ka1 = (H^+)(HC2O4^-)/(H2C2O4)
ka2 = (H^+)(C2O4^2-)/(HC2O4^-)
You get (H3O^+) from pH.
You're given C2O4^2- and H2C2O4. Plug and chug.
Oxalic acid, found in the leaves of rhubarb and other plants, is a diprotic acid.
H2C2O4 + H2O ↔ H3O+ + HC2O4-
Ka1= ?
HC2O4- + H2O ↔ H3O+ + C2O42-
Ka2 = ?
An aqueous solution that is 1.05 M H2C2O4 has pH = 0.67. The free oxalate ion concentration in this solution is [C2O42-] = 5.3 x 10-5 M. Determine Ka1 and Ka2 for oxalic acid.
2 answers
Ka1= 0.6381
Ka2= 5.565 x 10-5
Ka2= 5.565 x 10-5
1 answer