Question

There is an unknown amount of unlabelled monoprotic acid in an unknown amount of water titrated with a sample with a solution of NaOH of unknown molarity. After adding 10.0 mL of NaOH, the pH=5.0. The equivalence point is 32.22 mL of NaOH. What is the Ka?

I have been looking at this problem for quite a while now, and I am not sure where to begin.

Answer 1

Try this.
HA + NaOH ==> NaA + H2O

Ka = (H^+)(A^-)/(HA)
How many millimoles A- are formed when 10 mL of the base have been added? That will be 10 mL x M(whatever that is) and the concn is (10*M/V) where V is the total volume in liters.
How many millimoles of the HA will be left to be titrated? Since the equivalence point is reached at 32.22 mL of the base, then there are 22.22 x M millimoles of the acid not yet titrated and the concn of HA at that point is (22.22M/V). And the pH is 5.0. Substitute all of that into the Ka expression above to obtain.
Ka = (1E-5)(10M/V)/(22.22M/V)
M cancels, V cancels, and you can solve for Ka, the only unknown. Take a look at the value for Ka and see if that is a reasonable value. It looks ok to me.

Answer 2

Oh, ya know