Question

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`?

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.

Answer 1

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

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

Now, let's consider the equilibrium of acetate,

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

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`