Question

What is the pH of a solution that is prepared by adding 103 mole of Na2CO; to 1 liter of water (for closed system)? The temperature 15°C. (See Tables 4-2 and 4-7 below for equilibrium constants.) Compare this value with that obtained if the temperature is 25°C. TABLE 4-2 Ion Product of Water pH of a "neutral" solution K. pKw 15 20 25 30 40 0.12 x 10- 0.45 x 10-1 0.68 x 10- 1.01 x 10-14 1.47 x 10-4 2.95 x 10-14 14.93 14.35 14.17 14.00 13.83 13.53 7.47 7.18 7.08 7.00 6.92 6.76 TABLE 4-7 Temperature Dependence of Some Important Carbonate Equilibrium Constants Temperature, C 5 10 15 20 25 40 60 Reaction 1.20 1.27 1.34 1.41 1.47 1.64 1.8 6.52 6.46 6.42 6.38 6.35 6.30 6.30 10.56 10.49 10.43 10.38 10.33 10.22 10.14 8.09 8.15 8.22 8.28 8.34 8.51 8.74 -2.47 2.34 -2.21 2.10 1.891.711.40 5. CaCO3(s) + H+ ? Ca2+ + HCO,-;

Answer 1

CO_3^2- + H_2O(l) equilibrium HCO_3^- + OH^-, k_b_1 = K_w/K_a_2 k_b_1 = 0.45 times 10^- 14/10^- 10.43 = 1.21 times 10^- 4 at T = 15 degree C k_b_1 = 1.01 times 10^- 14/10^- 10.33 = 2.16 times 10^- 4 at T = 25 degree C since there is high difference between ka_1, and ka_2, so will be kb_1 and kb_2 of CO_3^2-. So, OH^- will mainly come from above equilibrium and kb_2 equilibrium contribution to OH^- can be ignored. Kb_1 = n^2/10^- 5 - x At T = 15 degree C, 1.21 times 10^- 4 = x^2/10^- 5 - x x^2 + 1.21 times 10^- 4 - 1.21 times 10^- 9 = 0 x = -1.21 times 10^- 4 + squareroot (1.21 times 10^- 4)^2 + 4 times 1.21 times 10^- 9/2 = 9.29 times 106 - 6 = [OH^-]