Answer:
Here's what I get
Explanation:
1. Write the chemical equation
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻; Kₐ = 2 × 10⁻⁵
Let's rewrite the equation as
A⁻ + H₂O ⇌ HA + OH⁻
2. Calculate Kb
[tex]K_{\text{b}} = \dfrac{K_{\text{w}}}{K_{\text{a}}} = \dfrac{1.00 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10}[/tex]
3. Set up an ICE table
A⁻ + H₂O ⇌ HA + OH⁻
I/mol·L⁻¹: 0.35 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.35-x x x
4. Solve for x
[tex]\dfrac{\text{[HA ][OH$^{-}$]}}{\text{[A$^{-}$]}} = \dfrac{x^{2}}{0.35-x} = 5 \times 10^{-10}[/tex]
Check for negligibility,
[tex]\dfrac{\text{[HA]}}{K_{\text{b}}} = \dfrac{0.35}{5 \times 10^{-10}} = 7 \times 10^{8}> 400\\\\\therefore x \ll 0.35\\\\\dfrac{x^{2}}{0.35} = 5 \times 10^{-10}\\\\x^{2} = 0.35 \times 5 \times 10^{-10} = 1.8\times 10^{-10}\\\\x = \sqrt{1.8\times 10^{-10}} = \mathbf{1 \times 10^{-5}}[/tex]
5. Calculate the pOH
[OH⁻] = 1 × 10⁻⁵ mol·L⁻¹
pOH = -log[OH⁻] = -log(1 × 10⁻⁵) = 4.88
6. Calculate the pH.
pH + pOH = 14.00
pH + 4.88 = 14.00
pH = 9.12
Note: The answer differs from that given by Silberberg because you used only one significant figure for the Kₐ of acetic acid.
