[Latsabb]

The question is this:

A sample of 0.1276g of and unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0633M NaOH solution. The volume of the base required to reach the equivalence point was 18.4 mL. (a) Calculate the molar mass of the acid. (b) After 10.0 mL of base had been added to the titration, the pH was determined to be 5.87. What is the Ka of the unknown acid.

I started out by finding out how many moles of OH was added, which was 1.16472x10^-3. Since this is at the equivalence point, then the amount of OH must equal the amount of H. I then divided .1276 by 1.16472x10^-3 to determine the molar mass, which was 109.55 g/mol. But then I got to part (b) and things went down hill.

I started off by taking the pH of 5.87, and turning that into the concentration of H, which was 1.35x10^-6M. Unless I am mistaken, Ka can only be calculated at the equilibrium, and that concentration is not at equilibrium. I know the starting concentration of the acid, as that was 1.16472mol/25ml, or .0333M. I do no, however, understand how I am supposed to tie all of this together to get the Ka... I have the initial, and a partial change, not the actual change, so my attempts at an ICE table have also been unsuccessful. Can someone give me some hints? Thanks!

[Borek]

I started off by taking the pH of 5.87, and turning that into the concentration of H, which was 1.35x10^-6M. Unless I am mistaken, Ka can only be calculated at the equilibrium, and that concentration is not at equilibrium.

This solution IS at equilibrium. Think if you can use the information given (initial amount of HA, amount of base added) to calculate how much HA and A^- are present in the solution - note that the situation is dominated by the stoichiometry (in other words, you can ignore HA dissociation).

[Latsabb]

Ok, I am still a bit confused... I see now why it would be in equilibrium. So from here I should try to backtrack, and see how much H there originally was, based on the amount of H that I have now, and the amount of base added? (therefore showing how much there was before the base consumed the H) And then with that, I could figure that all the H that was there before the base was introduced came from the HA, and then subtract that amount from the original concentration of HA, and then just put it into a Ka equation? Or am I overlooking some steps?

[Borek]

Think about it this way: initially solution contained HA only (almost, dissociation produced some H⁺ and A^-, but their concentrations were much lower). Adding NaOH you neutralize HA producing A^-. You can safely assume concentrations of HA and A^- are those forced on the solution by the neutralization (that is, amount of A^- equals amount of NaOH added, amount of HA is initial amount of HA minus amount of NaOH added. These two concentrations are forced upon the solution, and H⁺ concentration changes to make teh solution reach equilibrium. So you know H⁺ (because you were given pH), you can calculate HA and A^- (as explained above), just plug all three numbers into Ka formula.

This is an approximation only, but it works quite good for weak acids.

[Latsabb]

Perfect, thank you. I see where I went wrong. I had the HA correct, and the H+ correct, but I had equal values of H+ and A- since I was basing it off of the balanced equation. But now that I pinned A- to HA, my value is correct. Thanks again!