Question

Question:
A solution of H2X of unknown concentration is given.
pH of the solution is 4.

ka1=2*(10)^-6 M

ka2=1*(10)^-11 M
Find the concentrations of [H2X] ,[HX-] and X^2- in the solution.

My thoughts on the question:

The temperature which the solution is kept is not given.So we don't know the kW value.So from what I know we can't consider hydrolysis as kh=kW/(ka1/ka2) ,we can't calculate the constant of hydrolysis.

So can we neglect ka2 and continue calculations?
(Considering pH=4 so [H+]=(10)^-4 M , and all those H+ ions come from H2X being dissociting to H+ and HX-?)



ka1=[H+]^2/[H2X]
So [H2X]=((10)^-4)^2)/(2*(10)^-6)=5*(10)^-3 M

Answer 1

I always assume that if T is not given that we assume Kw = 1E-14. Yes, you know (H^+) = 1E-4 and since ka2 is so much smaller than ka1 you ignore the H^+ from ka2. And you're right that H2X dissociats to H^+ and HX^- but you're incorrect after that.
Let Y = (H2X)
.......H2X ==> H^+ + HX^-
I.......Y......0.....0
C......-x......x.....x
E......Y-x.....x.....x

So you know x - 1E-4 and you plug in the numbers to Ka1.
Ka1 = (1E-4)(1E-4)/(Y-1E-4) = ? and solve for Y. That gives you (H^+) and (HX^-) and H2X. That leaves only X^2-. You get that from Ka2.
Ka2 = (H^+)(X^2-)/(HX^-)
But from Ka1 calculation, you know (H^+) = (HX^-). (H^+) in the numerator cancels with (HX^-) in the denominator of Ka2 and leave (X^2-) = Ka2.