Question

12. Calculate the pH of a buffer solution made from 0.30 M hydrofluoric acid and 0.70 M sodium fluoride after the addition of 0.08 mol of NaOH to 1.0 L of this solution.  Assume no change in volume.  The Ka for HF is 3.5 • 10-4.

Can someone guide me through this question? The answer is below but I can figure out how they got this answer.

Answer: pH = 4.01

Answer 1

Consider the dissociation of hydrofluoric acid, HF as below:

HF (aq) <=====> H⁺ (aq) + F^- (aq)

K_a = [H⁺][F^-]/[HF] = 3.5*10^-4

===> pK_a = -log K_a = -log (3.5*10^-4) = 3.45593 (I will keep a few guard digits extra)

Consider the neutralization reaction. We are given the moles of NaOH added as 0.08 mol. The volume of the solution is 1.0 L.

Moles HF = (molarity of HF)*(volume of solution) = (0.30 mol/L)*(1.0 L) = 0.30 mol.

Moles NaF = (molarity of NaF)*(volume of solution) = (0.70 mol/L)*(1.0 L) = 0.70 mol.

Set up the ICE chart (in terms of number of moles of reactants and products).

HF (aq) + NaOH (aq) --------> NaF (aq) + H₂O (l)

initial                      0.30          0.08                          0.70

change                 -0.08          -0.08                       +0.08

equilibrium            0.22           0                               0.78

[HF]_eq = (moles HF)_eq/(volume of solution) = (0.22 mol)/(1.0 L) = 0.22 M

[NaF]_eq = (moles NaF)_eq/(volume of solution) = (0.78 mol)/(1.0 L) = 0.78 M

The pH of the solution can be found by the Henderson-Hasslebach equation as

pH = pK_a + log [NaF]/[HF] = 3.45593 + log (0.78 M)/(0.22 M) = 3.45593 + log (3.54545) = 3.45593 + 0.54967 = 4.0056 ≈ 4.01 (rounding off to two sig. figures).

Ans: The pH of the buffer solution is 4.01 .