Page 1 : Class–XI–CBSE-Chemistry, , Equilibrium, , CBSE NCERT Solutions for Class 11 Chemistry Chapter 7, Back of Chapter Questions, , 1. A 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?, , Solution:, (a) If the volume of the container is suddenly increased, then the vapour pressure, would decrease initially. This is because the amount of vapour remains the same, but the, volume increases suddenly. As a result, the same amount of vapour is distributed in a, larger volume., (b) Since the temperature is constant, the rate of evaporation also remains constant., When the volume of the container is increased, the density of the vapour phase decreases., As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of, condensation decreases initially., (c) When equilibrium is restored finally, the rate of evaporation becomes equal to the, rate of condensation. In this case, only the volume changes while the temperature remains, constant. The vapour pressure depends on temperature and not on volume. Hence, the, final vapour pressure will be equal to the original vapour pressure of the system., , 2. What is K c 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) ⟶ 2SO2 (g), , Solution:, Given: [SO2 ] = 0.60M,, [O2 ] = 0.82M, [SO3 ] = 1.90M, [𝑝𝑟𝑜𝑑𝑢𝑐𝑡], , Formula : The equilibrium constant (K c ) = = [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡], 1, Practice more on Equilibrium, , www.embibe.com

Page 2 : Class–XI–CBSE-Chemistry, , Equilibrium, , [SO3 ]2, Kc =, [SO2 ]2 [O2 ], =, , (1.90)2 M 2, (0.60)2 (0.821)M 3, , = 12.239 M −1, Hence, K for the equilibrium is = 12.239 M −1., , 3. At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I, atoms, I2 (g) ⇌ 2I (g), Calculate K p for the equilibrium., , Solution:, Given: total pressure of 105Pa, I atoms =40% by volume, Partial pressure=mole fraction × total pressure, 40, , Partial pressure of I atoms (p1) = 100 × ptotal, =, , (∴ 𝑣𝑜𝑙𝑢𝑚𝑒 ∝ 𝑚𝑜𝑙𝑒 ), , 40, × 105, 100, , = 4 × 104 Pa, Partial pressure of I2 molecules,, p𝐼2 =, =, , 60, × ptotal, 100, , 60, × 105, 100, , = 6 × 104 Pa, (partial pressure of product), , Formula : K p = (partial pressure of reactant), Kp =, , (pI )2, (P12 ), , 2, Practice more on Equilibrium, , www.embibe.com

Page 4 : Class–XI–CBSE-Chemistry, , Equilibrium, , 5. Find out the value of K c for each of the following equilibria from the value of K p :, (i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g); K p = 1.8 × 10−2 at 500 K, (ii) CaCO3 (s) ⇌ CaO(s) + CO2 (g); K p = 167 at 1073 K, , Solution:, Formula : The relation between K p and K c is given as:K p = K c (RT)Δn, (a) (i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g); K p = 1.8 × 10−2 at 500 K, Δn = 3 − 2 = 1, R = 0.0831 barLmol−1 K −1, T = 500 K, K P = 1.8 × 10−2, Put all values in the formula of K P = K c (RT)Δn, ⇒ 1.8 × 10−2 = K c (0.0831 × 500)1, ⇒ Kc =, , 1.8 × 10−2, 0.0831 × 500, , = 4.33 × 10−4 (approximately), (b) (ii) CaCO3 (s) ⇌ CaO(s) + CO2 (g); K p = 167 at 1073 K, Δn = 2 − 1 = 1, R = 0.0831 barLmol−1 K −1 T, T = 1073 K, K p = 167, Now,put all values in a formula K p = K c (RT)Δn, ⇒ 167 = K c (0.0831 × 1073)Δn, ⇒ Kc =, , 167, 0.0831 × 1073, , = 1.87 (approximately), , 6. For the following equilibrium, K c = 6.3 × 1014 at 1000 K, 4, Practice more on Equilibrium, , www.embibe.com

Page 5 : Class–XI–CBSE-Chemistry, , Equilibrium, , NO (g) + O3 (g) ⇌ NO2 (g) + O2 (g), Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions., What is K ,c for the reverse reaction?, , Solution:, [𝑝𝑟𝑜𝑑𝑢𝑐𝑡], , formula :The equilibrium constant (K c ) = = [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡], NO (g) + O3 (g) ⇌ NO2 (g) + O2 (g), It is given that Kc for the forward reaction is K c =, Then, K ′c for the reverse reaction =, K ′c =, =, , [NO2 (g)]2 [O2 (g)], [NO (g)][O3 (g)], , =6.3 × 1014, , [NO (g)][O3 (g)], [NO2 (g)]2 [O2 (g)], , 1, Kc, , 1, 6.3 × 1014, , = 1.59 × 10−15, , 7. Explain why pure liquids and solids can be ignored while writing the equilibrium constant, expression?, Solution:, For a pure substance (both solids and liquids),, Active mass of [Pure substance] =, =, , Number of moles Mass/molecular moles, =, Volume, Volume, , Mass, , = Volume×Molecular mass, , Density, Molecular mass, , Now, the molecular mass and density (at a particular temperature) of a pure substance is, always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure, substances (solid and liquid) are not mentioned in the equilibrium constant expression., , 8. Reaction between N2 and O2 takes place as follows:, 5, Practice more on Equilibrium, , www.embibe.com

Page 6 : Class–XI–CBSE-Chemistry, , Equilibrium, , 2N2 (g) + O2 (g) ⇌ 2N2 O (g), If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed, to form N2 O at a temperature for which K c = 2.0 × 10−37 determine the composition of the, equilibrium mixture., , Solution:, Given: 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 mol N2 = 0.482, initial mol O2 = and 0.933, the volume of container =10 L, Let the concentration of 𝑁2 𝑂 at equilibrium be x., The given reaction is:, , 2N2 (g) + O2 (g) ⇌ 2N2 O(g), Initial conc., 0.482 mol, 0.933 mol, 0, At equilibrium (0.482 − x) mol (1.933 − x) mol x mol, Therefore, at equilibrium, in the 10 L vessel:, [N2 ] =, , 0.482, 10, , = 0.0482 mol L−1 and [O2 ] =, , 0.933, 10, , x, , = 0.0933molL−1 [N2 O] = 10 molL−1, , [𝑝𝑟𝑜𝑑𝑢𝑐𝑡], , Formula: The equilibrium constant (K c ) = = [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡], Kc =, , [N2 O(g)]2, [N2 (g)]2 [O2 (g)], , ⇒ 2.0 × 10−37, ⇒, , x 2, (10), =, (0.0482)2 (0.0933), , x2, = 2.0 × 10−37 × (0.0482)2 × (0.0933), 100, , ⇒ x 2 = 43.35 × 10−40, ⇒ x = 6.6 × 10−20, [N2 O] =, , x, 6.6 × 10−20, =, = 6.6 × 10−21, 10, 10, , 6, Practice more on Equilibrium, , www.embibe.com

Page 7 : Class–XI–CBSE-Chemistry, , Equilibrium, , 9. Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction is 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 a constant, temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO, and Br2 ., , Solution:, Given: initial mol of NO =0.087, Initial mol of Br2 =0.0437, mol of NOBr is obtained at equilibrium=0.0518, The given reaction is:, 2NO (g) + Br2 (g) ⇋ 2NOBr (g), 2 mol, , 1 mol, , 2 mol, , According to stoichiometry , 2 mol of NOBr are formed from 2 mol of NO and 1 mole of Br2, react, , 2 mol of NOBr is formed, Therefore, 0.0518 mol of NOBr is formed from, , 0.0518, 2, , mol of Br, or 0.0259 mol of NO., , The amount of NO and Br present initially is as follows:, [NO] = 0.087 mol [Br2 ] = 0.0437 mol, Therefore, the amount of NO present at equilibrium = 0.087 – 0.0518 = 0.0352 mol, And, the amount of Br2 present at equilibrium = 0.0437 – 0.0259 = 0.0178 mol, , 10. At 450K, Kp = 2.0 × 1010 bar−1 for the given reaction at equilibrium., 2SO2 (g) + O2 (g) ⇌ 2SO3 (g) What is K c at this temperature?, , Solution:, given :, T = 450 K, K p = 2.0 × 1010 bar −1, , We know Δn = 2 − 3 = −1, 7, Practice more on Equilibrium, , www.embibe.com

Page 8 : Class–XI–CBSE-Chemistry, , Equilibrium, , R = 0.0831 bar L bar K −1 mol−1, Formula : K p = K c (RT)Δn, Put all values in the above formula, ⇒ 2.0 × 1010 bar −1 = K c (0.0831L bar K −1 mol−1 × 450K)−1, ⇒ Kc =, , 2.0 × 1010 bar −1, (0.0831 L bar K −1 mol−1 × 450K)−1, , = (2.0 × 1010 bar −1 )(0.0831LbarK −1 mol−1 × 450K), = 74.79 × 1010 L mol−1, = 7.48 × 1011 L mol−1, = 7.48 × 1011 M−1, , 11. A sample of HI(g) is placed in the flask at a pressure of 0.2 atm. At equilibrium, the partial, pressure of HI(g) is 0.04 atm. What is K p for the given equilibrium?, 2HI (g) ⇌ H2 (g) + I2 (g), , Solution:, Given: pressure = 0.2 atm, equilibrium the partial pressure of HI(g) = 0.04 atm, The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04, atm. Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction, is:, 2HI (g), Initial conc., At equilibrium, , ⇌ H2 (g) +, , 0.2 atm,, 0.04atm, , 0, 0.16, 2, = 0.08 atm, , I2 (g), 0, 2.15, 2, = 0.08atm, , 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑝𝑒𝑟𝑜𝑑𝑢𝑐𝑡, 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡, , formula for K p = 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, Kp =, , pH2 × pI2, (PHI )2, , 8, Practice more on Equilibrium, , www.embibe.com

Page 9 : Class–XI–CBSE-Chemistry, , =, , 0.08 × 0.08, (0.04)2, , =, , 0.0064, 0.0016, , Equilibrium, , = 4.0, Hence, the value of K p for the given equilibrium is 4.0., , 12. A 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, K c for the reaction, N2 (g) + 3H2 (g) ⇌ 2NH3 (g) is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what, is the direction of the net reaction?, Solution:, Given: mol of N2 = 1.57, mol of H2 = 1.92, mol of NH3 =8.13, the volume of vessel =20 Liter, temperature = 500 K, K c = 1.7 × 102, The given reaction is:, N2 (g) + 3H2 (g) ⇌ 2NH3 (g), The given concentration of various species is, [N2 ] =, , 1.57, mol L−1, 20, , [H2 ] =, , 1.92, 20, , mol L−1, , [NH3 ] =, , 8.13, 20, , mol L−1, , Now, reaction quotient Q c is:, Qc =, , [NH3 ]2, [N2 ][H2 ]3, , (8.13) 2, ( 20 ), =, 1.57 1.92 3, ( 20 ) ( 20 ), = 2.4 × 103, 9, Practice more on Equilibrium, , www.embibe.com

Page 10 : Class–XI–CBSE-Chemistry, , Equilibrium, , Since, Q c ≠ K c , the reaction mixture is not at equilibrium., Again, Q c > K c . Hence, the reaction will proceed in the reverse direction., , 13. The equilibrium constant expression for a gas reaction is,, Kc =, , [NH3 ]4 [O2 ]5, [NO]4 [H2 O]6, , Write the balanced chemical equation corresponding to this expression., , Solution:, 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑙𝑎𝑟 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑡, Kc =, 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑙𝑎𝑟 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡, The balanced chemical equation corresponding to the given expression can be written as:, 4NO (g) + 6H2 O(g) ↔ 4NH3 (g) + 5O2 (g), , 14. One mole of H2 O and one mole of CO is taken in 10 L vessel and heated to 725 K. At, equilibrium, 40% of water (by mass) reacts with CO according to the equation,, H2 O (g) + CO(g) ⇌ H2 (g) + CO2 (g), Calculate the equilibrium constant for the reaction., , Solution:, The given reaction is:, H2 O (g), , + CO(g) ⇌, , H2 (g) + CO2 (g), , 1, 1, 0, 0, M, M, 10, 10, 1 − 0.4, 1 − 0.4, 0.4, 0.4, At, M, M, M, M, equilibrium, 10, 10, 10, 10, = 0.06 M = 0.06 M = 0.04 M = 0.04 M, 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑙𝑎𝑟 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑡, Therefore, the equilibrium constant for the reaction = 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑙𝑎𝑟 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡, Initial conc, , 10, Practice more on Equilibrium, , www.embibe.com

Page 11 : Class–XI–CBSE-Chemistry, , [H ][CO ], , Equilibrium, , [0.04][0.04], , 2, K c = [H 2O][CO], = [0.06][0.06] = 0.444 (approximately), 2, , 15. At 700 K, the 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?, , Solution:, Given: equilibrium constant K c for the reaction H2 (g) + I2 (g) ⇌ 2HI (g) is 54.8., temperature =700 K, equilibrium constant K c for the reaction, , [H2 (g)] [I2 (g)], [HI (g)]2, , =54.8, , Therefore, at equilibrium, the equilibrium constant K ′c for the reaction, 2HI (g) ⇌ H2 (g) + I2 (g), K ′c = [H, , [HI (g)]2, 1, =54.8, (g)], (g)], [I, 2, 2, , 2HI (g) ⇌ H2 (g) + I2 (g), x, , 0.5, , x, , moles of HI(g) is present at equilibrium =0.5 mol L−1, [HI] = 0.5mol L−1 will be, Let the concentrations of hydrogen and iodine at equilibrium be x molL−1, [H2 ] = [I2 ] = x mol L−1, K ′C =, , x×x, 1, =, (0.5)2 54.8, , ⇒ x2 =, , 0.25, 54.8, , ⇒ x = 0.06754, x = 0.068 mol L−1, Hence, at equilibrium, [H2 ] = x = 0.068 mol L−1 ; [I2 ] = x = 0.068 mol L−1, 11, Practice more on Equilibrium, , www.embibe.com

Page 13 : Class–XI–CBSE-Chemistry, , Equilibrium, , Solution:, Let p is the partial pressure exerted by ethene and hydrogen gas (each) at equilibrium. Now,, according to the reaction,, C2 H6 (g) ⇌ C2 H4 (g) + H2 (g), Initial conc., 4.0 atm, At equilibrium, 4.0 − p, pC2H4 ×pH2, Formula : p, = Kp, , 0, p, , 0, p, , C 2 H6, , ⇒, , p×p, = 0.04, 4.0 − p, , ⇒ p2 = 0.16 − 0.04p, ⇒ p2 + 0.04p − 0.16 = 0, Now, p =, , −0.04±√(0.04)2 −4×1×(−0.16), 2×1, , =, , −0.04 ± 0.80, 2, , =, , 0.76, 2, , (Taking positive value), , = 0.38, , 18. Ethyl acetate is formed by the reaction between ethanol and acetic acid, and the equilibrium is, represented as:, CH3 COOH (l) + C2 H5 OH (l) ⇌ CH3 COOC2 H5 (l) + H2 O (l), (i) Write the concentration ratio (reaction quotient), Q c , 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 some time. Has equilibrium been reached?, , Solution:, (i), , water is not in excess and is not a solvent in this reaction so all will be considered in, the equilibrium expression, , 13, Practice more on Equilibrium, , www.embibe.com

Page 14 : Class–XI–CBSE-Chemistry, , Equilibrium, , [CH COOC2 H5 ][H2 O], 3 COOH][C2 H5 OH], , Reaction quotient ( Q c ) = [CH 3, , given : temperature = 293 K,, initial mol of acetic acid= 1.00, initial mol of ethanol= 0.18, mol of ethyl acetate=0.171, Let the volume of the reaction mixture be V. Also, here we will consider that water is, a solvent and is present in excess., (ii), , The given reaction is:, CH3 COOH(l) + C2 H5 OH(l) ⇌ CH3 COOC2 H5 (l) + H2 O(l), 1, 0.18, 0, 0, M, M, V, V, 1 − 0.171, 0.18 − 0.171, 0.171, 0.171, At equilibrium, M, M, V, V, V, V, 0.829, 0.009, =, M, =, M, V, V, Therefore, the equilibrium constant for the given reaction is:, Initial conc., , [CH COOC2 H3 ][H2 O], 3 COOH][C2 H5 OH], , K c = [CH 3, , 0.171 0.171, ×, V, 0.009, ×, V, V, , V, = 0.829, , = 3.919 ≈ 3.92, , (ii) temperature = 293 K,, Initial mol of acetic acid = 1, 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 mol of ethanol = 0. 5, mol of ethyl acetate formed = 0.214, (i), , Let the volume of the reaction mixture be V., , CH3 COOH(l) + C2 H5 OH(l) ⇌ CH3 COOC2 H5 (l) + H2 O(l), 1.0, 0.5, 0, 0, M, M, V, V, 10 − 0.214, 0.214, 0.214, 0.5 − 0.214, After some, M, M, time, V, V, V, V, 0.786, 0.286, =, M, =, M, V, V, Therefore, the reaction quotient is,, Initial conc., , Qc =, , [CH3 COOC2 H5 ], [CH3 COOH][C2 H5 OH], , 0.214 0.214, × V, = V, 0.786 0.286, V × V, = 0.2037, = 0204 (approximately), 14, Practice more on Equilibrium, , www.embibe.com

Page 15 : Class–XI–CBSE-Chemistry, , Equilibrium, , Since Qc < K c , equilibrium has not been reached., , 19. A 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 × 10−1 mol L−1 . If value of K c is 8.3 × 10−3,, what are the concentrations of PCl3 and Cl2 at equilibrium?, PCl5 (g) ⇌ PCl3 (g) + Cl2 (g), , Solution:, Given : temperature =473, After equilibrium concentration of PCl5 = 0.5 × 10−1 mol L−1, K c = 8.3 × 10−3, Let the concentrations of both PCl3 and Cl2 at equilibrium be x molL−1 . The given reaction, is:, At equilibrium, , PCl5 (g), 0.5 × 10−1 mol L−1, , ↔, , PCl3 (g), x mol L−1, , + Cl2 (g), x mol L−1, , It if given that the value of the equilibrium constant, K c is 8.3 × 10−3 ., Now we can write the expression for equilibrium as:, [PCl3 ][Cl2 ], = Kc, [PCl5 ], ⇒, , x×x, = 8.3 × 10−3, 0.5 × 10−1, , ⇒ x 2 = 4.15 × 10−4, ⇒ x = 2.04 × 10−2, = 0.0204 ≈ 0.02, Therefore, at equilibrium,, [PCl3 ] = [Cl2 ] = x = 0.02 mol L−1, , 20. One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II), oxide by carbon monoxide to give iron metal and CO2 ., FeO (s) + CO (g) → Fe (s) + CO2 (g); Kp = 0.265 atm at 1050 K, 15, Practice more on Equilibrium, , www.embibe.com

Page 16 : Class–XI–CBSE-Chemistry, , Equilibrium, , What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures, are: PCO = 1.4 atm and = 0.80 atm?, , Solution:, Given: Kp = 0.265 atm, Temperature = 1050K, initial partial pressures of CO (PCO ) = 1.4 atm, initial partial pressures of CO2 (PCO2 ) = 0.80 atm, For the given reaction,, FeO(s) + CO (g) ⇌ Fe (s) + CO2 (g), Initially,, , Qp =, =, , 1.4 atm, , 0.80 atm, , pCO2, PCO, , 0.80, 1.4, , = 0.571, It is given that K p = 0.265., Since Q c > K p , the reaction will proceed in the backward direction., Therefore, we can say that the pressure of CO will increase while the pressure of CO2 will, decrease., Now, let the increase in pressure of CO = decrease in pressure of CO2 be p. Then,, we can write,, FeO(s) + CO (g) ↔ Fe (s) + CO2 (g), Initially,, , Kp =, , 1.4 atm, , 0.80 atm, , pCO2, PCO, , ⇒ 0.265 =, , 0.80 − p, 1.4 + p, , ⇒ 0.371 + 0.265p = 0.80 − p, ⇒ 1.265p = 0.429, ⇒ p = 0.339 atm, Therefore, equilibrium partial pressure of CO2 (PCO2 ) = 0.80 − 0.339 = 0.461 atm, And, equilibrium partial pressure of CO(PCO ) = 1.4 + 0.339 = 1.739 atm, 16, Practice more on Equilibrium, , www.embibe.com

Page 17 : Class–XI–CBSE-Chemistry, , Equilibrium, , 21. Equilibrium constant, K c 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?, , Solution:, given: 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 = 500 K, K c =0.061, composition of N2 = 3.0 mol L−1, composition of H2 = 2.0 mol L−1, composition of NH3 = 0.5 mol L−1 ., , N2 (g) + 3H2 (g), 3.0 mol L−1, 2.0 mol L−1, , At a particular time:, , Formula: Q c =, , [product ], [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 ], , [NH3 ]2, 3, 2 ][H2 ], , = [N, , ↔, , 2NH3 (g), 0.5 mol L−1, , (0.5)2, , = (3.0)(2.0)3, , = 0.0104, It is given that K c = 0.061, Since Q c ≠ K c , the reaction is not at equilibrium., Since Q c < K c , the reaction will proceed in the forward direction to reach equilibrium., , 22. Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:, 2BrCl (g) ⇌ Br2 (g) + Cl2 (g) for which K c = 32 at 500 K. If initially pure BrCl is present at a, concentration of 3.3 × 10−3 mol L−1 , what is it's molar concentration in the mixture at, equilibrium?, , Solution:, Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:, 17, Practice more on Equilibrium, , www.embibe.com

Page 18 : Class–XI–CBSE-Chemistry, , Equilibrium, , 2BrCl(g), −3, , 3.3 × 10, 3.3 × 10−3 − 2x, , Initial conc,, At equilibrium, Now, we can write,, Kc =, , Cl2 (g), 0, x, , [Br2 ][Cl2 ], [BrCl]2, , ⇒ 32 =, ⇒, , ↔ Br2 (g) +, 0, x, , x×x, (3.3×10−3 −2x)2, , by taking root both side, , x, = 5.66, 3.3 × 10−3 − 2x, , ⇒ x = 18.678 × 10−3 − 11.32x, ⇒ 12.32x = 18.678 × 10−3, ⇒ x = 1.5 × 10−3, Therefore, at equilibrium,, [BrCl] = (3.3 × 10−3 − 2x) = 3.3 × 10−3 − (2 × 1.5 × 10−3 ) = 3.3 × 10−3 − 3.0 × 10−3, = 0.3 × 10−3 = 3.0 × 10−4 mol L−1, , 23. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon, has 90.55% CO by mass, C (s) + CO2 (g) ⇌ 2CO (g), Calculate K c for this reaction at the above temperature., , Solution:, Given : CO by mass =90.55%, Temperature =1127 k, Pressure = 1 atm, We know 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 CO2 = 44, 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 CO = 28, Let the total mass of the gaseous mixture is 100 g., Mass of CO = 90.55 g, And, the mass of CO2 = (100 – 90.55) = 9.45 g, , 18, Practice more on Equilibrium, , www.embibe.com

Page 20 : Class–XI–CBSE-Chemistry, , Equilibrium, , Solution:, For the given reaction,, 𝑁𝑂 (𝑔) +, , 1, 𝑂 (𝑔) ⇌ 𝑁𝑂2 (𝑔), 2 2, , ∆1 Go (NO2 ) = 52.0 kJ/mol, ∆Go (NO) = 87.0 kJ/mol, ∆Go (O2 ) = 0 kJ/mol, Formula : ΔGo = ΔGo (products) −ΔGo (Reactants), ΔGo = 52.0 − {87.0 + 0}, = −35.0 kJ mol−1, (b) formula : ΔGo = −RT ln K c, ΔGo = −2.303 RT log K c, Put all values in above formula, −35.0×10−3, , log K c = −2.303×8.314×298, log K c = 6.134, ∴ K c = antilog (6.134), K c = 1.36 × 106, Hence, the equilibrium constant for the given reaction K c is 1.36 × 106, , 25. Does the number of moles of reaction products increase, decrease or remain the 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) + 4H2 O (g) ⇌ Fe3 O4 (s) + 4H2 (g), , Solution:, (a) The number of moles of reaction products will increase. According to Le Chatelier’s, principle, if pressure is decreased, then the equilibrium shifts in the direction in which, the number of moles of gases is more. In the given reaction, the number of moles of, gaseous products is more than that of gaseous reactants. Thus, the reaction will, proceed in the forward direction. As a result, the number of moles of reaction products, 20, Practice more on Equilibrium, , www.embibe.com

Page 21 : Class–XI–CBSE-Chemistry, , Equilibrium, , will increase., (b) The number of moles of reaction products will decrease., (c) The number of moles of reaction products remains the same., , 26. Which 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) + 2H2 S (g), (iii) CO2 (g) + C (s) ⇌ 2CO (g), (iv) 2H2 (g) + CO (g) ⇌ CH3 OH (g), (v) CaCO3 (s) ⇌ CaO (s) + CO2 (g), (vi) 4 NH3 (g) + 5O2 (g) ⇌ 4NO (g) + 6H2 O(g), , Solution:, (i) COCl2 (g) ⇌ CO (g) + Cl2 (g), , ∆𝑛𝑔 = 1, , (ii) CH4 (g) + 2S2 (g) ⇌ CS2 (g) + 2H2 S (g) ∆𝑛𝑔 = 0, (iii) CO2 (g) + C (s) ⇌ 2CO (g) ∆𝑛𝑔 = 1, (iv) 2H2 (g) + CO (g) ⇌ CH3 OH (g) ∆𝑛𝑔 = −2, (v) CaCO3 (s) ⇌ CaO (s) + CO2 (g) ∆𝑛𝑔 = 1, (vi) 4 NH3 (g) + 5O2 (g) ⇌ 4NO (g) + 6H2 O(g) ∆𝑛𝑔 = 1, The reaction is given in (ii) ∆𝑛𝑔 = 0 no effect of pressure and no effect on equilibrium because of, the number of moles of gaseous reactants is the same that of gaseous products., (i), (iii), (iv), (v), and (vi) affected by pressure, The reaction is given in (iv) ∆𝑛𝑔 = −2, will proceed in the forward direction because of the number of moles of gaseous reactants is more, than that of gaseous products., The reactions are given in (i), (iii), (v), and (vi) in 𝑎𝑙𝑙 ∆𝑛𝑔 = + 1 will shift in the backward, direction because the number of moles of gaseous reactants is less than that of gaseous products., , 27. The equilibrium constant for the following reaction is 1.6 × 105 at 1024 K, 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 1024 K., , 21, Practice more on Equilibrium, , www.embibe.com

Page 22 : Class–XI–CBSE-Chemistry, , Equilibrium, , Solution:, K p for the reaction H2 (g) + Br2 (g) ⇌ 2HBr(g) = 1.6 × 105 ., [HBr]2, , 𝐾𝑝 = [H, , 2 [Br2 ], , Therefore, for the reaction 2HBr(g) ⇌ H2 (g) + Br2 (g), the equilibrium constant will be,, K ′p =, , [H2 [Br2 ], [HBr]2, , =, , 1, Kp, , 1, , = 1.6×105 = 6.25 × 10−6, , Now, let p be the pressure of both H2 and Br2 at equilibrium., , Now, we can write,, pH2 × pBr2, = K ′p, pHBr, p×p, = 6.25 × 10−6, (10 − 2p)2, p, = 2.5 × 10−3, 10 − 2p, p = 2.5 × 10−2 − (5.0 × 10−3 )p, p + (5.0 × 10−3 )p = 2.5 × 10−2, (10.5 × 10−3 )p = 2.5 × 10−2, p = 2.49 × 10−2 bar ≈ 2.5 × 10−2 bar, Therefore, at equilibrium,, [H2 ] = [Br2 ] = p = 2.49 × 10−2 bar, [HBr] = 10 − 2p = 10 − 2 × (2.49 × 10−2 ) bar, = 9.95 bar ≈ 10 bar, , 28. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following, endothermic reaction:, 22, Practice more on Equilibrium, , www.embibe.com

Page 23 : Class–XI–CBSE-Chemistry, , Equilibrium, , CH4 (g) + H2 O (g) ⇌ CO (g) + 3H2 (g), (a) Write as the expression for Kp for the above reaction., (b) How will the values of K p and composition of equilibrium mixture be affected by, (i) increasing the pressure., (ii) increasing the temperature., (iii) using a catalyst?, , Solution:, (a) For the given reaction,, Kp =, , pCO × [PH 2 ]2, pCH4 × pH2 O, , (b), (i), CH4 (g) + H2 O (g) ⇌ CO (g) + 3H2 (g) ; ∆𝑛𝑔 = 2, K p ∝ 𝑃2, According to Le Chatelier’s principle,on increasing pressure, the equilibrium will shift in the, backward direction., (ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will, shift in the forward direction., (iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only, increases the rate of a reaction. Thus, equilibrium will be attained quickly., , 29. Describe the effect of:, (A) addition of H2, (B) addition of CH3 OH, (C) removal of CO, (D) removal of CH3 OH on the equilibrium of the reaction:, 2H2 (g) + CO (g) ⇌ CH3 OH (g), , Solution:, Given the reaction : 2H2 (g) + CO (g) ⇌ CH3 OH (g) for this reaction ∆𝑛𝑔 = −2, 23, Practice more on Equilibrium, , www.embibe.com

Page 24 : Class–XI–CBSE-Chemistry, , Equilibrium, , (a) According to Le Chatelier’s principle, on the addition of H2 (reactant ), the equilibrium of the, given reaction will shift in the forward direction to maintain the equilibrium constant reactant has, to decrease ., (b) On addition of CH3 OH, the equilibrium will shift in the backward direction to maintain the, equilibrium constant reactant has to increase, (c) On removing CO, the equilibrium will shift in the backward direction to maintain the, equilibrium constant reactant has to increase, (d) On removing CH3 OH, the equilibrium will shift in the forward direction to maintain the, equilibrium constant reactant has to decrease, , 30. At 473 K, equilibrium constant K c for decomposition of phosphorus pentachloride,, PCl5 is 8.3 × 10−3 . If decomposition is depicted as,, PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) ∆r H e = 124.0 kJ mol−1, (a) write an expression for K c for the reaction., (b) what is the value of K c 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?, , Solution:, (a) for the reaction PCl5 (g) ⇌ PCl3 (g) + Cl2 (g), Kc =, , [PCl3 (g)][Cl2 (g)], [PCl5 (g)], , (a) Value of K c for the reverse reaction at the same temperature is, PCl3 (g) + Cl2 (g) ⇌ PCl5 (g), [PCl5 ], K ′c =, [PCl3 ][Cl2 ], , K ′c =, , 1, 1, =, = 1.2048 × 102 = 120.48, K c 8.3 × 10−3, , (b) (i) K c =, , [PCl3 (g)][Cl2 (g)], [PCl5 (g)], , 24, Practice more on Equilibrium, , www.embibe.com

Page 25 : Class–XI–CBSE-Chemistry, , Equilibrium, , more PCl5 is added then 𝐾𝑐 𝑤𝑖𝑙𝑙 𝑛𝑜𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 because of the equilibrium constant does, not change by change mole , concentration , pressure. It depends only on temperature and, stoichiometry of the reaction ., (ii) K c is constant ata constant temperature. Thus, in this case, K c would not change., (iii), PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) ∆r H o = 124.0 kJ mol−1, The reaction is endothermic , the value of K c increases with an increase in temperature., Since the given reaction in an endothermic reaction, the value of K c will increase if the, temperature is increased., Or 𝑙𝑛, , 𝑘2, 𝑘1, , 1, 𝑇1, , 1, 𝑇2, , = ∆𝐻[ − ], , As for endothermic reaction ∆𝐻 = +𝑣𝑒 ; 𝑇2 > 𝑇1 ; 𝐾2 > 𝐾1, , 31. Dihydrogen 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) + H2 O (g) ⇌ CO2 (g) + H2 (g), If a reaction vessel at 400o C is charged with an equimolar mixture of CO and steam such that, pco = pH2 O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? K p =, 10.1 at 400°C, , Solution:, Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction, is:, , It is given that K p = 10.1, Now,, pCO2 × pH2, = Kp, pCO × pH2 O, , 25, Practice more on Equilibrium, , www.embibe.com

Page 26 : Class–XI–CBSE-Chemistry, , Equilibrium, , p×p, , ⇒ (4.0−p)(4.0−p) = 10.1 by talking root both side we get, ⇒, , (p), = 3.178, (4.0 − p), , ⇒ p = 12.712 − 3.178p, ⇒ 4.178p = 12.712, ⇒ p = 3.04, Hence, at equilibrium, the partial pressure of H2 =P= 3.04 bar., , 32. Predict which of the following reaction will have appreciable concentration of reactants and, products:, a) Cl2 (g) ⇌ 2Cl (g), , K c = 5 × 10−39, , b) Cl2 (g) + 2NO (g) ⇌ 2NOCl (g), , K c = 3.7 × 108, , c) Cl2 (g) + 2NO2 (g) ⇌ 2NO2 Cl (g), , K c = 1.8, , Solution:, a) Cl2 (g) ⇌ 2Cl (g), , K c = 5 × 10−39, , b) Cl2 (g) + 2NO (g) ⇌ 2NOCl (g), , K c = 3.7 × 108, , c) Cl2 (g) + 2NO2 (g) ⇌ 2NO2 Cl (g), , K c = 1.8, , for (a) the values ofK c = 5 × 10−39 is very less, i.e. reactant is very more, b) Cl2 (g) + 2NO (g) ⇌ 2NOCl (g), K c = 3.7 × 108 is very high i.e. product is very more, (c) Cl2 (g) + 2NO2 (g) ⇌ 2NO2 Cl (g) K c = 1.8, If the value of K c lies between 10−3 and 103 , a reaction has appreciable concentration of, reactants and products. Thus, the reaction given in (c) will have appreciable concentration, of reactants and products., , 33. The value of K c for the reaction 3O2 (g) ⇌ 2O3 (g) is 2.0 × 10−50 at 25°C. If the equilibrium, concentration of O2 in air at 25°C is 1.6 × 10−2 , what is the concentration of O3 ?, , Solution:, Given: K c = 2.0 × 10−50, 26, Practice more on Equilibrium, , www.embibe.com

Page 27 : Class–XI–CBSE-Chemistry, , Equilibrium, , [O2 (g)] = 1.6 × 10−2 ., The given reaction 3O2 (g) ⇌ 2O3 (g), Kc =, , product of the molar concentration of product, product of the molar concentration of reactant, , [O (g)]2, , = [O3 (g)]3, 2, , Put all the given values in the expression of 𝐾𝑐, 2.0 × 10−50 =, , [O3 (g)]2, [1.6 × 10−2 ]3, , ⇒ [O3 (g)]2 = 2.0 × 10−50 × (1.6 × 10−2 )3, ⇒ [O3 (g)]2 = 8.192 × 10−56, ⇒ [O3 (g)]2 = 2.86 × 10−28 M, Hence, the concentration of O3 = 2.86 × 10−28 M, , 34. The reaction, CO(g) + 3H2 (g) ⇌ CH4 (g) + H2 O(g) is at equilibrium at 1300 K in a 1L flask. It, also contains 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2 O and an unknown amount of, CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant,, K c for the reaction at the given temperature is 3.90., Solution:, Given :Temperature = 1300 K, mol of CO = 0.30, mol of H2 = 0.10, mol of H2 O = 0.02, amount of CH4 = unknown, Let the concentration of methane at equilibrium be x., , It is given that K c = 3.90., , 27, Practice more on Equilibrium, , www.embibe.com

Page 28 : Class–XI–CBSE-Chemistry, , Equilibrium, , Therefore,, [CH4 (g)][H2 O(g)], = Kc, [CO(g)][H2 (g)]3, ⇒, , x × 0.02, = 3.90, 0.3 × (0.1)3, , ⇒x=, =, , 3.90 × 0.3 × (0.1)3, 0.02, , 0.00117, 0.02, , = 0.0585 M, = 5.85 × 10−2 M, Hence, the concentration of CH4 at equilibrium is 5.85 × 10–2 M., , 35. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following, species:, 2−, HNO2 , CN − , HClO4 , F − , OH − , CO2−, 3 , and S, , Solution:, A conjugate acid-base pair is a pair that differs only by one proton., 𝐴𝑐𝑖𝑑 (𝐻𝐴) → 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒(𝐴− ) + 𝐻 +, 𝑏𝑎𝑠𝑒 + 𝐻 + → 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑎𝑐𝑖𝑑, The conjugate acid-base for the given species is mentioned in the table below., , Species, HNO2, CN−, HClO4, F−, OH −, OH −, CO2−, 3, 𝑆 2−, , Conjugate acid-base, NO−, 2 (Conjugate - base), HCN (Conjugate - acid), ClO−, 4 (Conjugate - base), HF (Conjugate - acid), H2 O (Conjugate - acid), O2− (Conjugate - base), HCO−, 3 (Conjugate - acid), HS − (Conjugate - acid), , 28, Practice more on Equilibrium, , www.embibe.com

Page 30 : Class–XI–CBSE-Chemistry, , Equilibrium, , 39. The species: H2 O, HCO−3 , HSO−4 and NH3 can act both as Brönsted acids and bases. For each case, give the corresponding conjugate acid and base., Solution:, A conjugate acid-base pair is a pair that differs only by one proton., 𝐴𝑐𝑖𝑑 (𝐻𝐴) → 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒(𝐴− ) + 𝐻 +, 𝑏𝑎𝑠𝑒 + 𝐻 + → 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑎𝑐𝑖𝑑, The table below lists the conjugate acids and conjugate bases for the given species., Species, H2 O, HCO−, 3, HSO−, 4, NH3, , Conjugate acid, H3 O+, H2 CO3, H2 SO4, NH4+, , Conjugate base, OH −, CO2−, 3, SO2−, 4, NH2−, , 40. Classify 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, , Solution:, Lewis acids: species which can accept a pair of electrons. They are electron-deficient species, and having vacant orbital, Lewis base: species which can donate a pair of electrons. They are electron-efficient species, (a) OH − is a Lewis base since it can donate its lone pair of electrons., (b) F − is a Lewis base since it can donate a pair of electrons., (c) H + is a Lewis acid since it can accept a pair of electrons., (d) BCl3 is a Lewis acid since it can accept a pair of electrons., , 41. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10−3 M. what is its pH?, , 30, Practice more on Equilibrium, , www.embibe.com

Page 31 : Class–XI–CBSE-Chemistry, , Equilibrium, , Solution:, Given,, [H + ] = 3.8 × 10−3, ∴ pH value of soft drink = − log[H + ], = − log(3.8 × 10−3 ), , (∴ 𝑙𝑜𝑔𝑀 × 𝑁 = 𝑙𝑜𝑔𝑀 + 𝑙𝑜𝑔𝑁), , = −log(3.8 − log103 ), , (∴ 𝑙𝑜𝑔𝑀𝑁 = 𝑁𝑙𝑜𝑔𝑀), , = −log(3.8 + 3), = −0.58 + 3, = 2.42, , 42. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it., Solution:, Given, pH = 3.76, It is known that,, pH = − log[H + ], ⇒ log[H + ] = −pH, ⇒ [H + ] = antilog(−pH), = antilog(−3.76), = 1.74 × 10−4 M, Hence, the concentration of hydrogen ion [H + ] in the given sample of vinegar is 1.74 × 10−4 M., , 43. The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10−4 , 1.8 × 10−4 and 4.8 ×, 10−9 respectively. Calculate the ionization constants of the corresponding conjugate base., Solution:, Given : The ionization constant of HFat 298K is 6.8 × 10−4, The ionization constant of HCOOH at 298K is 1.8 × 10−4, 31, Practice more on Equilibrium, , www.embibe.com

Page 32 : Class–XI–CBSE-Chemistry, , Equilibrium, , The ionization constant of HCN at 298K is 4.8 × 10−9, Formula : K b K a =K w, Kb =, , Kw, Ka, , K a of HF = 6.8 × 10−4, Hence, K b of its conjugate base F − =, , Kw, Ka, , 10−14, , = 6.8×10−4 = 1.5 × 10−11, , K a of HCOOH = 1.8 × 10−4 Hence, K b of its conjugate base HCOO− =, , Kw, Ka, , 10−14, , = 1.8×10−4 = 5.6 ×, , 10−11, K a of HCN = 4.8 × 10−9Hence, K b of its conjugate base CN − =, , Kw, Ka, , =, , 10−14, 4.8×10−9, , = 2.08 × 10−6, , 44. The ionization constant of phenol is 1.0 × 1010 . 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?, Solution:, Given : Ionization of phenol =1.0 × 1010, , Ka =, , [C6 H5 O− ][H3 O+ ], [C6 H5 OH], x×x, , K a = 0.05−x =1.0 × 1010, As the value of the ionization constant is very less, x will be very small., We can ignore x in the denominator 0.05 − x ≈ 0.05, x×x, 0.05, , =1.0 × 1010, , ∴ x = √1 × 10−10 × 0.05, , 32, Practice more on Equilibrium, , www.embibe.com

Page 34 : Class–XI–CBSE-Chemistry, , Equilibrium, , Let the concentration of HS − = x M., , Then, K 𝑎1 =, , [H+ ][HS− ], [H2 S], , 9.1 × 10−8 =, , (x)(x), 0.1 − x, , (9.1 × 10−8 )(0.1 − x) = x 2, ∴ 𝑓𝑖𝑟𝑠𝑡 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑓 𝐻2 𝑆 = 9.1 × 10−8so x can be ignored and 0.1 − x M ≈ 0.1 M,, we have (9.1 × 10−8 )(0.1) = x 2 ., 9.1 × 10−9 = x 2, x = √9.1 × 10−9, = 9.54 × 10−5 M, ⇒ [HS]− = 9.54 × 10−5 M, Case b : (in the presence of HCl): concentration of H2 S solution = 0.1 M, In the presence of 0.1 M of HCl, let [HS − ] be y M, , Now, K a1 =, K aj =, , [HS− ][H+ ], [H2 S], , [y][0.1 + y], [0.1 − y], , 9.1 × 10−8 =, , y×0.1, 0.1, , (∵ 0.1 − y ≈ 0.1M ; 0.1 + y ≈ 0.1M ; ), , 9.1 × 10−8 = y, ⇒ [HS − ] = 9.1 × 10−8, (ii), , To calculate the concentration of [S 2− ], , Case c (in the absence of 𝟎. 𝟏 𝐌 𝐇𝐂𝐥):, 34, Practice more on Equilibrium, , www.embibe.com

Page 35 : Class–XI–CBSE-Chemistry, , Equilibrium, , HS − ⇌ H + + S 2−, [HS − ] = 9.54 × 10−5 M (From first ionization, case I), Let [S 2− ] be X., Also, [H + ] = 9.54 × 10−5 M, K a2 =, , [H + ][S 2− ], [HS − ], , K a2 =, , [9.54 × 10−5 ][X], [9.54 × 10−5 ], , (From first ionization, case I), , 1.2 × 10−13 = X = [S 2− ], Case d (in the presence of 0.1 M HCl):, Again, let the concentration of HS − be X′ M., [HS − ] = 9.1 × 10−8 M (From first ionization, case II), [H + ] = 0.1M, , (From HCl, case II), , [S − ] = X ′, HS −, , ⇌, , 9.1 × 10−8 M, Then, K a2 =, , H+ +, , S 2−, X′, , 0.1, , [H+ ][S2− ], , 1.2 × 10−13 =, , [HS− ], , (0.1)(X ′ ), 9.1 × 10−8, , 10.92 × 10−21 = 0.1X ′, 10.92 × 10−21, = X′, 0.1, X′ =, , 1.092 × 10−20, 0.1, , = 1.092 × 10−19 M, ⇒ K a1 = 1.74 × 10−5, , 46. The ionization constant of acetic acid is 1.74 × 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., , 35, Practice more on Equilibrium, , www.embibe.com

Page 36 : Class–XI–CBSE-Chemistry, , Equilibrium, , Solution:, Given: The ionization constant of acetic acid = 1.74 × 10−5, Concentration of acetic acid solution = 0.05 M solution, Method 1, 1) CH3 COOH ⇌ CH3 COO− + H + K a = 1.74 × 10−5, 2) H2 O + H2 O ⇌ H3 O− + OH − K w = 1.0 × 10−14, Since K a >> K w ,:, , K, , 1.74×10−5, =, 0.05, , α = √ ca = √, , 0.018 or 18 % so α can not be neglected so quadratic equation have to, , solve, Ka =, , (0.05α)(0.05α), (0.05 − 0.05α), , =, , (0.05α)(0.05α), 0.05(1 − α), , =, , 0.05α2, 1−α, , 1.74 × 10−5 =, , 0.05α2, 1−α, , 1.74 × 10−5 − 1.74 × 10−5 α = 0.05α2, 0.05α2 + 1.74 × 10−5 α − 1.74 × 10−5 =0, α=, =, , −b±√b2 −4𝑎𝑐, =, 2a, , −1.74×10−5 ±√(1.74×10−5 )2 −4(0.05)(−1.74×10−5 ), 2×0.05, , α =, , α=, , −1.74×10−5 ±√(1.74×10−5 )2 +(0.348 ×10−5 ), 2×0.05, , −1.74 × 10−5 ± (1.86 × 10−3 ), 2 × 0.05, , 36, Practice more on Equilibrium, , www.embibe.com

Page 37 : Class–XI–CBSE-Chemistry, , Equilibrium, , α=, , −1.74 × 10−5 + 1.86 × 10−3 ), 2 × 0.05, , α=, , 1.84×10−3, =1.84 ×, 0.1, , 10−2, , [H + ] = cα = 0.05 × 1.86 × 10−3, =, , 0.93 × 10−3, 1000, , = .000093, Method 2, Degree of dissociation,, Ka, a=√, c, c = 0.05 M, K a = 1.74 × 10−5, Then, α = √, , 1.74×10−5, .05, , α = √34.8 × 10−5, α = √3.48 × 10−4, α = 1.86 × 10−2, CH3 COOH ⇌ CH3 COO− + H +, Thus, concentration of CH3 COO− = c. α, = 0.05 × 1.86 × 10−2, = 0.093 × 10−2, = .00093M, Since [OAc − ] = [H + ],, [H + ] = 0.00093 = 0.093 × 10−2, pH = −log[H + ], = − log(0.093 × 10−2 ), ∴ pH = 3.03, Hence, the concentration of acetate ion in the solution is 0.00093 M and its pH is 3.03., 37, Practice more on Equilibrium, , www.embibe.com

Page 44 : Class–XI–CBSE-Chemistry, , Equilibrium, , 𝑝𝑂𝐻 = − log(. 065 × 105 ), = 6.187, 𝑝𝐻 = 7.813, the ionization constant of the conjugate acid of aniline, 𝐾𝑎 × 𝐾𝑏 = 𝐾𝑤, ∴ 4.27 × 10−10 × 𝐾𝑎 = 𝐾𝑤, 10−14, 𝐾𝑎 =, 4.27 × 10−10, = 2.34 × 10−5, Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10−5 ., , 53. Calculate 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) 0.01M𝑖𝑛 𝐻𝐶𝑙, (b) 0.1M in HCl?, , Solution:, 𝑐 = 0.05 𝑀, 𝑝𝐾𝑎 = 4.74, 𝑝𝐾𝑎 = − log(𝐾𝑎 ), 𝐾𝑎 = 1.82 × 10−5, 𝐾𝑎, 𝐾𝑎 = 𝑐𝛼 2 𝛼 = √, 𝑐, 1.82 × 10−5, 𝛼=√, = 1.908 × 10−2, 5 × 10−2, When 𝐻𝐶𝑙 is added to the solution, the concentration of 𝐻 + ions will increase. Therefore,, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will, decrease., Case I: When 0.01 𝑀 𝐻𝐶𝑙 is taken., Let x be the amount of acetic acid dissociated after the addition of 𝐻𝐶𝑙., , 44, Practice more on Equilibrium, , www.embibe.com

Page 46 : Class–XI–CBSE-Chemistry, , 𝛼=, =, , Equilibrium, , Amount of acid dissociated, Amount of acid taken initially, , 1.82 × 10−4 × 0.05, 0.05, , = 1.82 × 10−4, , 54. The ionization constant of dimethylamine is 5.4 × 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?, , Solution:, K b = 5,4 × 10−4, c = 0.02 M, K, , Then, α = √ cb = √, , 5.4×10−4, 0.02, , = 0.643, , Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes, complete ionization., NaOH(aq) ↔ Na+ (aq) + OH − (aq), 0.1 M, , 0.1 M, , (𝐶𝐻3 )2 𝑁𝐻 + 𝐻2 𝑂 ⇌ (𝐶𝐻3 )2 𝑁𝐻2+ + 𝐻𝑂−, At equi., , (0.02-x), , x, , x+0.1, , Then, [(CH3 )2 NH2+ ] = x, (𝐶𝐻3 )2 𝑁𝐻 = 0.02 − x ≈ 0.02, [OH − ] = x + 0.1 ≈ 0.1, ⇒ Kb =, , [(CH3 )2 NH2+ ][OH − ], [(CH3 )2 NH], , 5.4 × 10−4 =, , x × 0.1, 0.02, , x = 0.0054, It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated., 46, Practice more on Equilibrium, , www.embibe.com

Page 47 : Class–XI–CBSE-Chemistry, , Equilibrium, , 55. Calculate 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., , Solution:, (a) Human muscle fluid 6.83:, pH = 6.83, pH = − log[H + ], ∴ 6.83 = − log[H + ], [H + ] = 1.48 × 10−7 M, (b) Human stomach fluid, 1.2:, pH = 1.2, pH = − log[H + ], 1.2 = − log[H + ], ∴ [H + ] = 0.063, (c) Human Blood, 7.38, pH = 7.38, pH = − log[H + ], pH = 7.38 = − log[H + ], ∴ [H + ] = 4.17 × 10−8 M, (d) Human saliva, 6.4:, pH = 6.4, pH = − log[H + ], 6.4 = −log[H + ], [H + ] = 3.98 × 10−7, , 47, Practice more on Equilibrium, , www.embibe.com

Page 48 : Class–XI–CBSE-Chemistry, , Equilibrium, , 56. The 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., , Solution:, The hydrogen ion concentration in the given substances ca be calculated by using the given, relation: pH = − log[H + ], (i) the pH of milk = 6.8, Since pH = − log[H + ], 6.8 = − log[H + ], [H + ] = antilog(−6.8), [H + ] = 1.5 × 19−7 M, , (ii) the pH of black coffee = 5.0, Since pH = − log[H + ], 5.0 = − log[H + ], Log [H + ] = −5.0, [H + ] = antilog(−5.0) = 10−5 M, , (iii) the pH of tomato juice = 4.2, Since pH = − log[H + ], 4.2 = − log[H + ] log, [H + ] = antilog(−4.2) = 6.31 × 10−5 M, , (iv) the pH of lemon juice = 2.2, Since pH = − log[H + ], 2.2 = − log[H + ] log, [H + ] = −2.2, [H + ] = antilog(−2.2) = 6.31 × 10−3 M, (v) the pH of egg white = 7.8, , 48, Practice more on Equilibrium, , www.embibe.com

Page 49 : Class–XI–CBSE-Chemistry, , Equilibrium, , Since pH = − log[H + ], 7.8 = − log[H + ], Log [H + ] = −7.8, [H + ] = antilog(−7.8) = 1.58 × 10−8 M, , 57. If 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?, Solution:, We know : molecular weight of KOH=39+17=56, moles of KOH=0.561/56, solute, , molar concentration of [KOH(aq)] = moles of volume of solution(liter) =, , 0.561, 56, , (200/1000)L, , 𝑀=, , 0.05 M, KOH(aq) → K + (aq) + OH − (aq), [OH − ] = 0.05 M = [K + ], [H + ][OH − ] = K w, [H + ] =, , Kw, [OH − ], , [H + ] =, , 10−14, = 2 × 10−13 M, 0.05, , ∴ pH = − log[𝐻 + ]=, -log(2 × 10−13), =-[log2 + log10−13, = −[𝑙𝑜𝑔2 − 13log10] = 12.70, , 49, Practice more on Equilibrium, , www.embibe.com

Page 50 : Class–XI–CBSE-Chemistry, , Equilibrium, , 58. The 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., , Solution:, Solubility of Sr(OH)2 = 19.23 g⁄L, 19.23, , Then, concentration of Sr(OH)2 = 121.63 M = 0.1581 M, Sr(OH)2 (aq) → Sr 2 (aq) + 2(OH − )(aq), ∴ [Sr 2+ ] = 0.5181 M, [OH − ] = 2 × 0.1581M = 0.3126 M, Now,, K w = [OH − ][H + ], 10−14, = [H + ], 0.3126, ⇒ [H + ] = 3.2 × 10−14, 𝑝𝐻 = − log[𝐻 + ], = - [log0.32 + log10−14 ], = −[𝑙𝑜𝑔0.32 − 14log10], pH = 13.495 ≈ 13.50, , 59. The ionization constant of propanoic acid is 1.32 × 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?, , Solution:, Let the degree of ionization of propanoic acid be α, . Then, representing propionic acid as HA, we have:, 𝐻𝐴, , + 𝐻2 𝑂, , ⇌, , 0.05-0.05α ≈ 0.05, , 𝐻3 𝑂+ +, 0.05α, , 𝐴−, , 0.05α, , 50, Practice more on Equilibrium, , www.embibe.com

Page 52 : Class–XI–CBSE-Chemistry, , Equilibrium, , − log[H + ] = 2.34, [H + ] = antilog(−2.34), [H + ] = 4.5 × 10−3, Also,, [H + ] = cα, 4.5 × 10−3 = 0.1 × α, 4.5 × 10−3, =α, 0.1, α = 45 × 10−3 = 0.045, Then,, K a = cα2, = 0.1 × (45 × 10−3 )2, = 202.5 × 10−6, = 2.02 × 10−4, , 61. The ionization constant of nitrous acid is 4.5 × 10−4 . Calculate the pH of 0.04 M sodium nitrite, solution and also its degree of hydrolysis., , Solution:, NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2 )., −, NO−, 2 + H2 O ⇋ HNO2 + OH, , Kh =, , [HNO2 ][OH− ] [HNO2 ][OH− ][𝐻 + ], = [NO−][𝐻+], [NO2− ], 2, , ∴ 𝑘𝑤 = [OH− ][𝐻+ ] = 10−14 ; 𝐾𝑎 =, , [NO2− ][𝐻 + ], [HNO2 ], , Kw, 10−14, ⇒, =, = 0.22 × 10−10, K a 4.5 × 10−4, Now, If x moles of the salt undergoes hydrolysis, then the concentration of various species, present in the solution will be:, [NO−, 2 ] = 0.04 − x ≈ 0 . 04, [HNO2 ] = x, [OH− ] = x, , 52, Practice more on Equilibrium, , www.embibe.com

Page 53 : Class–XI–CBSE-Chemistry, , Equilibrium, , x2, Kh =, = 0.22 × 10−10, 0.04, x 2 = 0.0088 × 10−10, x = 0.093 × 10−5, ∴ [OH− ] = 0.093 × 10−5 M, [H3 O+ ] =, , Kw, 10−14, =, = 10.75 × 10−9 M, [HO− ] 0.093 × 10−5, , ⇒ pH = − log(10.75 × 10−9 ), pH = −[log(10.75 + log(10−9 )], pH = −[log(10.75) − 9log10 ), = 7.96, x, , Therefore, degree of hydrolysis (h)= 0.04 =, , 0.093×10−5, 0.04, , = 2.325 × 10−5, , 62. A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant, of pyridine., , Solution:, pyridinium hydrochloride (𝐶5 𝐻7 𝐶𝑙𝑁 + ) is salt of a weak base(𝐶5 𝐻5 𝑁) and strong acid(HCl), pH = 3.44, We know that,, pH =– log [H + ], -3.44= log [H + ], ∴ [H + ] = 3.63 × 10−4, 𝑝𝑦𝑟𝑖𝑑𝑖𝑛𝑖𝑢𝑚 ℎ𝑦𝑑𝑟𝑜𝑐ℎ𝑙𝑜𝑟𝑖𝑑𝑒 + 𝐻2 𝑂 ⇌ 𝑝𝑦𝑟𝑖𝑑𝑖𝑛𝑖𝑢𝑚 + 𝐻𝑂 −, , 2, , Then, K h =, , (3.63×10−4 ), 0.02, , (∵ concentration = 0.02M), , ⇒ K h = 6.6 × 10−6, , 53, Practice more on Equilibrium, , www.embibe.com

Page 54 : Class–XI–CBSE-Chemistry, , Equilibrium, , Kw, Ka, , Now, K h =, , Kw, 10−14, =, = 1.51 × 10−9, K h 6.6 × 10−6, , ⇒ Ka =, , 63. Predict if the solutions of the following salts are neutral, acidic or basic:, NaCl, KBr, NaCN, NH4 NO3 , NaNO2 and KF, Solution:, (i) NaCl, , , , NaCl, , H 2O, , , , NaOH, , HCl, , Strong base, , Strong acid, , It is a salt of strong acid and strong base .Therefore it is a neutral solution. And its pH=7, (ii) KBr, , KBr, , , , H2O, , KOH, , , , HBr, , Strong base, , Strong acid, , It is a salt of strong acid and strong base .Therefore it is a neutral solution. And its pH=7, (iii) NaCN:, , NaCN, , , , H 2O, , HCN, , , , Weak acid, , NaOH, Strong base, , It is a salt of a weak acid and strong base. Therefore, it is a basic solution.its pH > 7, (iv) NH4 NO3, , , , NH 4 NO3, , H 2O, , , , NH 4 OH, Weak base, , HNO, , Strong acid, , It is a salt of a strong acid and weak base.Therefore, it is an acidic solution.its pH < 7, (v) NaNO2, , NaNO2, , , , H 2O, , , , NaOH, Strong base, , HNO 2, Weak acid, , It is a salt of a weak acid and strong base. Therefore, it is a basic solution.its pH > 7, 54, Practice more on Equilibrium, , www.embibe.com

Page 56 : Class–XI–CBSE-Chemistry, , Equilibrium, , = 0.740 × 10−11, x2, , Also, K h = 0.1 (where x is the concentration of OH − and ClCH2 COOH), 0.740 × 10, , −11, , x2, =, 0.1, , 0.074 × 10−11 = x 2, ⇒ x 2 = 0.74 × 10−12, x = 0.86 × 10−6, [OH− ] = 0.86 × 10−6, ∴ [H + ] =, =, , Kw, 0.86 × 10−6, , 10−14, 0.86 × 10−6, , [H + ] = 1.162 × 10−8, pH = − log[H + ], =-[log(1.162 × 10−8)], =-[log1.162-(8log10)], =log1.162+8log10, = 7.94, , 65. Ionic product of water at 310 K is 2.7 × 10−14 . What is the pH of neutral water at this, temperature?, Solution:, Ionic product,, K w = [H + ][OH− ], Let [H + ] = x, Since [H + ] = [OH− ], K w = x 2 ., ⇒ K w at 310K is 2.7 × 10−14, ∴ 2.7 × 10−14 = x 2, 56, Practice more on Equilibrium, , www.embibe.com

Page 57 : Class–XI–CBSE-Chemistry, , Equilibrium, , ⇒ x = 1.64 × 10−7, ⇒ [H + ] = 1.64 × 10−7, ⇒ pH = − log[H + ], = − log[1.64 × 10−7 ], = −[log[1.64 + log10−7 ], = -log1.64+7log10, = 6.78, Hence, the pH of neutral water is 6.78., , 66. Calculate 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 H2 SO4 + 10 mL of 0.01M Ca(OH)2, (c) 10 mL of 0.1M H2 SO4 + 10 mL of 0.1M KOH, , Solution:, (a) Moles of H3 O+ = molarity × volume (litre) =, Moles of OH− = molarity × volume (litre) =, Total volume=, , 10+25, 𝑙𝑖𝑡𝑒𝑟, 1000, , 25×0.1, 1000, , 10×0.2×2, 1000, , = 0.0025 mol, , = 0.0040 mol, , = 35 × 10−3 𝑙𝑖𝑡𝑒𝑟, , Thus, [OH− ] > [H + ], [OH− ] =, =, , Moles of OH − − Moles of H3 O+ 0.0040 − 0.1125, =, total volume, 35 × 10−3, , 0.0015, mol⁄L = 0.0428, 35 × 10−3, , pOH = − log[ OH− ] = - log(0.0428), = 1.36, pH = PK w − pOH = 14 − 1.36, = 12.63 (not matched), , 57, Practice more on Equilibrium, , www.embibe.com

Page 58 : Class–XI–CBSE-Chemistry, , (b) Moles of H3 O+ =, Moles of OH− =, , Equilibrium, , 2×10×0.01, 1000, , 2×10×.01, 1000, , = 0.0002 mol, , = 0.0002 mol, , Since there is neither an excess of H3 O+ or OH − so the pH of the solution =7, (c) Moles of H3 O+ =, Moles of OH− =, Total volume=, , 2×10×0.1, 1000, , 10×0.1, 1000, , = 0.002 mol, , = 0.001 mol, , 10+10, 𝑙𝑖𝑡𝑒𝑟, 1000, , = 20 × 10−3 𝑙𝑖𝑡𝑒𝑟, , H3 O+ > OH−, [H3 O+ ] =, , Moles of (OH − ) − Moles of (H3 O+ ) 0.002 − 0.001, 0.001, 10−3, =, =, =, total volume, 20 × 10−3, 20 × 10−3 20 × 10−3, = 0.05, , ∴ pH = − log(0.05), = 1.30, The solution is acidic., , 67. Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride, and mercurous iodide at 298K from their solubility product constant are 1.1 × 10−12 , 1.2 ×, 10−10, 1.0 × 10−38 ,1.6 × 10−5 ,4.5 × 10−29 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦 . Determine also the molarities of, individual ions., Solution:, (1) Silver chromate:, Ag2 CrO4 → 2Ag+ + CrO2−, 4, Then,, K sp = [Ag+ ]2 [CrO2−, 4 ], Let the solubility of Ag 2 CrO4 be s., ⇒ [Ag+ ] = 2s and [CrO2−, 4 ]=s, Then,, K sp = (2s)2 × s = 4s 3, 58, Practice more on Equilibrium, , www.embibe.com

Page 59 : Class–XI–CBSE-Chemistry, , Equilibrium, , ⇒ 1.1 × 10−12 = 4s 3, . 275 × 10−12 = s 3, s = 0.65 × 10−4 M, Molarity of Ag + = 2s = 2 × 0.65 × 10−4 = 1.30 × 10−4 M, −4, Molarity of CrO2−, 4 = s = 0.65 × 10 M, , (2) Barium chromate:, BaCrO4 → Ba2+ + CrO2−, 4, K sp = [Ba2+ ][CrO2−, 4 ], Let solubility of BaCrO4 be s., 2, So, [Ba2+ ] = s and [CrO2−, 4 ] = s ⇒ K SP = s, , ⇒ 1.2 × 10−10 = s 2, ⇒ s = 1.09 × 10−5 M, −5, Molarity of Ba+ = Molarity of CrO2−, 4 = s = 1.09 × 10 M, , (3) Ferric hydroxide:, Fe(OH)3 →Fe3+ + 3OH−, K sp = [Fe3+ ][OH− ]3, Let s be the solubility of Fe(OH)3 ., Thus,[Fe3+ ] = s and [OH − ] = 3s, ⇒ K SP = s × (3s)3, = s × 27s3, K SP = 27s 4, 1.0 × 10−38 = 27s 4, 1.0 × 10−38 = 27s 4, 0.037 × 10−38 = s 4, 0. 00037 × 10−36 = s 4 ⇒ 1.39 × 10−10 M = S, Molarity of Fe3+ = s = 1.39 × 10−10 M, Molarity of OH− = 3s = 4.17 × 10−10 M, , 59, Practice more on Equilibrium, , www.embibe.com

Page 61 : Class–XI–CBSE-Chemistry, , Equilibrium, , Solution:, Let s be the solubility of Ag 2 CrO4., Then,Ag 2 CrO4 ⇌ 2Ag + + CrO2−, 4, K sp = (2s)2 s = 4s 3, 1.1 × 10−12 = 4s 3, s = 6.5 × 10−5 M, Let s´ be the solubility of AgBr., AgBr ⇌ Ag+ + Br−, s ′2, , s′, , K sp = s ′2 = 5.0 × 10−13, ∴ s ′ = 7.07 × 10−7 M, Therefore, the ratio of the molarities of their saturated solution is, s, 6.5 × 10−5 M, =, = 91.9, s ′ 7.07 × 10−7 M, , 69. Equal 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 K sp = 7.4 × 10−8 )., Solution:, When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the, molar concentrations of both solutions are reduced to half i.e., 0.001 M. Then,, , NalO3  Na   IO30.00IM, 0.00IM, Cu  ClO3 2, 0.00lM, ,  Cu 2  2ClO3, 0.00IM, , Now, the solubility equilibrium for copper iodate can be written as:, Cu(IO3 )2 → Cu2+ (aq) + 2I𝑂3− (aq), Ionic product of copper iodate:, 2, = [Cu2+ ][IO−, 3], , = (0.001)(0.001)2, 61, Practice more on Equilibrium, , www.embibe.com

Page 65 : Class–XI–CBSE-Chemistry, , [𝑆 2− ] =, , Equilibrium, , 1.0 × 10−19 × 10, = 6.67 × 10−20 𝑀, 15, , [𝑀2+ ] =, , 0.04 × 5, = 1.33 × 10−2 𝑀, 15, , Ionic product = [𝑀2+ ][𝑆 2− ] = (1.33 × 10−2 )(6.67 × 10−20 ) = 8.87 × 10−22, This ionic product exceeds the 𝐾𝑠𝑝 of 𝑍𝑛𝑆 and 𝐶𝑑𝑆. Therefore, precipitation will occur in 𝐶𝑑𝐶𝑙2, and 𝑍𝑛𝐶𝑙2 solutions., , 65, Practice more on Equilibrium, , www.embibe.com