pH value of a 0. 011 m naf solution is 8.57.
Solution
This problem uses the relationship between Kb and the dissociation constants which is expressed as Kw = KaKb. Calculations are as follows:
Kb = KaKb
1.00 x 10^-14 = 7.2 x 10^-4(x)
x = 1.39 x 10^-11
We now need to calculate the [OH¯] using the Kb expression:
1.39 x 10^-11 = x^2 / (0.30 - x)
The denominator can be neglected. Thus, x is 3.73 x 10^-6.
pOH = -log 3.73 x 10^-6 = 5.43
pH = 14-5.43 = 8.57
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