Computation of pH when an acid and base are mixed in solution

Question

I'm doing a basic chemistry course, and we are currently learning how to compute $\text{pH}$ from the acid dissociation constant (using $\left[\text{H}^{+}_{(\text{aq})}\right]=\sqrt{K_{a}\left[\text{HA}_{(\text{aq})}\right]}$) along with computing the $\text{pH}$ of strong bases by assuming full dissociation into $\text{OH}^{-}$ ions, and then using the ionic product of water to calculate the concentration of protons.

The question I have been given is:

Calculate the pH of the solution obtained when $14.9\text{ cm}^{3}$ of $0.100\text{ mol dm}^{-3}$ sodium hydroxide solution has been added to $25.0\text{ cm}^{3}$ solution of methanoic acid of concentration $0.100\text{ mol dm}^{-1}$ ($K_{a}=1.60\times 10^{-4}\text{ mol dm}^{-3}$)

I'm not sure how I should go about solving this problem, I can calculate concentrations of hydrogen and hydroxide ions for each of the solutions but I'm unsure how to combine them (as presumably some of the $\text{OH}^{-}$ ions will react with the $\text{H}^{+}$ ions to form $\text{H}_{2}\text{O}$?).

Thanks in advance!

Answer 1

You are headed in the right direction. Pretend that the $\ce{H+}$ ions and the $\ce{OH-}$ ions neutralize each other 1 for 1. That will leave you with only one of those ions left.

The water ionization won't affect your problem as that ionization is repressed by the excess ion present in the solution.

Now you should be able to solve the problem.