Answer:
pH = 5.493
Explanation:
It is possible to find pH of a buffer using H-H equation:
pH = pKa + log₁₀ [A⁻] / [HA]
Where [A⁻] is concentration of sodium butanoate and [HA] is concentration of butanoic acid.
pKa is -log Ka. As Ka of butanoic acid is 1.52×10⁻⁵, pKa is 4.818
Before reaction, moles of butanoic acid and sodium butanoate are:
butanoic acid: 1.50L × (0.118mol / L) = 0.177 moles butanoic acid
sodium butanoate: 1.50L × (0.318mol / L) = 0.477 moles sodium butanoate
NaOH reacts with butanoic acid thus:
NaOH + butanoic acid → sodium butanoate + water.
Butanoic acid decreases concentration and sodium butanoate increases concentration
Thus, after reaction, moles of butanoic acid and sodium butanoate are:
butanoic acid: 0.177mol - 0.063mol = 0.114mol
sodium butanoate: 0.477mol + 0.063mol = 0.540mol
As total volume is 1.50L, concentrations are:
[A⁻] [sodium butanoate] = 0.540mol / 1.50L = 0.360M
[HA] [butanoic acid] = 0.114mol / 1.50L = 0.076M
Replacing in H-H equation:
pH = 4.818 + log₁₀ [0.360M] / [0.076M]
pH = 5.493
The pH of the 1.50 L buffer solution consisting of 0.118 M butanoic acid and 0.318 M sodium butanoate after the addition of 0.063 moles of NaOH is 5.49.
The pH of the solution can be calculated with the Henderson-Hasselbalch equation:
[tex]pH = pKa + log(\frac{[C_{3}H_{7}COONa]}{[C_{3}H_{7}COOH]})[/tex] (1)
Where:
[C₃H₇COOH]: is the butanoic acid concentration
[C₃H₇COONa]: is the sodium butanoate concentration
The pKa is given by:
[tex] pKa = -log(Ka) = - log(1.52 \cdot 10^{-5}) = 4.818 [/tex]
After the addition of NaOH, we have:
C₃H₇COOH + NaOH ⇄ C₃H₇COONa + H₂O (2)
The addition of NaOH will reduce the acid concentration ([C₃H₇COOH]) and will increase the salt concentration ([C₃H₇COONa]).
To find the concentrations C₃H₇COOH and C₃H₇COONa after the reaction, we need to calculate the initial number of moles of C₃H₇COOH and C₃H₇COONa.
[tex] n_{{C_{3}H_{7}COOH}_{i}} = C__{{C_{3}H_{7}COOH}_{i}}*V = 0.118 mol/L*1.50 L = 0.177 \:moles [/tex]
[tex] n_{{C_{3}H_{7}COONa}_{i}} = C__{{C_{3}H_{7}COONa}_{i}}*V = 0.318 mol/L*1.50 L = 0.477 \:moles [/tex]
After the addition of 0.063 moles of NaOH, the number of moles of C₃H₇COOH and C₃H₇COONa in solution is:
[tex] n_{{C_{3}H_{7}COOH}_{f}} = n_{{C_{3}H_{7}COOH}_{i}} - n_{NaOH} = (0.177 - 0.063) \:moles = 0.114 \:moles [/tex]
[tex] n_{{C_{3}H_{7}COONa}_{f}} = n_{{C_{3}H_{7}COONa}_{i}} + n_{NaOH} = (0.477 + 0.063) \:moles = 0.540 \: moles [/tex]
Now, since the volume of the solution does not change after the addition of NaOH, the concentrations of C₃H₇COOH and C₃H₇COONa is:
[tex][C_{3}H_{7}COOH] = \frac{n_{{C_{3}H_{7}COOH}_{f}}}{V} = \frac{0.114 \:moles}{1.50 L} = 0.076 mol/L[/tex]
[tex][C_{3}H_{7}COONa] = \frac{n_{{C_{3}H_{7}COONa}_{f}}}{V} = \frac{0.540 \:moles}{1.50 L} = 0.360 mol/L[/tex]
Finally, the pH of the solution is (eq 1):
[tex]pH = pKa + log(\frac{[C_{3}H_{7}COONa]}{[C_{3}H_{7}COOH]}) = 4.818 + log(\frac{0.360}{0.076}) = 5.49[/tex]
The initial pH of the buffer solution (before the addition of NaOH) is:
[tex]pH_{i} = pKa + log(\frac{[C_{3}H_{7}COONa]_{i}}{[C_{3}H_{7}COOH]_{i}}) = 4.818 + log(\frac{0.318}{0.118}) = 5.25[/tex]
We can see that the addition of NaOH produces a slight increase in the pH of the initial solution.
Therefore, the pH of the solution following the addition of NaOH is 5.49.
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