Kc = [NO2]^4[O2]; (iii) Kc = [CH3COOH][C2H5OH]/[CH3COOC2H5]; (iv) Kc = 1/([Fe3+][OH-]^3); (v) Kc = [IF5]^2/[F2]^5.

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Q.6.5Find out the value of Kc for each of the following equilibria from the value of Kp: (i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g); Kp= 1.8 x 10^-2 at 500 K (ii) CaCO3 (s) ⇌ CaO(s) + CO2(g); Kp= 167 at 1073 Kv
Solution

Use Kp = Kc(RT)^Δn with R = 0.0831 L bar K^-1 mol^-1. For (i), Δn = 3 - 2 = 1, so Kc = 1.8 x 10^-2/(0.0831 x 500) = 4.33 x 10^-4. For (ii), Δn = 1, so Kc = 167/(0.0831 x 1073) = 1.87, approximately 1.90.

Answer:

(i) Kc = 4.33 x 10^-4; (ii) Kc = 1.90.

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Q.6.6For the following equilibrium, Kc= 6.3 x 10^14 at 1000 K NO (g) + O3 (g) ⇌ NO2 (g) + O2 (g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?v
Solution

The equilibrium constant for the reverse reaction is the reciprocal of that for the forward reaction. Thus Kc(reverse) = 1/(6.3 x 10^14) = 1.59 x 10^-15.

Answer:

Kc for the reverse reaction = 1.59 x 10^-15.

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Q.6.7Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?v
Solution

For a pure solid or pure liquid, the amount present may change but its active mass does not change appreciably at a fixed temperature. That constant factor is absorbed into the numerical value of the equilibrium constant, so pure solids and pure liquids are omitted from K expressions.

Answer:

Because their concentrations or activities remain constant at a fixed temperature.

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Q.6.9Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below: 2NO (g) + Br2 (g) ⇌ 2NOBr (g) When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.v
Solution

Formation of 0.0518 mol NOBr consumes 0.0518 mol NO and 0.0518/2 = 0.0259 mol Br2 from the stoichiometry 2NO + Br2 ⇌ 2NOBr. Therefore NO left = 0.087 - 0.0518 = 0.0352 mol, and Br2 left = 0.0437 - 0.0259 = 0.0178 mol.

Answer:

NO = 0.0352 mol; Br2 = 0.0178 mol.

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Q.6.10At 450K, Kp= 2.0 x 10^10/bar for the given reaction at equilibrium. 2SO2(g) + O2(g) ⇌ 2SO3 (g) What is Kc at this temperature ?v
Solution

For 2SO2(g) + O2(g) ⇌ 2SO3(g), Δn = 2 - 3 = -1. Hence Kp = Kc(RT)^-1, so Kc = KpRT = (2.0 x 10^10)(0.0831)(450) = 7.48 x 10^11.

Answer:

Kc = 7.48 x 10^11.

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Q.6.11A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium ? 2HI (g) ⇌ H2 (g) + I2 (g)v
Solution

HI pressure decreases from 0.20 atm to 0.04 atm, so 0.16 atm of HI decomposes. From 2HI ⇌ H2 + I2, pH2 = pI2 = 0.16/2 = 0.08 atm. Therefore Kp = (pH2 pI2)/(pHI)^2 = (0.08 x 0.08)/(0.04)^2 = 4.

Answer:

Kp = 4.

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Q.6.12A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g) is 1.7 x 10^2. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?v
Solution

Concentrations are [N2] = 1.57/20 = 0.0785 M, [H2] = 1.92/20 = 0.096 M and [NH3] = 8.13/20 = 0.4065 M. Qc = [NH3]^2/([N2][H2]^3) = 2.38 x 10^3. Since Qc > Kc (1.7 x 10^2), there is excess product and the reaction shifts backward to form N2 and H2.

Answer:

The mixture is not at equilibrium; the net reaction proceeds in the reverse direction.

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Q.6.14One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation, H2O (g) + CO (g) ⇌ H2 (g) + CO2 (g) Calculate the equilibrium constant for the reaction.v
Solution

40% of 1 mol water reacts, so x = 0.40 mol. At equilibrium, H2O = CO = 0.60 mol and H2 = CO2 = 0.40 mol. Volume cancels because Δn = 0. Therefore Kc = ([H2][CO2])/([H2O][CO]) = (0.40 x 0.40)/(0.60 x 0.60) = 0.44.

Answer:

Kc = 0.44.

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Q.6.15At 700 K, equilibrium constant for the reaction: H2 (g) + I2 (g) ⇌ 2HI (g) is 54.8. If 0.5 mol L^-1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?v
Solution

For H2 + I2 ⇌ 2HI, Kc = [HI]^2/([H2][I2]). Starting with HI only gives equal H2 and I2 at equilibrium, say x each. Thus 54.8 = (0.5)^2/x^2, so x = sqrt(0.25/54.8) = 0.0675 M.

Answer:

[H2] = [I2] = 0.0675 mol L^-1.

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Q.6.16What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl (g) ⇌ I2 (g) + Cl2 (g); Kc = 0.14v
Solution

Let [I2] = [Cl2] = x at equilibrium. Then [ICl] = 0.78 - 2x. Kc = x^2/(0.78 - 2x)^2 = 0.14. Taking square root, x/(0.78 - 2x) = sqrt(0.14). Solving gives x = 0.167 M and [ICl] = 0.446 M.

Answer:

[ICl] = 0.446 M; [I2] = [Cl2] = 0.167 M.

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Q.6.17Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium? C2H6 (g) ⇌ C2H4 (g) + H2 (g)v
Solution

Let x atm of C2H6 decompose. Then pC2H4 = pH2 = x and pC2H6 = 4.0 - x. Kp = x^2/(4.0 - x) = 0.04, giving x = 0.380 atm and pC2H6 = 3.62 atm. Using p = CRT, C = 3.62/(0.0821 x 899) = 0.049 mol L^-1.

Answer:

p(C2H6) = 3.62 atm, corresponding to [C2H6] = 0.049 mol L^-1 at 899 K.

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Q.6.18Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: CH3COOH (l) + C2H5OH (l) ⇌ CH3COOC2H5 (l) + H2O (l) (i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction) (ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?v
Solution

Water is part of the reacting mixture, so it is included. In (ii), ester = water = 0.171 mol, acetic acid = 1.00 - 0.171 = 0.829 mol and ethanol = 0.18 - 0.171 = 0.009 mol. Kc = (0.171 x 0.171)/(0.829 x 0.009) = 3.92. In (iii), Qc = (0.214 x 0.214)/[(1.0 - 0.214)(0.5 - 0.214)] = 0.204, which is less than Kc, so equilibrium has not been reached and the forward reaction continues.

Answer:

(i) Qc = [CH3COOC2H5][H2O]/([CH3COOH][C2H5OH]); (ii) Kc = 3.92; (iii) equilibrium has not been reached, and the reaction proceeds forward.

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Q.6.19A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5 x 10^-1 mol L^-1. If value of Kc is 8.3 x 10^-3, what are the concentrations of PCl3 and Cl2 at equilibrium? PCl5 (g) ⇌ PCl3 (g) + Cl2(g)v
Solution

For PCl5 ⇌ PCl3 + Cl2, Kc = [PCl3][Cl2]/[PCl5]. Starting with pure PCl5 gives [PCl3] = [Cl2] = x. Thus 8.3 x 10^-3 = x^2/(0.5 x 10^-1), so x = sqrt(8.3 x 10^-3 x 0.05) = 2.04 x 10^-2 M.

Answer:

[PCl3] = [Cl2] = 2.04 x 10^-2 mol L^-1.

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Q.6.21Equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g) at 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L^-1 N2, 2.0 mol L^-1 H2 and 0.5 mol L^-1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?v
Solution

Qc = [NH3]^2/([N2][H2]^3) = (0.5)^2/[3.0 x (2.0)^3] = 0.0104. Since Qc < Kc = 0.061, the mixture has too little product, so the reaction proceeds forward to form more NH3.

Answer:

The reaction is not at equilibrium; it proceeds in the forward direction.

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Q.6.22Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) ⇌ Br2 (g) + Cl2 (g) for which Kc = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 x 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium?v
Solution

Let [Br2] = [Cl2] = x at equilibrium. Then [BrCl] = 3.3 x 10^-3 - 2x. Kc = x^2/(3.3 x 10^-3 - 2x)^2 = 32. Taking square root and solving gives x = 1.516 x 10^-3 M, so [BrCl] = 3.3 x 10^-3 - 2x = 2.68 x 10^-4 M.

Answer:

[BrCl] = 2.68 x 10^-4 mol L^-1.

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Q.6.24Calculate a) ∆G° and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K NO (g) + 1/2 O2 (g) ⇌ NO2 (g) where ∆fG° (NO2) = 52.0 kJ/mol ∆fG° (NO) = 87.0 kJ/mol ∆fG° (O2) = 0 kJ/molv
Solution

For NO(g) + 1/2 O2(g) ⇌ NO2(g), ∆rG° = ∆fG°(NO2) - [∆fG°(NO) + 1/2∆fG°(O2)] = 52.0 - 87.0 = -35.0 kJ mol^-1. From ∆G° = -RT ln K, ln K = 35000/(8.314 x 298) = 14.13. Therefore K = e^14.13 = 1.37 x 10^6.

Answer:

∆G° = -35.0 kJ mol^-1; K = 1.37 x 10^6.

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Q.6.25Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume? (a) PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) (b) CaO (s) + CO2 (g) ⇌ CaCO3 (s) (c) 3Fe (s) + 4H2O (g) ⇌ Fe3O4 (s) + 4H2 (g)v
Solution

Decreasing pressure favours the side with more gaseous moles. In (a), products have two gas moles versus one on the reactant side, so products increase. In (b), the gaseous side is the reactant side, so products decrease. In (c), both sides have four gaseous moles, so the amount of products remains essentially unchanged.

Answer:

(a) Increase; (b) decrease; (c) remain same.

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Q.6.26Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction. (i) COCl2 (g) ⇌ CO (g) + Cl2 (g) (ii) CH4 (g) + 2S2 (g) ⇌ CS2 (g) + 2H2S (g) (iii) CO2 (g) + C (s) ⇌ 2CO (g) (iv) 2H2 (g) + CO (g) ⇌ CH3OH (g) (v) CaCO3 (s) ⇌ CaO (s) + CO2 (g) (vi) 4 NH3 (g) + 5O2 (g) ⇌ 4NO (g) + 6H2O(g)v
Solution

Increasing pressure favours the side with fewer gaseous moles. In (i), 1 gas mole is on the reactant side and 2 on the product side, so the shift is backward. In (ii), gaseous moles are equal, so pressure has no effect. In (iii), 1 gas mole shifts backward from 2 gas moles. In (iv), 3 gas moles shift forward to 1 gas mole. In (v), the backward side has no gas. In (vi), 9 gas moles on the reactant side are favoured over 10 gas moles on the product side.

Answer:

(i) backward; (ii) not affected; (iii) backward; (iv) forward; (v) backward; (vi) backward.

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Q.6.27The equilibrium constant for the following reaction is 1.6 x10^5 at 1024K H2(g) + Br2(g) ⇌ 2HBr(g) Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.v
Solution

Starting with HBr only, let pH2 = pBr2 = x at equilibrium. Then pHBr = 10.0 - 2x. Kp = (pHBr)^2/(pH2 pBr2) = (10.0 - 2x)^2/x^2 = 1.6 x 10^5. Taking square root, (10.0 - 2x)/x = 400, so x = 10.0/402 = 0.0249 bar and pHBr = 9.95 bar.

Answer:

pH2 = pBr2 = 0.0249 bar and pHBr = 9.95 bar.

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Q.6.28Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: CH4 (g) + H2O (g) ⇌ CO (g) + 3H2 (g) (a) Write as expression for Kp for the above reaction. (b) How will the values of Kp and composition of equilibrium mixture be affected by (i) increasing the pressure (ii) increasing the temperature (iii) using a catalyst?v
Solution

For the reaction CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g), Kp is the product partial pressures divided by reactant partial pressures with stoichiometric powers. Increasing pressure favours fewer gas moles, so the equilibrium shifts to CH4 and H2O, but Kp is unchanged at fixed temperature. Since the forward reaction is endothermic, increasing temperature favours products and increases Kp. A catalyst only helps equilibrium be reached faster; it does not change Kp or the equilibrium composition.

Answer:

Kp = pCO(pH2)^3/(pCH4 pH2O). Pressure increase shifts the composition backward without changing Kp; temperature increase raises Kp and favours products; catalyst does not change Kp or equilibrium composition.

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Q.6.29Describe the effect of: a) addition of H2 b) addition of CH3OH c) removal of CO d) removal of CH3OH on the equilibrium of the reaction: 2H2(g) + CO (g) ⇌ CH3OH (g)v
Solution

By Le Chatelier's principle, adding a reactant shifts equilibrium towards product, while adding product shifts it towards reactants. Removing a reactant shifts backwards to replace it, and removing product shifts forwards to replace the product. Therefore addition of H2 and removal of CH3OH favour methanol formation, while addition of CH3OH and removal of CO favour the reverse reaction.

Answer:

(a) forward shift; (b) backward shift; (c) backward shift; (d) forward shift.

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Q.6.30At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 x10^-3. If decomposition is depicted as, PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) ∆rH° = 124.0 kJ mol^-1 a) write an expression for Kc for the reaction. b) what is the value of Kc for the reverse reaction at the same temperature? c) what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased (iii) the temperature is increased ?v
Solution

For PCl5(g) ⇌ PCl3(g) + Cl2(g), Kc = [PCl3][Cl2]/[PCl5]. The reverse reaction has Kc = 1/(8.3 x 10^-3) = 1.20 x 10^2. At a fixed temperature, changing concentration or pressure may shift the equilibrium composition but does not change Kc. Since decomposition is endothermic, increasing temperature favours decomposition and increases Kc.

Answer:

(a) Kc = [PCl3][Cl2]/[PCl5]; (b) 1.20 x 10^2; (c) adding PCl5 or increasing pressure does not change Kc, while increasing temperature increases Kc.

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Q.6.31Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction, CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g) If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that pco = pH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°Cv
Solution

Let pH2 = pCO2 = x at equilibrium. Then pCO = pH2O = 4.0 - x. Kp = (pCO2 pH2)/(pCO pH2O) = x^2/(4.0 - x)^2 = 10.1. Taking square root, x/(4.0 - x) = sqrt(10.1). Solving gives x = 3.04 bar.

Answer:

pH2 = 3.04 bar.

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Q.6.32Predict which of the following reaction will have appreciable concentration of reactants and products: a) Cl2 (g) ⇌ 2Cl (g) Kc = 5 x10^-39 b) Cl2 (g) + 2NO (g) ⇌ 2NOCl (g) Kc = 3.7 x 10^8 c) Cl2 (g) + 2NO2 (g) ⇌ 2NO2Cl (g) Kc = 1.8v
Solution

When Kc is extremely small, reactants dominate; when Kc is extremely large, products dominate. Reaction (a) has Kc = 5 x 10^-39, so reactants dominate. Reaction (b) has Kc = 3.7 x 10^8, so products dominate. Reaction (c) has Kc close to 1, so both reactants and products are appreciably present.

Answer:

Only reaction (c) will have appreciable concentrations of both reactants and products.

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Q.6.33The value of Kc for the reaction 3O2 (g) ⇌ 2O3 (g) is 2.0 x10^-50 at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6 x10^-2, what is the concentration of O3?v
Solution

Kc = [O3]^2/[O2]^3. Hence [O3] = sqrt(Kc[O2]^3) = sqrt((2.0 x 10^-50)(1.6 x 10^-2)^3) = 2.86 x 10^-28 M.

Answer:

[O3] = 2.86 x 10^-28 mol L^-1.

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Q.6.34The reaction, CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.v
Solution

In the 1 L flask, the given moles are numerically the molar concentrations. Kc = [CH4][H2O]/([CO][H2]^3). Therefore [CH4] = Kc[CO][H2]^3/[H2O] = (3.90)(0.30)(0.10)^3/0.02 = 0.0585 M.

Answer:

[CH4] = 0.0585 mol L^-1.

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Q.6.36Which of the followings are Lewis acids? H2O, BF3, H+, and NH4+v
Solution

A Lewis acid accepts an electron pair. BF3 accepts an electron pair because boron has an incomplete octet. H+ accepts an electron pair to form a bond. NH4+ can act as an acid through its proton. H2O has lone pairs and normally donates an electron pair, so it is a Lewis base.

Answer:

BF3, H+ and NH4+ act as Lewis acids; H2O is a Lewis base.

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Q.6.37What will be the conjugate bases for the Brönsted acids: HF, H2SO4 and HCO3-?v
Solution

A conjugate base is obtained when an acid loses one proton. HF loses H+ to form F-. H2SO4 loses H+ to form HSO4-. HCO3- loses H+ to form CO3^2-.

Answer:

F-, HSO4- and CO3^2-, respectively.

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Q.6.38Write the conjugate acids for the following Brönsted bases: NH2-, NH3 and HCOO-.v
Solution

A conjugate acid is obtained by adding one proton to a base. NH2- + H+ → NH3. NH3 + H+ → NH4+. HCOO- + H+ → HCOOH.

Answer:

NH3, NH4+ and HCOOH, respectively.

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Q.6.39The species: H2O, HCO3-, HSO4- and NH3 can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.v
Solution

For each amphiprotic species, adding H+ gives its conjugate acid and removing H+ gives its conjugate base. H2O gives H3O+ and OH-. HCO3- gives H2CO3 and CO3^2-. HSO4- gives H2SO4 and SO4^2-. NH3 gives NH4+ and NH2-.

Answer:

H2O: H3O+ and OH-; HCO3-: H2CO3 and CO3^2-; HSO4-: H2SO4 and SO4^2-; NH3: NH4+ and NH2-.

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Q.6.40Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH- (b) F- (c) H+ (d) BCl3 .v
Solution

OH- and F- possess lone pairs and can donate an electron pair, so they are Lewis bases. H+ accepts an electron pair from a donor to form a covalent bond, so it is a Lewis acid. BCl3 has an electron-deficient boron atom with an empty orbital and accepts an electron pair, so it is also a Lewis acid.

Answer:

OH- and F- are Lewis bases; H+ and BCl3 are Lewis acids.

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Q.6.41The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10^-3 M. What is its pH?v
Solution

pH = -log[H+] = -log(3.8 x 10^-3) = 2.42.

Answer:

pH = 2.42.

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Q.6.42The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.v
Solution

[H+] = 10^-pH = 10^-3.76 = 1.74 x 10^-4 M.

Answer:

[H+] = 1.74 x 10^-4 M.

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Q.6.43The ionization constant of HF, HCOOH and HCN at 298K are 6.8 x 10^-4, 1.8 x 10^-4 and 4.8 x 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.v
Solution

For a conjugate acid-base pair, KaKb = Kw = 1.0 x 10^-14 at 298 K. Therefore Kb(F-) = 10^-14/(6.8 x 10^-4) = 1.47 x 10^-11; Kb(HCOO-) = 10^-14/(1.8 x 10^-4) = 5.56 x 10^-11; Kb(CN-) = 10^-14/(4.8 x 10^-9) = 2.08 x 10^-6.

Answer:

F-: 1.47 x 10^-11; HCOO-: 5.56 x 10^-11; CN-: 2.08 x 10^-6.

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Q.6.44The ionization constant of phenol is 1.0 x 10^-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?v
Solution

For phenol alone, [C6H5O-] ≈ sqrt(KaC) = sqrt((1.0 x 10^-10)(0.05)) = 2.24 x 10^-6 M. With 0.01 M sodium phenolate, [C6H5O-] is approximately 0.01 M from the salt. Ka = [H+][C6H5O-]/[C6H5OH], so [H+] = (1.0 x 10^-10)(0.05)/0.01 = 5.0 x 10^-10 M. Degree of ionization = [H+]/0.05 = 1.0 x 10^-8.

Answer:

[C6H5O-] = 2.24 x 10^-6 M in pure phenol solution; degree of ionization in 0.01 M sodium phenolate = 1.0 x 10^-8.

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Q.6.45The first ionization constant of H2S is 9.1 x 10^-8. Calculate the concentration of HS- ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H2S is 1.2 x 10^-13, calculate the concentration of S2- under both conditions.v
Solution

For H2S alone, [HS-] ≈ sqrt(Ka1 C) = sqrt((9.1 x 10^-8)(0.1)) = 9.54 x 10^-5 M. For the second ionization, Ka2 = [H+][S2-]/[HS-]; since [H+] ≈ [HS-] in the weak-acid solution, [S2-] ≈ Ka2 = 1.2 x 10^-13 M. In 0.1 M HCl, [H+] ≈ 0.1 M, so [HS-] = Ka1[H2S]/[H+] = (9.1 x 10^-8)(0.1)/0.1 = 9.1 x 10^-8 M. Then [S2-] = Ka2[HS-]/[H+] = (1.2 x 10^-13)(9.1 x 10^-8)/0.1 = 1.09 x 10^-19 M.

Answer:

In 0.1 M H2S: [HS-] = 9.54 x 10^-5 M and [S2-] ≈ 1.2 x 10^-13 M. In 0.1 M HCl: [HS-] = 9.1 x 10^-8 M and [S2-] = 1.09 x 10^-19 M.

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Q.6.46The ionization constant of acetic acid is 1.74 x 10^-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.v
Solution

For weak acetic acid, α ≈ sqrt(Ka/C) = sqrt((1.74 x 10^-5)/0.05) = 0.0187. Therefore [CH3COO-] = Cα = 0.05 x 0.0187 = 9.33 x 10^-4 M, equal to [H+]. pH = -log(9.33 x 10^-4) = 3.03.

Answer:

Degree of dissociation = 0.0187; [CH3COO-] = 9.33 x 10^-4 M; pH = 3.03.

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Q.6.47It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.v
Solution

[H+] = 10^-4.15 = 7.08 x 10^-5 M, so [A-] = 7.08 x 10^-5 M. For HA ⇌ H+ + A-, Ka = [H+][A-]/[HA] = x^2/(0.01 - x) = (7.08 x 10^-5)^2/(0.009929) = 5.05 x 10^-7. pKa = -log Ka = 6.30.

Answer:

[A-] = 7.08 x 10^-5 M; Ka = 5.05 x 10^-7; pKa = 6.30.

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Q.6.48Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOHv
Solution

For strong acids, pH = -log[H+]. For strong bases, pOH = -log[OH-] and pH = 14 - pOH. Thus HCl: pH = -log(0.003) = 2.52. NaOH: pOH = -log(0.005) = 2.30, so pH = 11.70. HBr: pH = -log(0.002) = 2.70. KOH: pOH = -log(0.002) = 2.70, so pH = 11.30.

Answer:

(a) 2.52; (b) 11.70; (c) 2.70; (d) 11.30.

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Q.6.49Calculate the pH of the following solutions: a) 2 g of TlOH dissolved in water to give 2 litre of solution. b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution. c) 0.3 g of NaOH dissolved in water to give 200 mL of solution. d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.v
Solution

(a) Molar mass of TlOH ≈ 221.4 g mol^-1; [OH-] = (2/221.4)/2 = 4.52 x 10^-3 M, pH = 11.65. (b) Moles Ca(OH)2 = 0.3/74.1 = 4.05 x 10^-3 in 0.500 L, so [OH-] = 2(4.05 x 10^-3)/0.500 = 0.0162 M and pH = 12.21. (c) [OH-] = (0.3/40.0)/0.200 = 0.0375 M, pH = 12.57. (d) [H+] = (13.6 x 0.001)/1.000 = 0.0136 M, pH = 1.87.

Answer:

(a) 11.65; (b) 12.21; (c) 12.57; (d) 1.87.

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Q.6.50The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.v
Solution

[H+] = Cα = 0.1 x 0.132 = 0.0132 M, so pH = -log(0.0132) = 1.88. Ka = Cα^2/(1 - α) = 0.1(0.132)^2/(1 - 0.132) = 2.01 x 10^-3. Therefore pKa = -log(2.01 x 10^-3) = 2.70.

Answer:

pH = 1.88; pKa = 2.70.

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Q.6.51The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.v
Solution

pOH = 14 - 9.95 = 4.05, so [OH-] = 10^-4.05 = 8.91 x 10^-5 M. For the weak base, [BH+] = [OH-] = x and [B] = 0.005 - x. Kb = x^2/(0.005 - x) = (8.91 x 10^-5)^2/(0.00491) = 1.62 x 10^-6. Hence pKb = -log Kb = 5.79.

Answer:

Kb = 1.62 x 10^-6; pKb = 5.79.

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Q.6.52What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.v
Solution

From Table 6.7, Kb(aniline) = 4.27 x 10^-10. For 0.001 M aniline, [OH-] ≈ sqrt(KbC) = sqrt((4.27 x 10^-10)(0.001)) = 6.53 x 10^-7 M. Therefore pOH = 6.185 and pH = 7.82. Degree of ionization = [OH-]/C = 6.53 x 10^-7/0.001 = 6.53 x 10^-4. For the conjugate acid, Ka = Kw/Kb = 10^-14/(4.27 x 10^-10) = 2.34 x 10^-5.

Answer:

pH = 7.82; degree of ionization = 6.53 x 10^-4; Ka of conjugate acid = 2.34 x 10^-5.

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Q.6.53Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1M in HCl ?v
Solution

Ka = 10^-4.74 = 1.82 x 10^-5. In pure acid, α ≈ sqrt(Ka/C) = sqrt((1.82 x 10^-5)/0.05) = 0.0191. In the presence of strong acid, ionization is suppressed and α ≈ Ka/[H+]. With 0.01 M HCl, α = 1.82 x 10^-5/0.01 = 1.82 x 10^-3. With 0.1 M HCl, α = 1.82 x 10^-5/0.1 = 1.82 x 10^-4.

Answer:

In pure 0.05 M acetic acid, α = 0.0191. In 0.01 M HCl, α = 1.82 x 10^-3; in 0.1 M HCl, α = 1.82 x 10^-4.

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Q.6.54The ionization constant of dimethylamine is 5.4 x 10^-4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?v
Solution

For dimethylamine, Kb = Cα^2/(1 - α). Substituting Kb = 5.4 x 10^-4 and C = 0.02 gives α = 0.151, about 15.1%. In 0.1 M NaOH, the common OH- ion suppresses ionization, and α ≈ Kb/[OH-] = (5.4 x 10^-4)/0.1 = 5.4 x 10^-3. Percentage ionized = 0.54%.

Answer:

Degree of ionization in 0.02 M solution = 0.151; in 0.1 M NaOH, 0.54% is ionized.

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Q.6.55Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below: (a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2 (c) Human blood, 7.38 (d) Human saliva, 6.4.v
Solution

Use [H+] = 10^-pH. Muscle fluid: 10^-6.83 = 1.48 x 10^-7 M. Stomach fluid: 10^-1.2 = 6.31 x 10^-2 M. Blood: 10^-7.38 = 4.17 x 10^-8 M. Saliva: 10^-6.4 = 3.98 x 10^-7 M.

Answer:

(a) 1.48 x 10^-7 M; (b) 6.31 x 10^-2 M; (c) 4.17 x 10^-8 M; (d) 3.98 x 10^-7 M.

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Q.6.56The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.v
Solution

Use [H+] = 10^-pH. For pH 6.8, [H+] = 1.58 x 10^-7 M. For pH 5.0, [H+] = 1.0 x 10^-5 M. For pH 4.2, [H+] = 6.31 x 10^-5 M. For pH 2.2, [H+] = 6.31 x 10^-3 M. For pH 7.8, [H+] = 1.58 x 10^-8 M.

Answer:

Milk: 1.58 x 10^-7 M; black coffee: 1.0 x 10^-5 M; tomato juice: 6.31 x 10^-5 M; lemon juice: 6.31 x 10^-3 M; egg white: 1.58 x 10^-8 M.

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Q.6.57If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?v
Solution

Moles of KOH = 0.561/56.1 = 0.0100 mol. Volume = 0.200 L, so [K+] = [OH-] = 0.0100/0.200 = 0.050 M. At 298 K, [H+] = Kw/[OH-] = 1.0 x 10^-14/0.050 = 2.0 x 10^-13 M. pOH = -log(0.050) = 1.30, so pH = 12.70.

Answer:

[K+] = 0.050 M; [OH-] = 0.050 M; [H+] = 2.0 x 10^-13 M; pH = 12.70.

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Q.6.58The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.v
Solution

Molar mass of Sr(OH)2 = 121.6 g mol^-1. Molar solubility = 19.23/121.6 = 0.158 M, so [Sr2+] = 0.158 M. Since Sr(OH)2 gives two hydroxide ions, [OH-] = 2 x 0.158 = 0.316 M. pOH = -log(0.316) = 0.500, so pH = 13.50.

Answer:

[Sr2+] = 0.158 M; [OH-] = 0.316 M; pH = 13.50.

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Q.6.59The ionization constant of propanoic acid is 1.32 x 10^-5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?v
Solution

For 0.05 M propanoic acid, α ≈ sqrt(Ka/C) = sqrt((1.32 x 10^-5)/0.05) = 0.0162. Thus [H+] = Cα = 0.05 x 0.0162 = 8.12 x 10^-4 M and pH = 3.09. In 0.01 M HCl, α ≈ Ka/[H+] = (1.32 x 10^-5)/0.01 = 1.32 x 10^-3.

Answer:

In pure solution, α = 0.0162 and pH = 3.09. In 0.01 M HCl, α = 1.32 x 10^-3.

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Q.6.60The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.v
Solution

[H+] = 10^-2.34 = 4.57 x 10^-3 M. Degree of ionization α = [H+]/C = (4.57 x 10^-3)/0.1 = 0.0457. Ka = x^2/(C - x) = (4.57 x 10^-3)^2/(0.1 - 0.00457) = 2.19 x 10^-4.

Answer:

Ka = 2.19 x 10^-4; degree of ionization = 0.0457.

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Q.6.61The ionization constant of nitrous acid is 4.5 x 10^-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.v
Solution

NO2- is the conjugate base of HNO2. Kb = Kw/Ka = 1.0 x 10^-14/(4.5 x 10^-4) = 2.22 x 10^-11. For 0.04 M sodium nitrite, [OH-] ≈ sqrt(KbC) = sqrt((2.22 x 10^-11)(0.04)) = 9.43 x 10^-7 M. pOH = 6.03, so pH = 7.97. Degree of hydrolysis = [OH-]/C = 9.43 x 10^-7/0.04 = 2.36 x 10^-5.

Answer:

pH = 7.97; degree of hydrolysis = 2.36 x 10^-5.

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Q.6.62A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.v
Solution

Pyridinium ion is the conjugate acid of pyridine. [H+] = 10^-3.44 = 3.63 x 10^-4 M. For 0.02 M pyridinium ion, Ka = x^2/(C - x) = (3.63 x 10^-4)^2/(0.02 - 3.63 x 10^-4) = 6.71 x 10^-6. Therefore Kb(pyridine) = Kw/Ka = 1.0 x 10^-14/(6.71 x 10^-6) = 1.49 x 10^-9.

Answer:

Kb for pyridine = 1.49 x 10^-9.

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Q.6.63Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH4NO3, NaNO2 and KFv
Solution

Salts of strong acid and strong base are neutral, so NaCl and KBr are neutral. Salts containing anions of weak acids with strong-base cations are basic, so NaCN, NaNO2 and KF are basic. NH4NO3 contains NH4+, the conjugate acid of weak base NH3, and NO3- from a strong acid, so its solution is acidic.

Answer:

NaCl and KBr are neutral; NaCN, NaNO2 and KF are basic; NH4NO3 is acidic.

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Q.6.64The ionization constant of chloroacetic acid is 1.35 x 10^-3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?v
Solution

For the acid, solve Ka = x^2/(0.1 - x) with Ka = 1.35 x 10^-3. This gives [H+] = x = 1.10 x 10^-2 M and pH = 1.96. For the sodium salt, Kb = Kw/Ka = 7.41 x 10^-12. [OH-] ≈ sqrt(KbC) = sqrt((7.41 x 10^-12)(0.1)) = 8.61 x 10^-7 M. pOH = 6.07, so pH = 7.93.

Answer:

pH of 0.1 M chloroacetic acid = 1.96; pH of 0.1 M sodium chloroacetate = 7.93.

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Q.6.65Ionic product of water at 310 K is 2.7 x 10^-14. What is the pH of neutral water at this temperature?v
Solution

For neutral water, [H+] = [OH-] = sqrt(Kw). Thus [H+] = sqrt(2.7 x 10^-14) = 1.64 x 10^-7 M. pH = -log(1.64 x 10^-7) = 6.78.

Answer:

pH = 6.78.

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Q.6.66Calculate the pH of the resultant mixtures: a) 10 mL of 0.2M Ca(OH)2+25 mL of 0.1M HCl b) 10 mL of 0.01M H2SO4+10 mL of 0.01M Ca(OH)2 c) 10 mL of 0.1M H2SO4+10 mL of 0.1M KOHv
Solution

(a) OH- equivalents from Ca(OH)2 = 2(0.010 x 0.2) = 0.0040 mol; H+ from HCl = 0.025 x 0.1 = 0.0025 mol. Excess OH- = 0.0015 mol in 0.035 L, so [OH-] = 0.0429 M and pH = 12.63. (b) H2SO4 provides 0.0002 mol H+ and Ca(OH)2 provides 0.0002 mol OH-, so the mixture is neutral and pH = 7.00. (c) H2SO4 provides 0.0020 mol H+ and KOH provides 0.0010 mol OH-. Excess H+ = 0.0010 mol in 0.020 L, so [H+] = 0.050 M and pH = 1.30.

Answer:

(a) pH = 12.63; (b) pH = 7.00; (c) pH = 1.30.

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Q.6.67Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.v
Solution

Using Table 6.9: Ksp(Ag2CrO4)=1.1 x 10^-12, so Ksp=(2S)^2S=4S^3 and S=6.50 x 10^-5 M. Ksp(BaCrO4)=1.2 x 10^-10, so S=sqrt(Ksp)=1.10 x 10^-5 M and both ions have this concentration. Ksp(Fe(OH)3)=1.0 x 10^-38, so Ksp=S(3S)^3=27S^4 and S=1.39 x 10^-10 M. Ksp(PbCl2)=1.6 x 10^-5, so Ksp=S(2S)^2=4S^3 and S=1.59 x 10^-2 M. Ksp(Hg2I2)=4.5 x 10^-29, so Ksp=S(2S)^2=4S^3 and S=2.24 x 10^-10 M.

Answer:

Ag2CrO4: S = 6.50 x 10^-5 M, [Ag+] = 1.30 x 10^-4 M, [CrO4^2-] = 6.50 x 10^-5 M. BaCrO4: S = 1.10 x 10^-5 M. Fe(OH)3: S = 1.39 x 10^-10 M, [OH-] = 4.16 x 10^-10 M. PbCl2: S = 1.59 x 10^-2 M, [Cl-] = 3.17 x 10^-2 M. Hg2I2: S = 2.24 x 10^-10 M, [Hg2^2+] = 2.24 x 10^-10 M, [I-] = 4.48 x 10^-10 M.

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Q.6.68The solubility product constant of Ag2CrO4 and AgBr are 1.1 x 10^-12 and 5.0 x 10^-13 respectively. Calculate the ratio of the molarities of their saturated solutions.v
Solution

For Ag2CrO4, Ksp = (2S)^2S = 4S^3, so S = (1.1 x 10^-12/4)^(1/3) = 6.50 x 10^-5 M. For AgBr, Ksp = S^2, so S = sqrt(5.0 x 10^-13) = 7.07 x 10^-7 M. Ratio = (6.50 x 10^-5)/(7.07 x 10^-7) = 92.

Answer:

Molarity of saturated Ag2CrO4 solution : molarity of saturated AgBr solution = 92 : 1 approximately.

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Q.6.69Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Ksp = 7.4 x 10^-8 ).v
Solution

On mixing equal volumes, each concentration is halved: [Cu2+] = 0.001 M and [IO3-] = 0.001 M. For Cu(IO3)2, Qsp = [Cu2+][IO3-]^2 = (0.001)(0.001)^2 = 1.0 x 10^-9. Since Qsp < Ksp = 7.4 x 10^-8, the solution is unsaturated with respect to copper iodate and no precipitate forms.

Answer:

No precipitation occurs.

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Q.6.70The ionization constant of benzoic acid is 6.46 x 10^-5 and Ksp for silver benzoate is 2.5 x 10^-13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?v
Solution

For a salt of a weak acid in acidic buffer, S_buffer/S_water = sqrt(([H+] + Ka)/Ka). At pH 3.19, [H+] = 10^-3.19 = 6.46 x 10^-4 M. Therefore ratio = sqrt((6.46 x 10^-4 + 6.46 x 10^-5)/(6.46 x 10^-5)) = sqrt(11) = 3.32. Ksp cancels in the ratio.

Answer:

Silver benzoate is about 3.32 times more soluble in the buffer.

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Q.6.71What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 x 10^-18).v
Solution

Let the concentration of each original solution be C. After equal-volume mixing, [Fe2+] = C/2 and [S2-] = C/2. For no precipitation, Qsp must not exceed Ksp: (C/2)(C/2) = C^2/4 = 6.3 x 10^-18 at the limit. Therefore C = 2sqrt(6.3 x 10^-18) = 5.02 x 10^-9 M.

Answer:

Maximum concentration of each original equimolar solution = 5.02 x 10^-9 M.

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Q.6.72What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 x 10^-6).v
Solution

For CaSO4 ⇌ Ca2+ + SO4^2-, Ksp = S^2. Thus S = sqrt(9.1 x 10^-6) = 3.02 x 10^-3 mol L^-1. Molar mass of CaSO4 ≈ 136.1 g mol^-1, so solubility in g L^-1 = (3.02 x 10^-3)(136.1) = 0.411 g L^-1. Volume needed for 1.0 g = 1.0/0.411 = 2.43 L.

Answer:

Minimum volume = 2.43 L.

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Q.6.73The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 x 10^-19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place?v
Solution

After mixing, [S2-] = (1.0 x 10^-19)(10/15) = 6.67 x 10^-20 M and [M2+] = (0.04)(5/15) = 1.33 x 10^-2 M. Thus Qsp = [M2+][S2-] = 8.89 x 10^-22. Compare with Ksp values from Table 6.9: FeS = 6.3 x 10^-18 and MnS = 2.5 x 10^-13, so Qsp < Ksp and no precipitate forms. For ZnS = 1.6 x 10^-24 and CdS = 8.0 x 10^-27, Qsp > Ksp, so ZnS and CdS precipitate.

Answer:

Precipitation will take place with ZnCl2 and CdCl2 only.

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