A 150.0 mL sample of 0.20 M HF is titrated with 0.10 M LiOH. Determine the pH of the solution after the addition of 600.0 mL of LiOH. The Ka of HF is 6.8 × 10-4.

pH = 12.6

Explanation:

The HF reacts with LiOH as follows:

HF + LiOH → LiF + H₂O

To solve this question we need to find the moles of each reactant:

Moles HF:

0.1500L * (0.20mol / L) = 0.030 moles HF

Moles LiOH:

0.600L * (0.10mol / L) = 0.060 moles LiOH

That means there is an amount of LiOH in excess, that is:

0.060 mol - 0.030 mol = 0.030 moles LiOH

In 600.0mL + 150.0mL = 750.0mL = 0.750L

The molarity of LiOH is:

0.030 moles LiOH / 0.750L =

0.040M LiOH = [OH⁻]

As:

Kw = 1x10⁻¹⁴ = [H⁺] [OH⁻]

1x10⁻¹⁴ = [H⁺] [0.040M]

2.5x10⁻¹³M = [H⁺]

As pH = -log [H⁺]

c, liquids evaporate, creating a gas