Answer to Question #87340 in Organic Chemistry for Kendall Howe

Question

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

Answer 1

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

K_a = [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.