Question

a) Calculate the acetate ion concentration in a solution prepared by dissolving 7.90×10^-3 mol of HCl(g) in 1.00 L of 9.00×10^-1 M aqueous acetic acid (K_a = 1.80×10^-5).

_____ mol/L
1pts

b) Calculate the pH of the above solution.

Give your answer to two decimal places.

_____

Answer 1

(a) Since HCl is strong acid then HCl dissociates completely;

Given moles of HCl=7.9x10^-3 mol and volume=1 L.

HCl -----> H^+ + Cl^-

the [H+] from HCl would be=mol/volume

[H^+]=7.9x10^-3 moles/1.00 L = 7.9x10^-3 M.

Given [CH3COOH]=9x10^-1 M=0.9 M. And Ka=1.8x10^-5

Since it is a weak acid then

CH3COOH <--------> CH3COO^- + [H^+]

Initial 0.9 0 . 7.9x10^-3

Change . - x . +x . +x

Equilibrium . 0.9-x. x. 7.9x10^-3+x

(Since here the solution initillin have some [H^+] from HCl.

Ka = [H+][CH3COO^-]/[CH3COOH] = 1.80 x 10^-5

Ka = 1.80x10^-5 = (x)(7.9x10^-3 + x)/(0.9- x)

1.62x10^-5 - (1.8x10^-5 x)=(7.9x10^-3 x+x^2)

x^2+7.882x10^-3x-1.62x10^-5=0

After solving this quadratic equation, x=0.001692

Therefore concentration of acetate ion=0.001692 M.

[H^+]=7.9x10^-3+x=0.0079+0.001692=0.009592 M.

(b)

Since pH = -log[H+]

pH=-log(9.592 x 10^-3) = 2.018

Therefore the pH of the given solution=2.02.

Please let me know if you have any doubt. Thanks and I hope you like it.