Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0136 M solution. The pH of the resulting solution is 2.45. Calculate the Ka for the acid.

Answer 1

HA ---> H+ + A-

[H+] = 10^(-2.45) = 3.54 x 10^-3 M
At equilibrium, the concentration of H+ and A- are the same. Originally your reactant's concentration was 0.0136 M, but at equilibrium, you know that 3.54 x 10^-3 M of products are formed, so your concentration of HA = 0.0136 M - 3.54 x 10^-3 M = 0.0101 M.
\[K_c = \frac {[H^+][A^-]}{[HA]} = \frac {3.54 \times 10^{-3 } M \times 3.54 \times 10^{-3 } M}{0.0101 M} = 1.25 \times 10^{-3}\]

Answer 2

Its supposed to be Ka, not Kc, although they're the same in this situation...