If the Ka of a monoprotic weak acid is 83 x106 what is the p

If the Ka of a monoprotic weak acid is 8.3 x106, what is the pH of a 0.38 M solution of this acid? Number

Solution

HA dissociates as:

HA          ----->     H+   + A-
0.38                 0         0
0.38-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 ((8.3*10^-6)*0.38) = 1.776*10^-3

since c is much greater than x, our assumption is correct
so, x = 1.776*10^-3 M



So, [H+] = x = 1.776*10^-3 M


use:
pH = -log [H+]
= -log (1.776*10^-3)
= 2.7506
Answer: 2.75