Question

Calculate the acetate ion concentration in a solution prepared by dissolving 5.40×10^-3 mol of HCl(g) in 1.00 L of 1.30 M aqueous acetic acid (K_a = 1.80×10^-5).

Assume that the volume of the solution does not change upon dissolution of the HCl.

Answer 1

HCl disscoiates into H+ and Cl^- so the concentration of H+ ion obtained form HCl is 5.4 x 10^-3 / 1 L = 5.4 x 10^-3 M

Acetic acid dissociation:

K_a = [H+]*[OAc-] / [HOAc]

It starts with 1.3 M HOAc

HOAc decreases by X = 1.3 - X

OAc- increases by X = X

H+ increases by x = 5.4 x 10^-3 + X

Plug these into the equation for K_a :

K_a = 1.8 x 10^-5 = [5.4 x 10^-3 + X ] [X] / [1.3 - X]

2.34 x 10^-5 - 1.8 x 10^-5 X = 5.4 x 10^-3 X + X²

X² + 5.418 x10^-3 X - 2.34 x 10^-5 = 0

X = 0.002835 = 2.835 x 10^-3

H+ = 5.4 x 10^-3 + 2.835 x 10^-3 = 8.235 x 10^-3

pH = - log [ H+] = - log [8.235 x 10^-3 ] = 2.08

Concentration of the Acid = 2.08