what volume of 0.120 M NaOH must be added to 100 ml of 0.100 M NaHC2O4 to reach ph of 4.70?

H2C2O4 Ka1 = 5.9 x 10-2

HC2O4^- Ka2 = 5.1 x 10-5

The answer is 60 ml but I don't know how to find it.
Can you please show me the steps and how to do it please

3 answers

Stoichiometry and Henderson-Hasselbalch equation. You have 100 mL of 0.1M NaHC2O4 = 10 millimols sodium bioxalate

..HC2O4^- + OH^- ==> C2O4^2- + H2O
I..10........0........0
add..........x..............
C..-x.......-x........x
E..10-x......0........x

Plug the E line into the HH equation.pKa2 is -log Ka2
4.70 = 4.29 + log (10-x)/x
Solve for x. I get approximately 7.2
Then M = millimols/mL
You know M of the NaOH is 0.120 M, millimols is approx 7.2, solve for mL. I ran through the calculations and obtained 59.99 which rounds to 60 mL. Viola!

I have not understood