Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0113 M solution. The pH of the resulting solution is 2.33. Calculate the Ka for the acid.

Answer 1

http://chemwiki.ucdavis.edu/Physical_Chemistry/Acids_and_Bases/Ionization_Constants/Calculating_A_Ka_Value_From_A_Measured_Ph

Answer 2

this should help

Answer 3

pH= - log [H+] => [H+] = 10^(-pH)

\[inonization.percentage= \frac{ [H ^{+}] }{ C _{a} }\times 100\]

the % of dissociation of the acid is > 5% we can not assume the x is too small and state the equilibrium use the formula
\[Ka=\frac{ [A ^{-} ][H ^{+}]}{ C _{a}-[H ^{+}] }\] [A-]=[H+]
\[K _{a}= \frac{ [H ^{+}]^{2} }{ C _{a} -[H ^{+}]}\]