Question

(12) Consider the titration of 25.0 mL, of 0.060 M Na CO, with 0.1000 M HCI. Determine the pH of the titration solution after the addition of 10.00 mL, 15.00mL, and 30.00 mL of HCl. Kbl n 2.0 x 10", K62 = 2.51 x 10-8)

Answer 1

No. of moles of Co^2-_3 = 25 times 0.06 = 1.5 a) No. of moles of HCl = 10 times 0.1 = 1.0 Co^2-_3 + HCl rightarrow HCo^-_3 1.5 1.0 0 0.5 0 1.0 PH= Pka_2 + log [salt]/[acid] because Pka_2 = -log ka_2 Ka_2 = K_w/K_B_2 = 1 times 10^-14/2.51 times 10^-8 rightdoublearrow K_a_2 = 0.398 times 10^-6 Pka_2=-log(0.398 times 10^-6) = 6.40 rightdoublearrow P^H = 6.40 + log [0.5]/[1.0] P^H= 6.40 - 0.30 = 6.10 b) No. of moels of HCl = 15 times 0.1 = 1.5 Co^2-_3 + Hcl rightarrow HCO^-_3 1.5 1.5 0 0 0 1.5 rightdoublearrow P^H = PKa_1+PKa_2/2 P^H=10.30+6.40/2=13.5 because PK_a_1 = log Ka_1 = - log(K_w/K_b) =-log(1 times 10^-14/2 times 10^-4) = - log(0.5 times 10^-10) = 10.30 c) No. of moles of HCl = 30 times 0.1 = 3 CO^2-_3 + Hcl rightarrow HCO^-_3 1.5 3.0 0 0 1.5 1.5 Here strong acid remains. [Hcl]=1.5/30+25 = 0.0273 M PH = -log(0.0273)= 1.56