## anonymous one year ago 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.

• This Question is Open
1. RamiroCruzo

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^+)

2. Rushwr

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 $K _{a}= \frac{ [H ^{+}]^{2} }{ [HCl] }$