Answer :
The pH of the resulting buffer solution is approximately 4.32.
To solve this problem, we first need to determine the moles of weak acid and moles of NaOH added.
Then, we calculate the concentration of the weak acid and its conjugate base after the reaction with NaOH. Finally, we use the Henderson-Hasselbalch equation to find the pH.
1. Calculate the moles of weak acid (HA) and NaOH added:
- Moles of HA = volume (L) × concentration (mol/L)
- Moles of HA = 0.595 L × 0.250 mol/L = 0.14875 mol HA
- Moles of NaOH = volume (L) × concentration (mol/L)
- Moles of NaOH = 0.500 L × 0.160 mol/L = 0.080 mol NaOH
2. Determine the limiting reactant and the excess reactant:
- Since NaOH is a strong base, it will react completely with HA. Therefore, NaOH is the limiting reactant, and HA is in excess.
3. Calculate the moles of the resulting buffer components:
- Moles of A- (conjugate base of HA) = moles of NaOH added = 0.080 mol
- Moles of HA remaining = initial moles of HA - moles of NaOH reacted
- Moles of HA remaining = 0.14875 mol - 0.080 mol = 0.06875 mol
4. Calculate the concentrations of HA and A- in the resulting buffer solution:
- Concentration of HA = moles of HA remaining / total volume (L)
- Concentration of HA = 0.06875 mol / (0.595 L + 0.500 L) = 0.06875 mol / 1.095 L ≈ 0.0628 M
- Concentration of A- = moles of A- / total volume (L)
- Concentration of A- = 0.080 mol / 1.095 L ≈ 0.0731 M
5. Use the Henderson-Hasselbalch equation to calculate the pH of the buffer:
- pH = pKa + log([A-]/[HA])
- pH = -log(Ka) + log([A-]/[HA])
- pH = -log(1.69 × 10^-5) + log(0.0731/0.0628)
- pH = -(-4.77) + log(1.16)
- pH = 4.77 + 0.064 = 4.32
Therefore, the pH of the resulting buffer solution is approximately 4.32.
