Chemistry reaction (NAOH + HF --> NAF + H2O)

Question

Experts, i'll need some help over this question please:

Calculate the pH of a solution obtained by adding 4.0 g of NaOH (s) to 1.0 L of 0.1 M HF solution (ignore volume changes).

A. 13.0

B. 9.8

C. 8.1

D. 7.5

E. 22.1

Answer 1

NaOH + HF ==> NaF + H2O

mols HF = 1 L x 0.1 mol/L = 0.1 mols HF

mols NaOH = 4.0 g x 1 mol / 40 g = 0.1 mols NaOH

After the reaction, no NaOH and No HF remain (equivalence point), and we have only 0.1 mol NaF in 1 L.

If you prefer to use an ICE table, it looks like this

NaOH + HF ==> NaF + H2O

0.1mol.....0.1mol.....0.............Initial

-0.1.........-0.1..........+0.1........Change

0................0............0.1.........Equilibrium

So, to determine pH of solution, we must consider the hydrolysis of NaF, or simply of F^-

F^- + H2O ==> HF + OH^-

F^- is acting as a base in the above hydrolysis reaction, so we need the Kb for F^-

Looking up the Ka for HF, I find it to be 7.2x10^-4

KaKb = 1x10^-14

Kb = 1x10^-14 / Ka = 1x10^-14 / 7.2x10^-4

Kb for F- = 3.70x10^-11

Kb = 3.70x10^-11 = [HF][OH^-] / [F^-]

3.70x10^-11 = (x)(x) / 0.1

x² = 3.70x10^-12

x = 1.92x10^-6 M = [OH-]

pOH = -log 1.9x10^-6 = 5.72

pH = 14 - 5.72

pH = 8.28 (close to 8.1 option C) ... difference may be due to rounding or math error but most probably is due to use of a different value for the Kb)