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.

Question

australopithecus · Answer

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

australopithecus · Answer

this should help

cuanchi · Answer

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