A 100.0 mL sample of 0.20 mol L^-1 HF is titrated with 0.10 mol L^-1 KOH. Determine the pH of the solution after the addition of 100.0 mL of KOH. The K_a of HF is 3.5 × 10^-4. A) 2.08 B) 3.15 C) 4.33 D) 3.46 E) 4.15

Question

Quizplus · Accepted Answer

First, let's calculate the amount of moles of HF present in the initial solution:

moles of HF = (0.20 mol/L) x (0.100 L) = 0.0200 mol

When the KOH is added, it will react with the HF to form water and KF:

HF + KOH → KF + H2O

The balanced chemical equation shows a 1:1 stoichiometry between HF and KOH, meaning that all the moles of KOH added will react with the moles of HF present. Therefore, the amount of moles of KOH added is also 0.0200 mol.

Now we need to calculate the concentration of the resulting KF solution:

moles of KF = 0.020 mol
volume of KF solution = (100.0 mL + 100.0 mL) / 1000 mL/L = 0.200 L

concentration of KF = moles of KF / volume of KF solution = 0.100 mol/L

To calculate the pH, we need to use the acid dissociation constant (Ka) of HF:

Ka = [H+][F-] / [HF]

We know the value of Ka (3.5 × 10-4) and the initial concentration of HF (0.20 mol/L). At equilibrium, the concentration of HF will be decreased by the amount that reacted with the KOH, but the concentration of F- and H+ will increase due to the formation of KF and H2O. Let's define x as the amount of moles of HF that reacts with the KOH, and use the stoichiometry of the balanced chemical equation to calculate the concentrations of F- and H+:

HF + KOH → KF + H2O

Initial:	0.0200 mol	0 mol	0 mol	0 mol
Change:	-x mol	-x mol	+x mol	+x mol
Equilibrium:	0.0200 - x mol	0 - x mol	+ x mol	+ x mol

Substituting these values into the equation for Ka:

Ka = [H+][F-] / [HF]
3.5 × 10-4 = x2 / (0.0200 - x)

This is a quadratic equation that can be solved for x using the quadratic formula:

x = [HF]eq = 2.4 × 10-3 mol

Since we now know the concentration of H+ at equilibrium, we can calculate the pH:

pH = -log[H+]
pH = -log(2.4 × 10-3)
pH = 2.62

However, we need to remember that the question asks for the pH after the addition of 100.0 mL of KOH, which has a concentration of 0.10 mol/L. Using the stoichiometry of the balanced chemical equation, we can calculate the final concentration of KF and convert it to moles:

KF concentration = 0.100 mol/L
moles of KF = (0.100 mol/L) x (0.100 L) = 0.010 mol

This means that the initial 0.0200 mol of HF is completely consumed, and we have a remaining 0.010 mol of KF. We can use the concentration of KF to calculate the concentration of F- and the concentration of H+ using the equilibrium values in the table:

[F-] = 0.030 mol/L
[H+] = 0.030 mol/L

Now we can calculate the pH:

pH = -log[H+]
pH = -log(0.030)
pH = 3.46

Therefore, the answer is D) 3.46.

A 1000 ML Sample of 0