The volumes of acids and alkali solutions that react with each other can be measured by titration using a suitable indicator.

The results of a titration can be used to calculate the concentration of a solution, or the volume of solution needed.

In a titration, 25.0 cm^{3} of 0.100 mol/dm^{3} sodium hydroxide solution is exactly neutralised by 20.00 cm^{3} of a dilute solution of hydrochloric acid. Calculate the concentration of the hydrochloric acid solution.

Volume of sodium hydroxide solution = 25.0 ÷ 1,000 = 0.0250 dm^{3}

Rearrange:

Concentration in mol/dm^{3} =

Amount of solutein mol = concentration in mol/dm^{3} × volume in dm^{3}

Amount of sodium hydroxide = 0.100 × 0.0250

= 0.00250 mol

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H_{2}O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.00250 mol of NaOH reacts with 0.00250 mol of HCl

Volume of hydrochloric acid = 20.00 ÷ 1000 = 0.0200 dm^{3}

Concentration in mol/dm^{3} =

Concentration in mol/dm^{3} =

= 0.125 mol/dm^{3}

Relative formula mass of HCl = 1 + 35.5 = 36.5

Mass = relative formula mass × amount

Mass of HCl = 36.5 × 0.125

= 4.56 g

So concentration = 4.56 g/dm^{3}

- Question
In a titration, 25.00 cm

^{3}of 0.200 mol/dm^{3}sodium hydroxide solution is exactly neutralised by 22.70 cm^{3}of a dilute solution of hydrochloric acid.NaOH(aq) + HCl(aq) → NaCl(aq) + H

_{2}O(l)Calculate the concentration of the hydrochloric acid.

Volume of sodium hydroxide solution = 25.00 ÷ 1000 = 0.0250 dm

^{3}Amount of sodium hydroxide = 0.200 × 0.0250 = 0.005 mol

From the equation, 0.005 mol of NaOH reacts with 0.005 mol of HCl

Volume of hydrochloric acid = 22.70 ÷ 1000 = 0.0227 dm

^{3}Concentration of hydrochloric acid = 0.005 mol ÷ 0.0227

= 0.220 mol/dm

^{3}

25.00 cm^{3} of 0.300 mol/dm^{3} sodium hydroxide solution is exactly neutralised by 0.100 mol/dm^{3} sulfuric acid. Calculate the volume of sulfuric acid needed.

Volume of sodium hydroxide solution = 25.0 ÷ 1000 = 0.0250 dm^{3}

Amount of sodium hydroxide = concentration × volume

Amount of sodium hydroxide = 0.300 mol/dm^{3} × 0.0250 dm^{3}

= 0.00750 mol

The balanced equation is:

2NaOH(aq) + H_{2}SO_{4}(aq) → Na_{2}SO_{4}(aq) + 2H_{2}O(l)

So the mole ratio NaOH:H_{2}SO_{4} is 2:1.

Therefore 0.00750 mol of NaOH reacts with (0.00750 ÷ 2) = 0.00375 mol of H_{2}SO_{4}

Rearrange:

Concentration in mol/dm^{3} =

Volume in dm^{3} =

Volume in dm^{3} =

= 0.0375 dm^{3} (37.5 cm^{3})

- Question
25.00 cm

^{3}of 0.100 mol/dm^{3}sodium hydroxide solution is exactly neutralised by 0.125 mol/dm^{3}hydrochloric acid.NaOH(aq) + HCl(aq) → NaCl(aq) + H

_{2}O(l)Calculate the volume of hydrochloric acid needed.

Volume of sodium hydroxide solution = 25.00 ÷ 1000 = 0.0250 dm

^{3}Amount of sodium hydroxide = 0.100 × 0.0250 = 0.00250 mol

From the equation, 0.00250 mol of NaOH reacts with 0.00250 mol of HCl

Volume of hydrochloric acid = 0.00250 ÷ 0.125

= 0.020 dm

^{3}(20 cm^{3})