Question

Calculate the pH of a 0.67M oxalic acid solution and calculate the pKb of the resulting ionic base that will form. (Look up Ka of Oxalic acid)

Please show all work as I would like to learn how to do the problem and not just have the answer. Thank you.

Answer 1

Oxalic acid = 0.67M

                              H2C2O4   -------------------- HC2O4- + H+

Initial                                0.67                                     0                0

change                               -x                                      +x                +x

equilibrium                    0.67 -x                                +x                    +x

Ka of oxalic acid= 5.9x10^-2

                     Ka= [HC2O4-][H+]/[H2C2O4]

                      5.9x10^-2 = x*x/(0.67-x)

for solving the equation

x= 0.171

[H+]=0.171M

-log[H+]= -log(0.171)

PH= 0.767

PH= 0.78

KaxKb= Kw                   where Kw= ionic product of water = 1.0x10^-14

Kb= Kw/Ka = 1.0x10^-14/5.9x10^-2= 1.69 x10^-13

Kb= 1.69x10^-13

-log(Kb) = 1.69x10^-13

PKb= -log( 1.69x10^-13)

PKb=12.77