What is the ph of a 1.288 m solution of sodium acetate? ka of acetic acid is 1.8*10^-5. (2 decimal places)

The pH of the solution is 9.43

Explanation:

sodium acetate = CH3COONa

after removing the spectator ion Na

CH3COO⁻ + H2O ⇄ CH3COOH + H3O⁺

since sodium acetate is a weak base, it will not react completely with water. Thus a value "x" will be used to make the products.

At equilibrium:

[CH3COO⁻] = 1.288 - x

[CH3COOH] = x

[H3O⁺] = x

Since a base is reacting with H2O, Kb is used at equilibrium: Kb=(1*10^-14)/(1.8*10^-5) = 5.6*10^-10)

Kb = [CH3COOH][H3O⁺]/[CH3COO⁻]

Kb = (x²)/(1.288-x) since the Kb is significantly lower than 1.288 we can make a mathematical assumption that x is significantly lower than 1.288.

5.6*10⁻¹⁰ = x²/1.288

x = [OH⁻] = 2.675*10⁻⁵ M

pOH = -log[OH⁻] = 4.57

pH=14-pOH = 14 - 4.57 = 9.43

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