what is the ph of a 1.8 m solution of a weak acid ha, with a ka of 9.55×10−10? the equilibrium expression is: ha(aq) h2o(l)⇋h3o

Question

2 years ago

12

(aq) a−(aq)

1 answer:

sesenic [268] · Answer 1 · 2023-10-14T12:08:54+03:00

The pH of the solution is 4.38.

The first thing that we must do is to set up the ICE table for the reaction as shown;

Given that;

HA = 1.8 M

HA(aq) + H2O(l) ⇔ H3O^+(aq) + A^-(aq)

I 1.8 0 0

C - x +x +x

E 1.8 - x x x

K = [H3O^+] [A^-]/[HA]

9.55 × 10^−10 = x^2/0.0158 - x

9.55 × 10^−10(1.8 - x ) = x^2

1.719 * 10^-9 - 9.55 * 10^-10x = x^2

x^2 + 9.55 * 10^-10x - 1.719 * 10^-9 = 0

x=0.000041 M

pH = -log(0.000041 M)

pH = 4.38

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