Answer:
5.30
Explanation:
The pH of a buffer given by the Henderson-Hasselbach equation can be expressed as:
[tex]pH = pKa + log \dfrac{[salt]}{[Acid]}[/tex]
Given that:
Volume of the solution = 100.0 mL
To liters; the volume of the solution be 0.1 L
The concentration of [tex]C_4H_9COONa[/tex] = 0.20 M
The concentration of [tex]C_4H_9COONa[/tex] = molarity × volume
The concentration of [tex]C_4H_9COONa[/tex] = 0.20 mol/L × 0.1 L
The concentration of [tex]C_4H_9COONa[/tex] = 0.02 mol
The concentration of [tex]C_4H_9COOH[/tex] = 0.20 M
The concentration of [tex]C_4H_9COOH[/tex] = 0.20 mol/L × 0.1 L
The concentration of [tex]C_4H_9COOH[/tex] = 0.02 mol
However, the number of moles of NaOH added = 0.01 moles
Now; The ICE table can be computed as:
C₄H₉COOH + OH⁻ ⇄ C₄H₉COO⁻ + H₂O
Initial 0.02 0.01 0.02
Change - 0.01 -0.01 +0.01 +0.01
Equilibrium 0.01 - 0.03 0.01
Recall that the pH of [tex]C_4H_9COOH[/tex] = 4.82
∴
[tex]pH = 4.82 + log \begin {pmatrix} \dfrac{\dfrac{0.03}{0.1} }{ \dfrac{0.01}{0.1} } \end {pmatrix}[/tex]
[tex]pH = 4.82 + log \begin {pmatrix} \dfrac{0.3 }{0.1 } \end {pmatrix}[/tex]
pH = 4.82 + log ( 3 )
pH = 4.82 + 0.4771
pH = 5.2971
pH ≅ 5.30
