What is the pH of a solution prepared by mixing 55.0 mL of 0.183 M $KOH$ and 70.0 mL of 0.145 M $HC_2H_3O_2$ ?

Question

A solution is prepared by mixing:
$55.0mL$ of $0.183M$ $KOH$
$70.0mL$ of $0.145M$ $CH_3COOH$

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.

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Answer 1 · 2018-05-28T00:24:27

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

$CH_3COOH + OH^(-) to CH_3COO^(-) + H_2O$

puu.sh puu.sh

Now, let's consider the equilibrium of acetate,

$CH_3COO^(-) rightleftharpoons CH_3COOH + OH^(-)$

puu.sh puu.sh

where,

$K_"b" = ([CH_3COOH][OH^-])/([CH_3COO^-]) approx 5.6*10^-10$

Let's derive the equilibrium concentration of hydroxide ions,

$=> K_"b" = (x^2)/(1.01*10^-2 - x) =5.6*10^-10$

$=> x = [OH^-]_"eq" approx 2.38*10^-6"M"$

Hence,

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