If the ka of a monoprotic weak acid is 1.2 × 10-6, what is the ph of a 0.40 m solution of this acid?

First, we write out a balanced equation.
HA <--> H(+) + A(-)

Next, we create an ICE table

        HA     <-->      H+     +    A-
[]i     0.40M            0M           0M
Δ[]    -x                   +x            +x
[]f     0.40-x             x              x

Next, we write out the Ka expression.

Ka = [H+][A-]/[HA]

Ka = x*x/(0.40-x) 

However, because Ka is less than 10^-3, we can assume the amount of dissociation is negligible. Thus,

Assume 0.40-x ≈ 0.40

Therefore, 1.2x10^-6 = x^2/0.40

Then we solve for the [H+] concentration, or x



x=6.93x10^-4

Next, to find pH we do 

pH = -log[H+]

pH = -log[6.93x10^-4]

pH = 3.2

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