Determine the pH of the solution.

Question

Determine the pH (to two decimal places) of the solution that is produced by mixing 1.21 mL of 3.65×10^-3 M HCl with 69.3 mL of 9.43×10^-3 M KH.

Answer 1

HCl + H2O ==>H3O+ + Cl-

KH + H2O ==> KOH + H2

Final volume after mixing = 1.21 ml + 69.3 ml = 70.51 ml = 0.0705 L

[H⁺] from HCl = (1.21 ml)(3.65x10^-3 M) = (7 0.51 ml)(x M) and x = 6.26x10^-5 M H⁺

[OH^-] from KH = (69.3 ml)(9.43x10-3 M) = (70.51 ml)(x M) and x = 0.00927 M OH^-

[OH^-] in excess = 0.00927 M - 6.26x10^-5 M = 0.00921 M OH^-

pOH = -log [OH^-] = -log 0.00921

pOH = 2.04

pH = 14 - pOH

pH = 11.96

Answer 2

C(H⁺) = (1.21x3.65×10^-3 + 69.3x9.43×10^-3) / 70.51 = 9.33×10^-3 M.

So pH = - log C(H⁺) = 2.03.