Question

What is the pH at the equivalence point in the titration of 0.20 L of 0.66 M HF with 0.33 M NaOH? (F- Kb = 1.4 x 10-11)

a. 7.00

b. 5.76

c. 8.24

d. 3.25

e. 10.75

Answer 1

NaOH + HF ----> Na+ + F- + H2O

At equivalence point all HF is converted to F-, NaF being salt of a strong base and a weak acid. It is completely dissociated.

So pH is controlled by F-

F- + H2O <-----> HF + OH-

Kb = [HF][OH-] / [F-]

use ICE table we can write: Kb = x^2 / (0.66-x)

we know HF is a weak acid and hence 0.66-x = 0.66.

x = ((1.4*10^-11)*0.66)^0.5 = (0.924*10^-11)^0.5 = 3.039*10^-6 M

Thus, pOH = -log(3.039*10^-6) = 5.5

pH = 14-5.5 = 8.5

We have used approximation at the point 0.66-x = 0.66.

Hence our answer should be 8.24.