Question : ka1 and ka2 values of H2X is given 1*(10)^-5 M and 1*(10)^-9 M.50cm^3 of 0.2 M KOH and 50 cm^3 of H2X of unknown concentration is mixed together and the pH of the final solution is given as 7.

Question

1)Find the concentration of H2X

2)Find the pH of the initial H2X solution(before being mixed with KOH)?

Work:
I took the concentration of H2X as c(moldm-3)

As KOH is a strong base I considered that it reacts completely with H2X.

When they are mixed together their volumes get double so their concentrations become half of the initial values.

H2X(aq) + KOH(aq) -----> KHX(aq) + H2O(l)

After the reaction the concentrations of H2X and KHX are respectively [ (C/2) - 0.1)] and (0.1) M

When applied to ka1 equation we get the C= 1*(10)^-3 M

2) as ka2 is much smaller than ka1 we can neglect ka2.

H2X <=====> H+ + HX- (all are in aqueous form)

After ionization , by applying values in ka1 equation again,we get pH of the initial solution as 4.

Am I correct in the process?

Answer 1

Follow along and confirm my conclusions. My rationale is as follows:
millimols KOH = 50 x 0.2 = 10; therefore, any H2X < 10 mmols will leave excess (xs) KOH in the solution and that will be more basic. So we must have at least 10 mmols H2X. Using exactly 10 mmols H2X will produce 10 mmols KHX and will leave neither an xs KOH nor an xs of H2X and the pH will be pH = (pKa1 + pKa2)/2 = (5.0+9.0)/2 = 7.0 and that's what you want. You will note that such a calculation of the pH that way has no c term in it so it can't be used to solve for c. Using the Henderson-Hasselbalch equation won't work either (you can try it if you wish) because leaving an xs H2X along with the KHX formed makes the solution more acid (actually <5) which is not what you want. So we want the (H2X) to be 0.2M (that's 10 mmols/50 mL = 0.2 M) and that number sounds much more reasonable to me than the 0.001 or 0.002 we discussed last evening. At 3 A.M. this morning I just wasn't thinking very clearly.
So that answers 1. I think you can do 2 on your own but my answer is about pH of 2.8 or so.
For your follow up question to Emily, most of the comments I made early this morning are OK. Any HCl added to the KHX will titrate the KHX to the methyl orange end point and will form KCl and H2X. Please feel free to respond with any questions or comments.

Answer 2

Thank you very much!
Now it sounds more reasonable!

And I have another question regarding this.

If we titrate the final solution with NaOH of same concentration what are the changes which should be done in this experiment?

Answer 3

Thank you very much!
Now it sounds more reasonable!

And I have another question regarding this.

If we titrate the final solution with NaOH of same concentration as HClmentioned in the last part what are the changes which should be done in this experiment?

Answer 4

I'm at a loss since I don't know what experiment, the details, etc. If the "final" solution is the KHX, adding NaOH will titrate the H and leave K2X as if more NaOH were added in the first place as in
H2X + 2KOH ==> K2X + 2H2O

Answer 5

So what are the changes which should be made,if we titrate the final solution with NaOH,not with HCl?
Change of indicators or molarities etc.