Question

ook at sample problem 18.12 in the 8th ed Silberberg book. Write a balanced chemical equation (salt hydrolysis). So acetate ion + water goes to acetic acid + OH-. Write the Kb expression. Notice it is a Kb since hydroxide ion is a product. Ka is given for acetic acid so calculate the Kb for the acetate ion. Set up an ICE chart. Find x which represents hydroxide ion concentration, find pOH, convert to pH and enter that to 2 decimal places. What is the pH of 0.35 M sodium acetate solution at 25°C? The Ka of acetic acid is 2x10-5.

Answer 1

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

$K_{\text{b}} = \dfrac{K_{\text{w}}}{K_{\text{a}}} = \dfrac{1.00 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10}$

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

$\dfrac{\text{[HA ][OH ^{-} ]}}{\text{[A ^{-} ]}} = \dfrac{x^{2}}{0.35-x} = 5 \times 10^{-10}$

Check for negligibility,

$\dfrac{\text{[HA]}}{K_{\text{b}}} = \dfrac{0.35}{5 \times 10^{-10}} = 7 \times 10^{8}> 400\\\\\therefore x \ll 0.35\\\\\dfrac{x^{2}}{0.35} = 5 \times 10^{-10}\\\\x^{2} = 0.35 \times 5 \times 10^{-10} = 1.8\times 10^{-10}\\\\x = \sqrt{1.8\times 10^{-10}} = \mathbf{1 \times 10^{-5}}$

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.