CHEMISTRY HELP PLEASE: Acids and Bases question???

Question

The pH of an 0.018 M solution of weak acid A is 2.55. The pH of an 0.042 M solution of weak acid B is 3.06. The pH of an 0.089 M solution of weak acid C is 5.10. Identify the three acids from among those listed below. Find Ka values online.

A) HNO2
B) HF
C) HCHO2
D) HC7H5O2
E) HC2H3O2
F) H2S
G) HClO
H) H3BO3
I) HCN
J) None of the above

Answer 1

A) HNO2, Ka = ??
B) HF, pKa = 3.15
C) HCHO2, pKa = 3.744
D) HC7H5O2, pKa = 4.21
E) HC2H3O2, pKa = 4.76
F) H2S, pKa = 6.89
G) HClO, pKa = 7.497
H) H3BO3, pKa = 9.24
I) HCN, pKa = 9.25

General approach: [H+] = 10^-(pH)
Ka = [H+]*[A-]/[HA] = [H+]^2/[HA]
Thus pKa = 2*pH + log([HA]) = 2*pH + log([HA(original)]-[H+])

For acid A, pKa = 2*2.55+log(0.018-10^-2.55) = 3.28
Very likely HF

For acid B, pKa = 2*3.06+log(0.042-10^-3.06) = 4.73
Most likely acitic acid HC2H3O2

For acid C, pKa = 2*5.10+log(0.089-10^-5.10) = 9.15
Possibly either H3BO3 or HCN
Note:
Since pH is quite high, for more precise calculation, we may now try to evaluate the contribution from water self-dissociation.
As I said before, [H+] = 10^-pH = 10^-5.10 = 7.94x10^-6 (M). Based on Kw = 10^-14 = [H+]*[OH-], we know:
[OH-] = Kw/[H+] = 10^-14/7.94x10^-6 = 1.26x10^-9 (M).
That means [H+] contributed by water-self-dissociation is also 1.26x10^-9 (M). Compared to the total [H+] of 7.94x10^-6 M, this small portion of H+ of water self-dissociation can indeed be omitted.