Question

The `pH` of a `0.01*mol*L^-1` solution of `"formic acid"` was `2.4`; what is `K_a` for this acid?

Answer 1

`K_a=2.63xx10^-3`

Explanation:

We interrogate the equilibrium.....

`HCO_2H(aq) + H_2O(l)rightleftharpoonsHCO_2^(-) + H_3O^+`

And so `K_a=([HCO_2^-][H_3O^+])/([HCO_2H])`

Now we are given that the `pH` of the solution was `2.4`, and thus by definition, `[H_3O^+]=10^(-2.4)*mol*L^-1`. But given the reaction `[H_3O^+]=[HCO_2^(-)]=10^(-2.4)*mol*L^-1`. We ASSUME that the STARTING concentration of `HCO_2H` was `0.01*mol*L^-1`.

And thus...... `K_a=(10^(-2.4))^2/(0.01-10^(-2.4))=2.63xx10^-3`

`pK_a=-log_10K_a=-log_10(2.63xx10^-3)=2.58`

This site lists the `pK_a` of formic acid as `3.77` so we are an order of magnitude out on the literature.