A 50mL of 0.50M hydrofluoric acid (Ka = 7.2 x 10^-4) is titrated with a 0.20M
sodium hydroxide solution. What is the pH of the solution when 35mL of base
has been added?
will the solution have a pH less that 7 or more than 7
Didn't I show you how to do one like this just a day or so ago. That one had a final solution that was a buffer and used the Henderson-Hasselbalch equation to calculate the pH.
In fact, I think it was the same problem.
To determine the pH of the solution when 35mL of 0.20M sodium hydroxide (NaOH) solution is added to 50mL of 0.50M hydrofluoric acid (HF), you need to calculate the moles of acid and base and use them to determine the remaining concentration of the acid and its associated pH.
1. Calculate the moles of acid:
Moles of HF = concentration (M) x volume (L)
Moles of HF = 0.50M x 0.050L
Moles of HF = 0.025 moles
2. Calculate the moles of base:
Moles of NaOH = concentration (M) x volume (L)
Moles of NaOH = 0.20M x 0.035L
Moles of NaOH = 0.007 moles
3. Determine the limiting reactant:
Since HF and NaOH react in a 1:1 ratio according to the balanced chemical equation: HF + NaOH → NaF + H2O, the moles of HF are in excess, and NaOH is the limiting reactant.
4. Calculate the remaining moles of HF:
Remaining moles of HF = initial moles of HF - moles of NaOH
Remaining moles of HF = 0.025 moles - 0.007 moles
Remaining moles of HF = 0.018 moles
5. Calculate the remaining concentration of HF:
Remaining concentration of HF = remaining moles of HF / volume of solution (L)
Remaining concentration of HF = 0.018 moles / 0.085L (total volume after addition)
Remaining concentration of HF = 0.212M
6. Calculate the pKa of HF:
pKa = -log(Ka)
pKa = -log(7.2 x 10^-4)
pKa ≈ 3.14
7. Calculate the pH of the remaining solution using the Henderson-Hasselbalch equation:
pH = pKa + log(Base/Acid)
pH = 3.14 + log(0.20M/0.212M)
pH ≈ 3.14 + log(0.943)
pH ≈ 3.14 - 0.026
pH ≈ 3.11
The pH of the solution after adding 35mL of 0.20M NaOH will be approximately 3.11. Since the pH is less than 7, the solution will be acidic.