Calculate the expected pH of a 50.0 ml solution of .150 M acetic acid that is be

Question

Calculate the expected pH of a 50.0 ml solution of .150 M acetic acid that is being titrated with a .200 M solution of NaOH at any point in region II of the titration. Ka of CH3COOH=1.8 x 10^-5

Choose any appropriate volume of .200M NaOH to use. Use the appropriate method to solve the problem.

Explanation / Answer

Ka = 1.8 x10^ -5 CH3COOH ---> CH3COO- + H+ 0.15 -x x x ka = [H+]{CH3COO-]/[CH3COOH] = 1.8 x10^ -5 = (x^2)/(0.15-x) x= [H+] = 0.00163 let we chose 10 ml of 0.2 M NaOH so [OH] = (0.2 x10/1000) = 0.002 net [OH] after neutralizing acid = 0.002-0.00163 = 0.00037 pOH = -log(0.00037) = 3.432 pH = 14-3.432 = 10.567

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