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.

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

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

Check for negligibility,

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.

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the answers would be

the first one and the second one