The Ka of a weak acid is 1.73 x 10-5. Assuming the "x is small" approximation is valid, what is the predicted pH of a 0.10 M solution of the weak acid?
Let the weak acid be written as HA
HA dissociates as:
HA -----> H+ + A-
0.1 0 0
0.1-x x x
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.73*10^-5)*0.1) = 1.315*10^-3
since c is much greater than x, our assumption is correct
so, x = 1.315*10^-3 M
so.[H+] = x = 1.315*10^-3 M
use:
pH = -log [H+]
= -log (1.315*10^-3)
= 2.88
Answer: 2.88
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