Answer:
[tex]\large \boxed{4.64}[/tex]
Explanation:
The equation for the buffer is
HC₇H₅O₂ + H₂O ⇌ H₃O⁺ + C₇H₅O₂⁻; Kₐ =6.3 × 10⁻⁵
Let's rewrite it as
HA + H₂O ⇌ H₃O⁺ + A⁻
1. Moles of sodium benzoate
[tex]\text{Moles} = \text{3.96 g} \times \dfrac{\text{1 mol}}{\text{144.10 g}} = \text{0.027 49 mol}[/tex]
2. Concentration of benzoate ion
[tex][\text{A}^{-}] = \dfrac{\text{0.027 49 mol}}{\text{1 L}} = \text{0.027 49 mol/L }[/tex]
3. Calculate pKₐ
[tex]\text{p}K_{\text{a}} = -\log(6.3 \times 10^{-5}) = 4.20[/tex]
4. Calculate the pH
We can use the Henderson-Hasselbalch equation to get the pH.
[tex]\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\& = & 4.20 +\log\dfrac{0.02749}{0.0100}\\\\& = & 4.20 + \log2.749 \\& = & 4.20 + 0.4390\\& = & 4.64\\\end{array}\\\text{The pH of the buffer is $\large \boxed{\mathbf{4.64}}$}[/tex]
