A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M 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, we need to determine the amount of moles of HF in the initial solution:

mol HF = concentration x volume in liters
mol HF = 0.20 M x 0.100 L
mol HF = 0.020

Next, we need to determine how much KOH is required to completely neutralize the HF.

1 mol KOH reacts with 1 mol HF, so the amount of moles of KOH required is also 0.020.

Now, we need to determine how much KOH is added during the titration. We know that 100.0 mL of 0.10 M KOH is added, so we can determine the amount of moles of KOH added:

mol KOH = concentration x volume in liters
mol KOH = 0.10 M x 0.100 L
mol KOH = 0.010

Since this is less than the amount of KOH required to fully neutralize the HF, we know that there will be some KOH left over after the titration.

The reaction between HF and KOH produces KF and water:

HF + KOH -> KF + H2O

Since KOH is a strong base, it will react completely with water to produce OH- ions:

KOH + H2O -> K+ + OH- + H2O

The OH- ions from the excess KOH will react with the HF to produce F- ions and water:

HF + OH- -> F- + H2O

Since the initial solution contains 0.020 moles of HF and the amount of KOH added only neutralizes half of this, we know that there will be 0.010 moles of HF left over at the end of the reaction. This means that there will be 0.010 moles of F- ions produced.

We can use an ICE table to determine the concentration of the F- ions:

I: HF + OH- -> F- + H2O
C: -x + x -> x
E: 0.010 - x x x

We know that the equilibrium constant (Ka) for HF is 3.5 x 10^-4, so we can set up an expression for the equilibrium constant:

Ka = [F-][H3O+]/[HF]
3.5 x 10^-4 = x^2/(0.020 - x)

Since the initial concentration of HF is much larger than the concentration of F- at equilibrium, we can assume that the change in the concentration of HF is approximately equal to the initial concentration of HF. This allows us to simplify the expression:

3.5 x 10^-4 = x^2/0.020

Solving for x gives us:

x = 4.70 x 10^-3

To find the pH, we need to determine the concentration of H3O+:

[H3O+] = Ka[H2F+]/[HF]
[H3O+] = (3.5 x 10^-4)(4.70 x 10^-3)/(0.020 - 4.70 x 10^-3)
[H3O+] = 9.15 x 10^-4

pH = -log[H3O+]
pH = -log(9.15 x 10^-4)
pH = 3.46

Therefore, the answer is D) 3.46.

A 1000 ML Sample of 0