Question

Calculate the [H_3O^+] and pH of the following polyprotic acid solution 0.340 M H_2C_2O_4. Calculate the pH of this solution

Answer 1

C)

H2C2O4 <--> H+ + HC2O4- Ka1 = 5.9*10^-2

HC2O4- <--> H+ + C2O4- Ka2 = 6.4*10^-5

note that there is no need to include the second ionization, sicne 1000*Ka2 < Ka1

Ka1 = [H+][HA-]/[H2A]

5.9*10^-2 = x*x/(0.34-x)

x = [H+] = 0.115

include 2nd ionization

Ka2 = [H+][A-2]/[HA-]

6.4*10^-5 = (x+y)(y)/(0.115-y)

solve for y

6.4*10^-5 = (0.115+y)(y)/(0.115)

0.115y+ y^2 = 0.115*(6.4*10^-5)

y^2 + 0.115- 7.36*10^-6 = 0

y = 6.39*10^-5

[H+] = 0.115+y =0.115+ 6.39*10^-5 = 0.11506 (as expected)

[H3O+] = [H+] = 0.11506 M

D)

pH = - log(0.11506 ) = 0.93907