What is pH when 0.2M acetic acid (Ka = 1.8 times 10^-5) is allowed to go...

Question

Question

Answer 1

Answer #1

lets construct the ICE table

AcH (aq) + H2O (l) <===> H3O+ + Ac-

I 0.2 0 0

C -x +x +x

E 0.2-x +x +x

Ka = [H3O+][Ac-] / [AcH]

1.8 x 10^-5 = [x][x] / [0.2-x]

x² + x 1.8 * 10^-5 - 3.6 * 10^-6 = 0

solve the quadratic equation

x = 0.0188 M = [H3O+]

pH = -log[0.0188]

pH = 1.72

Answer 2

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