Question

25.0 cm of sodium carbonate solution neutralises 27.9 cm of 0.500 mol/dmºnitric acid solution.
2 HNO3(aq) + Na tos(aq)
2 NaNO-(aq) + CO2(aq) + 2 H2O(l)
HE
Find the concentration of sodium carbonate solution in mol/dmº. Give your answer to 3 significant figures.​

Answer 1

Answer:

The specific gravity of sodium carbonate is 1.25 g/mL.

Hence, the mass of 25 mL of solution of sodium carbonate is 1.25×25=31.25 g.

The mass of HCl used for neutralization is

1000

32.9

×109.5=3.60 g.

The number of moles of HCl in 3.60 g is

36.5

3.60

=0.0987.

2 moles of HCl will neutralize 1 mole of sodium carbonate.

The number of moles of sodium carbonate neutralized by 0.0987 moles of HCl are

2

0.987

=0.04935

Thus, 31.25 g of sodium carbonate contains 0.04935 moles.

Hence, 125 g of sodium carbonate will contain

31.25

125

×0.04935=0.1974 moles.

They will neutrlaize 0.1974 moles of sulphuric acid which corresponds to 2×0.1974=0.3948 g eq of sulphuric acid.

The volume of 0.84 N sulphuric acid required will be

0.84

0.3948

=0.470 L or 470 mL.

Explanation:

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