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Question

Concentration of sodium carbonate solution=0.105M
Volume of each aliquot of sodium carbonate= 20.00mL
Recorded titres of hydrochloric acid (1)12.75mL (2)12.45mL (3)12.39mL (4)12.40mL
d)Calculate the number of mole of sodium carbonate in each aliquot
e)Using the mole ratio calculate the numbers of mole of hydrochloric acid in the average titre.
f)Therefore calculate the concentration of the acid.

Answer 1

d)
No. of moles of Na2CO3
= MV
= 0.105 x (20/1000)
= 0.0021 mol

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e)
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
Molar ratio Na2CO3 : HCl = 1 : 2

No. of moles of HCl in average titre
= 0.0021 x 2
= 0.0042 mol

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f)
Concentration of HCl
= mol / (vol in L)
= 0.0042 / (12.41/1000)
= 0.338 M