What is the pH of a solution prepared by mixing 55.0 mL of 0.183 M KOHKOH and 70.0 mL of 0.145 M HC_2H_3O_2HC2H3O2?

A solution is prepared by mixing:
55.0mL55.0mL of 0.183M0.183M KOHKOH
70.0mL70.0mL of 0.145M0.145M CH_3COOHCH3COOH

Calculate the pH.

The answer is approximately 3.2, but I am far off nearly every solve. Most of these problems are strong-strong, but this one is strong-weak.

1 Answer
Al E.
May 28, 2018

Let's assume that the hydroxide ions fully dissociate and neutralize one equivalent of acetic acid,

CH_3COOH + OH^(-) to CH_3COO^(-) + H_2OCH3COOH+OH−→CH3COO−+H2O

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Now, let's consider the equilibrium of acetate,

CH_3COO^(-) rightleftharpoons CH_3COOH + OH^(-)CH3COO−⇌CH3COOH+OH−

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where,

K_"b" = ([CH_3COOH][OH^-])/([CH_3COO^-]) approx 5.6*10^-10Kb=[CH3COOH][OH−][CH3COO−]≈5.6⋅10−10

Let's derive the equilibrium concentration of hydroxide ions,

=> K_"b" = (x^2)/(1.01*10^-2 - x) =5.6*10^-10⇒Kb=x21.01⋅10−2−x=5.6⋅10−10

=> x = [OH^-]_"eq" approx 2.38*10^-6"M"⇒x=[OH−]eq≈2.38⋅10−6M

Hence,

"pH" = 14 + log[OH^-]_"eq" approx 8.38pH=14+log[OH−]eq≈8.38

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