Frequently Asked Question

Here's the pH calculation for each point in the titration of HF with NaOH:
1. Initial pH
HF is a weak acid, so we need to use the Ka expression to find the initial pH.
HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq)
Ka = [H3O+][F-]/[HF] = 6.6 x 10^-4
Let x = [H3O+] = [F-], then [HF] = 0.3 - x
6.6 x 10^-4 = x^2 / (0.3 - x)
Since Ka is small, we can approximate 0.3 - x ≈ 0.3
6.6 x 10^-4 = x^2 / 0.3
x^2 = 1.98 x 10^-4
x = [H3O+] = 0.0141 M
Therefore, pH = -log[H3O+] = -log(0.0141) = 1.85
2. After adding 10mL of NaOH
The moles of HF initially present are:
(25 mL)(0.3 M) = 7.5 mmol HF
The moles of NaOH added are:
(10 mL)(0.3 M) = 3 mmol NaOH
The reaction consumes 3 mmol of HF, leaving:
7.5 mmol HF - 3 mmol NaOH = 4.5 mmol HF
This reaction also produces 3 mmol of F-. We now have a buffer solution.
The volume of the solution is 25 mL + 10 mL = 35 mL
[HF] = (4.5 mmol) / (35 mL) = 0.129 M
[F-] = (3 mmol) / (35 mL) = 0.086 M
Using the Henderson-Hasselbalch equation:
pH = pKa + log([F-]/[HF])
pH = -log(6.6 x 10^-4) + log(0.086/0.129)
pH = 3.18
3. After adding 12.50mL of NaOH
The moles of NaOH added are:
(12.50 mL)(0.3 M) = 3.75 mmol NaOH
This is exactly half the amount of HF initially present, which means we are at the half-equivalence point. At this point, pH = pKa.
Therefore, pH = -log(6.6 x 10^-4) = 3.18
4. After adding 25mL of NaOH
This is the equivalence point. All of the HF has reacted with NaOH, and we have a solution of the conjugate base, F-. To find the pH, we need to consider the hydrolysis of F-:
F-(aq) + H2O(l) ⇌ HF(aq) + OH-(aq)
Kb = Kw/Ka = 1.0 x 10^-14 / 6.6 x 10^-4 = 1.52 x 10^-11
[F-] = (7.5 mmol) / (50 mL) = 0.15 M
Let x = [OH-] = [HF]
Kb = [HF][OH-]/[F-] = x^2 / (0.15 - x)
Since Kb is small, we can approximate 0.15 - x ≈ 0.15
1.52 x 10^-11 = x^2 / 0.15
x^2 = 2.28 x 10^-12
x = [OH-] = 1.51 x 10^-6 M
pOH = -log[OH-] = -log(1.51 x 10^-6) = 5.82
Therefore, pH = 14 - pOH = 8.18
5. After adding 26mL of NaOH
We are now past the equivalence point. The excess NaOH will determine the pH.
The moles of NaOH added are:
(26 mL)(0.3 M) = 7.8 mmol NaOH
The moles of NaOH in excess are:
7.8 mmol NaOH - 7.5 mmol HF = 0.3 mmol NaOH
The total volume of the solution is 25 mL + 26 mL = 51 mL
[OH-] = (0.3 mmol) / (51 mL) = 0.0059 M
pOH = -log[OH-] = -log(0.0059) = 2.23
Therefore, pH = 14 - pOH = 11.77