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 $\text{H}_{2}\text{O}$?)

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.

Answer 2

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.

Answer 3

Count the protons. Figure out the amount of pure methanoic acid. Consider the addition of the dilute acid to your mix as that amount of methanoic but include the water in the total volume. Consider it as pure water. Do the same with the lye i.e. find out how much pure lye and how much pure water are being added to the mix.

Guess an answer pH. Calculate the concentration of OH- and H+ at that pH and determine the number of protons that you would have to add (or subtract) to your volume of pure water to reach that pH. If your guessed pH is < 7 the H+ concentration will be higher and you will have to add protons to establish that level and neutralize the 10^-7 concentration at pH 7. Call this need for protons the "proton deficit". If the guess pH is > 7 the OH- level will go up and the H+ concentration goes down and the H+ concentration of 10^-7 at pH 7 represents a surfeit.

Assume the NaOH is completely dissociated. Thus you will need as many protons as there are OH- ions in the added NaOH to convert the OH- to water and the eqivalents of OH- become part of the proton deficit.

Now the tricky part. Use Henderson - Hallelbalch to compute the charge on the formic acid ion at your guess pH. To do this compute the dissociation ratio r = 10^(pH - pK). This is the ratio of dissociated molecules to undissociated ones. If there was 1 mole of formic acid to start with and x remain undissociated the rx must be dissociated and x/(x + rx) = 1/(1 + r) is the fraction dissociated. The original acid is undissociated so in being added to a solution which has a certain pH it dissociates supplying C/(1 + r) protons where C is the number of moles of formic acid. This is a proton surfeit (with respect to the guessed pH).

So, assuming the guessed pH is < 7 1. Protons are required to neutralize the OH- in the pure water (deficit) 2. Protons are required to establish an H+ concentration in the water of 10^-pH (deficit) 3. Protons are required to neutralize the OH- ions from the NaOH (deficit) 4. Protons are released by the methanoic acid (surfeit)

Since no protons are generated or destroyed in this reaction the total of the deficits must equal the total of the surfeits when the reaction is complete. All you need to do is keep guessing pH's until you find the one that zeroes the sum. The easiest way to do this is to put your calculations into an Excel spreadsheet and use the Solver to find the pH. If you don't know about Solver, learn about it! If you don't want to it's pretty easy to zero in on the correct answer pretty quickly manually.