Take the 2-minute tour ×
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

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!

share|improve this question
add comment

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
add comment

The trick is: after the addition, a buffer solution is formed. Find the concentrations of the salt and acid and use Henderson's equation to arrive at the answer.

share|improve this answer
This is the start of a good answer. Perhaps you could show how the Henderson-Hasselbach equation can be applied to problems like this one. –  Ben Norris Jun 9 '13 at 0:18
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.