Answer:
[tex]\boxed{\text{D. 0.35 mol/L}}[/tex]
Explanation:
The equilibrium reaction is
CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺; Kₐ = 1.8 × 10⁻⁵
Let's rewrite the equation as
HA + H₂O ⇌ A⁻ + H₃O⁺
We can use the Henderson-Hasselbalch equation to calculate the concentrations in the buffer.
[tex]\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\ \\4.80 & = & -\log \left(1.8 \times 10^{-5}\right) + \log\dfrac{0.40}{\text{[HA]}} \\\\4.80 & = & 4.74 + \log\dfrac{0.40}{\text{[HA]}}\\\\0.06 & = & \log\dfrac{0.40}{\text{[HA]}}\\\\1.14 & = & \dfrac{0.40}{\text{[HA]}}\\\\1.14\text{[HA]} & = & 0.40\\\\\text{[HA]} & = & \dfrac{0.40}{1.14}\\\\ & = & \mathbf{0.35}\\\end{array}\\[/tex]
[tex]\text{The concentration of acetic acid must be $\boxed{\textbf{0.35 mol/L}}$}[/tex]
