Enough of a monoprotic acid is dissolved in water to produce a 0.0116 M solution. The pH of the resulting solution is 2.35. Calculate the Ka for the acid.

Question

ramirocruzo · Answer

Convert pH 2.68 to (H^+) with
pH = -log(H^+).
Then HA --> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
Substitute for (H^+) and (A^-). For (HA) substitute 0.0169-(H^+)

rushwr · Answer

monoprotic acid is an acid that can donate only one proton. So this acid is like HCl. 
$$HCl \rightarrow H ^{+} + Cl ^{-}$$
We know the concentration of HCl solution.
$$K _{a} = \frac{ [H ^{+}][Cl ^{-}] }{ [HCl] }$$
$$[H ^{+}] = [Cl ^{-}]$$
We can find H^+ concentration using pH
$$pH = -\log_{10} [H ^{+}]$$
HCl concentration is 0.0116M
So $$K _{a}= \frac{ [H ^{+}]^{2} }{ [HCl] }$$