8.3g of sodium carbonate of fomula na2 co3 was dissolved to make 100cm3 of solution?

Question

10 cm3 portions were titrated against 0.1 mol dm3 concentration of hcl and 18.2cm3 of the acid were needed for neutralisation. calculate the number of moles of acid used in this reaction? what is the number of moles of sodium carbonate used in the titration? calculate the concentration of the sodium carbonate solution

Answer 1

Number of moles of acid = volume (in litres) x molarity =0.0182 x 0.1 = 0.00182 Moles
The equation for the reaction shows twice as many moles of acid is needed as carbonate

2HCl + Na2CO3 = 2NaCl + H2O + CO2

So the number of moles of Na2CO3 = 0.00182 divided by 2 = 0.00091

This is how many moles are in 10cm3 of solution so in a litre there would be 100 x as much = 0.091 so the concentration is 0.091M
Concentration is also measured in g/litre and in this case it is 106 (the RMM of sodium carbonate) x 0.091 = 9.646g/litre

Answer 2

10 mL of carbonate took 18.2 cm3 of 0.1 M HCl
which is (0.1/1000) x 18.2 moles = 0.00182 moles acid used

reaction is 2 HCl to 1 carbonate, so moles of carbonate = moles HCl/2 = 0.00182/2 = 0.00091 moles carbonate in 10 cm3 (moles used in the titration)
which would give 0.091 moles carbonate in a litre or 0.091 M

Answer 3

10 mL of carbonate took 18.2 cm3 of 0.a million M HCl it is (0.a million/one thousand) x 18.2 moles = 0.00182 moles acid used reaction is two HCl to a million carbonate, so moles of carbonate = moles HCl/2 = 0.00182/2 = 0.00091 moles carbonate in 10 cm3 (moles used interior the titration) which might provide 0.091 moles carbonate in a litre or 0.091 M