If a 100 mL sample of sodium oxalate (Na2OOCCOO (aq)) has a pH of 9.23 at 25.0 °C, the initial concentration of the sample of sodium ?
sodium oxalate is?
- hcbiochemLv 710 months agoFavourite answer
Oxalic acid has two ionizable carboxyl groups. The Ka's for the two are 5.4X10^-2 and 5.4X10^-5. In a solution of sodium oxalate, only the second of these will be involved in a basic reaction. That reaction is:
C2O42- + H2O <--> HC2O4- + OH-
The expression for this Kb is:
Kb = [HC2O4-][OH-]/[C2O42-]
The Kb for that group will be:
Kb = 1X10^-14 / 5.4X10^-5 = 1.85X10^-10
Now, in that solution, [OH-] = [HC2O4-]
[OH-] can be calculated from the pH as:
pOH = 14 - 9.23 = 4.77
[OH-] = 10^-pOH = 1.7X10^-5 M = [HC2O4-]. So:
Kb = 1.85X10^-10 = (1.7X10^-5)^2 / [C2O42-]
[C2O42-] = 1.6 M = initial concentration of sodium oxalate