Question

If 20 mL of 0.10 M KOH is added to 40 mL of 0.050 M HCOOH, what is the pH of the mixed solution at 25°C? Ka for HCOOH is 1.8 × 10^–4 at 25°C.

Answer 1

Consider the equilibrium,

`HCOOH rightleftharpoons H^(+) + HCOO^(-)`

Moreover, let's assume that the hydroxide ions fully deprotonate one molar equivalent of formic acid.

Let's use a BCA table to organize the calculations,

`HCOOH +OH^(-) rightleftharpoons H_2O + HCOO^(-)`

It fully consumes the formic acid and it leaves us with only formate to equilibrate according to,

`HCOO^(-) rightleftharpoons HCOOH + OH^(-)`

where,

`K_"b" = ([HCOOH][OH^-])/([HCOO^-]) approx K_"w"/K_"a" approx 5.6*10^-11`

According to an ICE table,

`K_"b" = x^2/(0.002 - x) = 5.6*10^-11`

`=> x= [OH^-]_"eq" approx 3.35*10^-7`

and by extension,

`"pH" = 14 + log[OH^-] approx 7.52`