Can you please show me how to do this problem step by step?

Question

What is the pH of a 0.50 M solution of ascorbic acid?

vitamin C is H2C6H6O6

Kal = 7.9 x 10^-5, Ka2 = 1.6 x 10^-12

Παναγιωτης · Answer

H2C6H6O6<--->H{+} + HC6H6O6{-}

0,50-y................y.............y

Ka1=[H+][HC6H6O6-]/[H2C6H6O6] (1)

HC6H6O6{-}<--->H{+} + C6H6O6{2-}

y-z........................z............z

Ka2=[H+][C6H6O6(2-)]/[HC6H6O6(-)] (2)

H2O<--->H{+} + OH{-}

Kw=[H+][OH-] (3)

balance charge:[H+]=[OH-]+[HC6H6O6-]+2[C6H6O6(2-)] (4)

balance mass:C=[HC6H6O6-]+[C6H6O6(2-)]+[H2C6H6O6] (5)

Ka1 is larger than Ka2 meaning second stage of ionization is insignificant

also in normal concentrations [OH-] is insignificant

so you have (4)--->:[H+]=HC6H6O6-]=y--->

Ka1=[H+]^2/C-y--->Ka1=y^2/C-y

if you consider C>>y (*) then you have--->Ka1=y^2/C--->y=√(Ka1xC)--->

y=[H+]=6,28x10^-3 M--->pH=-log(6,28x10^-3)--->pH=2,20

(*) if you don't then you solve quadratic equation

Jan · Answer

pH = 0.5(pKa1 - lg(0.5)) = 2.2