Calculate the pH of a solution that results from mixing 30.5 mL of 0.155 M trimethylamine ((CH3)3N) with 23.2?

Question

Calculate the pH of a solution that results from mixing 30.5 mL of 0.155 M trimethylamine ((CH3)3N) with 23.2 mL of 0.177 M (CH3)3NHCl. The Kb value for (CH3)3N is 6.5 x 10-5.

Answer 1

0.155M (CH3)3N, is diluted by 30.5ml /53.7 ml total = 0.088035 molar

0.177 M (CH3)3NHCl, is diluted 23.2ml / 53.7 ml total = 0.07647 molar

(CH3)3N -->   (CH3)3NH+  & OH-

Kb = 6.5e-5 = [ (CH3)3NH+] [OH-] / [ (CH3)3N]

6.5e-5 = [ 0.07647] [OH-] / [ 0.088035]

[OH-] = 7.48e-5

pOH = 4.126

14 - pOH = pH of 9.87

your answer: 9.87