Find the pH of a mixture of .150 M HF(aq) solution and 0.100 M HClO2(aq)
HF + H2O <--> F- + H3O+
Ka= 3.5*10^-4
So I will show my attempt below...
[HF] [F-] [H3O+]
I .150 0 0
C -x +x +x
E .15-x x x
Ka= [F-][H3O+]/[HF]
3.5*10^-4= x^2/(.150-x)
assume x is small so (.150-x) = .150
7.25*10^-3 = x^2 Take square root of both sides
x= 7.25*10^-3
[H3O+] = 7.25*10^-3
pH=-log[H3O+]=2.14
Is this right?? If it is, why don't we do any work with the .100M HClO2 soution?
Thanks!
Technically, there is no way to get a firm solution to the problem because the problem doesn't specify the mixture's composition. Your calculation is correct, as far as it goes, and your observation is correct also. According to my tables, HClO2 is a stronger acid than HF and the H^+ content will be dictated more by the HClO2 than by the HF.
Your calculation for the pH of the mixture of HF and HClO2 is correct. However, you haven't done any work with the 0.100 M HClO2 solution because it is a weak acid, and its contribution to the overall acidity of the solution is negligible compared to the strong acid HF.
In this case, HF is the stronger acid and HClO2 is the weaker acid. The Ka value for HF is 3.5 * 10^-4, which indicates that it ionizes more in water compared to HClO2.
When two weak acids are mixed, it is typical to assume that the stronger acid will largely dissociate, while the weaker acid will only partially dissociate. In your calculation, you assumed that HF completely dissociates to F- and H3O+ ions, while ignoring the contribution from HClO2.
Since HClO2 is a weaker acid than HF, its contribution to the concentration of H3O+ ions will be small. Hence, the calculation was simplified by considering only the strong acid, HF.
In summary, the negligible contribution of the weaker acid, HClO2, to the overall concentration of H3O+ ions in the mixture allows us to disregard its impact in the pH calculation.