standardization/molarity?

Question

Sodium carbonate is a reagent that may be used to standardize acids. In such a standardization it was found that a .498g sample of sodium carbonate required 23.5mL of a sulfuric acid soln to reach the end pt for the rxn.

Na2CO3(aq) + H2SO4(aq)------->H2O(l) + CO2(g) + Na2SO4(aq)

what is the molarity of the H2SO4?

thanks for anyone's help:)

Answer 1

First of all lets find how many moles of Na2CO3 you had
moles= mass/MW =0.498/106 =4.7*10^-3

Na2CO3(aq) + H2SO4(aq) ->H2O(l) + CO2(g) + Na2SO4(aq)

From the stoichiometry
1 mole Na2CO3 reacts with 1 mole H2SO4
4.7*!0^-3 mole react with .. ..x mole

x=4.7*10^-3 mole H2SO4

M=mole/V = (4.7*10^-3)/(23.5*10^-3) = 4.7/23.5= 0.2 M

Using greqs would have been easier, but I don't know if you have learned about them yet.

Answer 2

If i did this right it should be .2M H2SO4

.498gNa2CO3 X 1 mil Na2CO3/105.99g X 1 mol H2SO4/1mol Na2CO3 = .00470 mol H2SO4

.00470 mol/.0235 L = .2MH2SO4