Calculate the pH of a solution that is 2.00 M HF, 1.00 M NaOH, and 0.704 M NaF. (Ka = 7.2 E -4)
You have two problems in one here.
First, the NaOH will neutralize part of the HF. Do that first and see how much HF is left. Then you will have a solution of NaCl, HF, and NaF. The NaCl is neutral and will not change the pH. The HF/NaF mixture is a buffered solution. Use the Henderson-Hasselbalch equation to solve for the pH. Post your work if you get stuck.
To calculate the pH of a solution that contains both a weak acid (HF) and its conjugate base (F-), you need to consider the dissociation of the weak acid and the hydrolysis of the conjugate base.
Here's how you can calculate the pH:
Step 1: Write the balanced equation for the dissociation of HF:
HF + H2O ⇌ H3O+ + F-
Step 2: Use the given concentrations of HF and NaF to determine the initial concentrations of HF, H3O+, and F-:
[H3O+] = 0 (since HF is a weak acid, we assume negligible dissociation at the beginning)
[F-] = 0.704 M (NaF is a strong electrolyte, so it completely dissociates)
[HF] = 2.00 M
Step 3: Set up the equilibrium expression for the dissociation of HF:
Ka = [H3O+][F-] / [HF]
Step 4: Substitute the initial concentrations into the equilibrium expression:
Ka = (x)(0.704) / (2.00 - x)
Step 5: Since x is the concentration of H3O+, which can be assumed to be small, we can approximate the term 2.00 - x as just 2.00. This is a reasonable approximation because the dissociation of HF will be relatively small compared to its initial concentration.
Step 6: Solve for x using the Ka value:
Ka = (x)(0.704) / (2.00)
Step 7: Rearrange the equation to solve for x:
x = Ka * 2.00 / 0.704
Step 8: Calculate x:
x = (7.2 E - 4) * 2.00 / 0.704
x = 0.00205 M
Step 9: Determine the pH using the concentration of H3O+:
pH = -log[H3O+]
pH = -log(0.00205)
pH ≈ 2.69
Therefore, the pH of the solution is approximately 2.69.