Chemistry Questions

A 250.0cm³ standard solution was made from 2.65g of powdered sodium carbonate. In a titration, 25.0cm³ of this solution required 20.00cm³ of hydrochloric acid solution for complete reaction in the presence of a suitable indicator.

a) Calculate the molarity of the standard sodium carbonate solution.

Molar mass of Na2CO3 = 23x2 + 12 + 16x3 = 106 g/mol
No. of moles of Na2CO3 used = 2.65/106 = 0.025 mol
Volume of the solution = 250 cm³ = 0.25 dm³
Molarity of the standard Na2CO3 solution = 0.025/0.25 = 0.1 M


b) Calculate the number of moles of sodium carbonate used to react with 20.00cm³ hydrochloric acid.

No. of moles of Na2CO3 used to react with 20.00 cm³ of HCl
= 0.1 x (25/1000)
= 0.0025 mol


c) Calculate the molarity of the hydrochloric acid

HCl + NaOH → NaCl + H2O
Mole ratio HCl : NaOH = 1 : 1

No. of moles of HCl used in titration = 0.0025 mol
No. of moles of HCl used in titration = 0.0025 mol
Volume of HCl used in titration = 20 cm³ = 0.02 dm³
Molarity of HCl = 0.0025/0.02 = 0.125 M