Asked by Goo
20.00 cm 3 of a solution containing 0.53 g of anhydrous Na2CO3 in 100 cm3 requires 25.00 cm3 of H2SO4 for complete neutralisation. The concentration of the acid solution in moles per dm is [H =1, C = 12, O = 16, Na =23, S = 32]
Answers
Answered by
DrBob222
Na2CO3 + H2SO4 ==> CO2 + H2O + Na2SO4
mols Na2CO3 = 0.53/106 = 0.005
M Na2CO3 = mols/dm^3 = 0.005/0.1 = 0.05 M
mols Na2CO3 used in the titration = M x L = 0.05 M x 0.020 = 0.001
M H2SO4 = mols/dm^3 = 0.001/0.025 = 0.04 M
mols Na2CO3 = 0.53/106 = 0.005
M Na2CO3 = mols/dm^3 = 0.005/0.1 = 0.05 M
mols Na2CO3 used in the titration = M x L = 0.05 M x 0.020 = 0.001
M H2SO4 = mols/dm^3 = 0.001/0.025 = 0.04 M