A certain weak acid HA has Ka=2.4 x 10-2. Calculate the pH of 0.150 M HA solutio

Question

A certain weak acid HA has Ka=2.4 x 10-2. Calculate the pH of 0.150 M HA solution.

Possible answers:

a. 1.31
b. 1.22
c. 1.62
d. 8.54 (obviously not since its a weak ACID solution)
e. 5.38

I've already tried solving this multiple ways. I tried using the quadratic formula with a=1, b = 2.4 x 10-2 and c= 3.6 x 10-3 but that doesn't seem possible. I would just assume the answer was E based on the fact that that pH is consistent with a weak acid pH. You cannot neglect x because [initial]/Ka is < 1000. Can someone please explain this step by step? Don't know why I'm struggling so much!!

Explanation / Answer

HA----< H+ + A- , at equ [HA] = 0.15-x, [H+]=[A-] = x , Ka = [H+][A-]/[HA],

2.4 x10^ -2 = x^2/(0.15-x) , solving we get x = 0.049= [H+] ,

pH = -log[H+] = -log(0..049) = 1.31 , hence answer is option a

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