A #40.16*mL# volume of sodium hydroxide solution with concentration of #0.1693*g*mol^-1# reached an endpoint with a quantity of oxalic acid. If the oxalic acid was initially dissolved in a #30.00*mL# what was the concentration of oxalic acid?

1 Answer
Nov 23, 2015

Answer:

Concentration of oxalic acid #=# #0.1133# #mol*L^-1#.

Explanation:

#HO(O=)C-C(=O)OH + 2NaOHrarr{C(=O)O}_2^(2-)Na_2^+ + 2H_2O#

Moles of #NaOH# for equivalence #=# #40.16xx10^-3Lxx0.1693# #mol*L^-1 =# #6.80xx10^-3# #mol#.

Thus there were #3.40xx10^-3# #mol# oxalic acid (because each equiv acid contributes 2 equiv #H^+#).

Since this quantity was dissolved in a #30.00# #mL# volume, concentration was #(3.40xx10^-3*mol)/(30.00xx10^(-3)L)# #=# #0.1133# #mol*L^-1# in oxalic acid.