Answer:
The pH of the solution following the addition of 0.075 moles of NaOH is 5.7.
Explanation:
The equation of the buffer solution is the following:
C₄H₇O₂H(aq) + H₂O(l) ⇄ C₄H₇O₂⁻(aq) + H₃O⁺(aq) (1)
The pH of the buffer solution can be found using the Henderson-Hasselbalch equation:
[tex] pH = pKa + log(\frac{[NaC_{4}H_{7}O_{2}]}{[C_{4}H_{8}O_{2}]}) [/tex] (2)
The NaOH added will react with butanoic acid tot produce sodium butanoate:
C₄H₇O₂H(aq) + OH⁻(aq) ⇄ C₄H₇O₂⁻(aq) + H₂O(l) (3)
[tex] n_{C_{4}H_{7}O_{2}H}_{i} = C*V = 0.107 M*1.30 L = 0.139 moles [/tex]
[tex] n_{C_{4}H_{7}O_{2}^{-}} = 0.345 M*1.30 L = 0.449 moles [/tex]
[tex] n_{NaOH} = 0.075 moles [/tex]
[tex] n_{C_{4}H_{7}O_{2}H} = 0.139 moles - 0.075 moles = 0.064 moles [/tex]
[tex] n_{C_{4}H_{7}O_{2}^{-}} = 0.449 moles + 0.075 moles = 0.524 moles [/tex]
[tex]C_{C_{4}H_{7}O_{2}H} = \frac{0.064 moles}{1.30 L} = 0.049 M[/tex]
[tex] C_{C_{4}H_{7}O_{2}^{-}} = \frac{0.524 moles}{1.30 L} = 0.403 M [/tex]
Now, from equation (1) we have:
C₄H₇O₂H(aq) + H₂O(l) ⇄ C₄H₇O₂⁻(aq) + H₃O⁺(aq)
0.049 - x 0.403 + x x
[tex] Ka = \frac{[C_{4}H_{7}O_{2}^{-}][H_{3}O^{+}]}{C_{4}H_{7}O_{2}H} [/tex]
[tex]1.52 \cdot 10^{-5} = \frac{(0.403 + x)x}{0.049 - x}[/tex]
[tex]1.52 \cdot 10^{-5}*(0.049 - x) - (0.403 + x)x = 0[/tex]
By solving the above equation for x we have:
x = 1.85x10⁻⁶ = [H₃O⁺]
So, the concentration of butanoic acid and sodium butanoate is:
[tex][C_{4}H_{7}O_{2}^{-}] = 0.345 + 1.85 \cdot 10^{-6} = 0.345 M[/tex]
[tex][C_{4}H_{7}O_{2}H] = 0.049 - 1.85 \cdot 10^{-6} = 0.049 M[/tex]
Finally, from equation (2) we have:
[tex] pH = pKa + log(\frac{[NaC_{4}H_{7}O_{2}]}{[C_{4}H_{8}O_{2}]}) [/tex]
[tex]pH = -log(1.52 \cdot 10^{-5}) + log(\frac{0.345}{0.049}) = 5.7[/tex]
Therefore, the pH of the solution following the addition of 0.075 moles of NaOH is 5.7.
I hope it helps you!
