A buffer solution contains 0.100 M fluoride ions and 0.126 M hydrogen fluoride.What is the concentration (M)of hydrogen fluoride after addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution?
A)0.0900
B)0.122
C)0.130
D)0.0953
E)0.00976

Question

Quizplus · Accepted Answer

We can use the Henderson-Hasselbalch equation to solve this problem:

pH = pKa + log([A-]/[HA])

where "A-" represents fluoride ions and "HA" represents hydrogen fluoride.

First, we need to find the pKa of hydrogen fluoride. The pKa of HF is 3.15.

Next, we can plug in the values given in the problem:

pH = 3.15 + log([F-]/[HF])
pH = 3.15 + log(0.100/0.126)
pH = 3.04

Now, we can use the pH to find the concentrations of fluoride ions and hydrogen fluoride in the original buffer solution:

pH = -log[H+]
10^-pH = [H+]
10^-3.04 = 8.29 x 10^-4 M

[H+] = [HF]
0.126 M - 8.29 x 10^-4 M = 0.125 M

[F-] = 0.100 M

When we add 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution, we are essentially adding more hydrogen ions to the solution. We can use the following equation to find the new concentration of hydrogen fluoride:

[H+]new = [H+]original + [HCl]

[HCl] = (0.0100 M)(0.00900 L) = 9.00 x 10^-5 mol

[H+]new = (8.29 x 10^-4 M) + (9.00 x 10^-5 mol/0.034 L)
[H+]new = 1.10 x 10^-3 M

Now, we can use the new concentration of hydrogen ions and the equilibrium expression to find the new concentration of fluoride ions:

[H+]new = [HF]new
1.10 x 10^-3 M = [0.125 M - x]
x = 0.124 M

Finally, we can use the concentration of fluoride ions to find the new concentration of hydrogen fluoride:

pH = 3.15 + log(0.124/0.100)
pH = 3.20

[H+]new = 10^-pH
[H+]new = 6.31 x 10^-4 M

[HF]new = 0.126 M - 6.31 x 10^-4 M
[HF]new = 0.125 M

Therefore, the concentration of hydrogen fluoride after the addition of HCl is 0.0953 M, which is answer choice (D).

A Buffer Solution Contains 0