calculate the concentration of acetate ions in a solution that is .150M CH3COOH and .05 M HI. Ka of CH3COOH is 1.75x10^-5

Md Ariful Islam

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V = 1L 

0.05M       X M

     HI   =   H+    +   I-

   1M        1M

[H+]HI = X = 0.05M.

Let Y moles of acetic acid be dissociated.

   Ymol            Ymol     Ymol

CH3COOH ⇄ CH3COO- + H+

   1mol            1mol    1mol

Ka = [H+][CH3COO-]/[CH3COOH] = ([H+]HI + [H+]CH3COOH) ×[CH3COO-]/[CH3COOH] = (0.05 + Y) × Y / (0.150 – Y) = 1.75×10-5.

0.05Y + Y2 = 0.2625×10-5 – 1.75×10-5Y

Y2 is very small, it can be neglected.

0.05Y + 1.75×10-5Y = 0.2625×10-5

Y = 0.2625×10-5 / (0.05+1.75×10-5) = 5.2482×10-5 ≈ 5.25×10-5.

[CH3COOH] = Y = 5.25×10-5M.

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