Question

FB1 is a solution containing a mixture of potassium carbonate and sodium hydrogen carbonate . 25cm^3 of this mixture was to titrated against 0.250mol/dm^3 solution of HCL. In one titration using phenolphthalein as indicator, 10.00cm^3 of the acid was needed. In another titration using methyl orange as indicator, 35.00cm^3 of acid was required.

Find the number of moles of potassium carbonate and sodium hydrogen carbonate present in the mixture.

I found the number of moles of potassium carbonate.
No of moles of HCL that reacted with potassium carbonate=No of moles of potassium carbonate = 0.01 x 0.250mol =0.00250mol

im not sure how to find moles of sodium hydrogencarbonate

Answer 1

Titration with phenolphthalein adds 1 H^+ to the CO3^2- to make HCO3^-. You are right with mols K2CO3. That titrates none of the HCO3^-

The methyl orange end point adds another H^+ to the HCO3^- formed from the CO3^2- in the first end point. It adds a H^+ to the HCO3^- that was there initially but was not titrated in the phenolphthalein end point.
35.00 mL = all of the HCO3^- + all CO3^2-
10 mL for CO3^2- to HCO3^- + another 10 mL for that HCO3^- to H2CO3.

Therefore, 35 - 20 = 15 mL for the HCO3^- by itself.