Question

A 0.100 M solution of a monoprotic weak acid has a pH of 3.00. Calculate the pKa

Answer 1

use
pH = -log [H+]
3 = -log [H+]
[H+] = 1*10^-3 M
HA dissociates as:

HA          ----->     H+   + A-
0.1                 0         0
0.1-x               x         x


Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 1*10^-3*1*10^-3/(0.1-1*10^-3)
Ka = 1.01*10^-5

use
pKa = -log Ka
= -log (1.01*10^-5)
= 5.00
Answer: 5.00

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