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Exercise · Q31

Q.Calculate the pH\text{pH} of a 0.1 M0.1\ \text{M} solution of sodium acetate (CH3COONa\text{CH}_3\text{COONa}), given Ka(CH3COOH)=1.8×10−5K_a(\text{CH}_3\text{COOH}) = 1.8 \times 10^{-5} and Kw=1.0×10−14K_w = 1.0 \times 10^{-14}.

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CH3COONa\text{CH}_3\text{COONa} is the salt of the weak acid CH3COOH\text{CH}_3\text{COOH} and the strong base NaOH\text{NaOH}. Na+\text{Na}^+ does not hydrolyze, but CH3COO−\text{CH}_3\text{COO}^- does: CH3COO−+H2O⇌CH3COOH+OH−\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-, with Kb(CH3COO−)=Kw/Ka=(1.0×10−14)/(1.8×10−5)≈5.56×10−10K_b(\text{CH}_3\text{COO}^-) = K_w/K_a = (1.0\times10^{-14})/(1.8\times10^{-5}) \approx 5.56\times10^{-10} (found earlier). Treating this like a weak base ionization, [OH−]=KbC=(5.56×10−10)(0.1)=5.56×10−11≈7.45×10−6 M[\text{OH}^-] = \sqrt{K_b C} = \sqrt{(5.56\times10^{-10})(0.1)} = \sqrt{5.56\times10^{-11}} \approx 7.45\times10^{-6}\ \text{M}. $\text{p …

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