Titration calculations - Higher

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

Calculating a concentration

Worked example

In a titration, 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution is exactly neutralised by 20.00 cm3 of a dilute solution of hydrochloric acid. Calculate the concentration of the hydrochloric acid solution.

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 25.0 ÷ 1,000 = 0.0250 dm3

Rearrange:

Concentration in mol/dm3 = \frac{\textup{amount~of~solute~in~mol}}{\textup{volume~in~dm}^3}

Amount of solutein mol = concentration in mol/dm3 × volume in dm3

Amount of sodium hydroxide = 0.100 × 0.0250

= 0.00250 mol

Step 2: Find the amount of hydrochloric acid in moles

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.00250 mol of NaOH reacts with 0.00250 mol of HCl

Step 3: Calculate the concentration of hydrochloric acid in mol/dm3

Volume of hydrochloric acid = 20.00 ÷ 1000 = 0.0200 dm3

Concentration in mol/dm3 = \frac{\textup{amount~of~solute~in~mol}}{\textup{volume~in~dm}^3}

Concentration in mol/dm3 = \frac{\textup{0.00250}}{\textup{0.0200}}

= 0.125 mol/dm3

Step 4: Calculate the concentration of hydrochloric acid in g/dm3

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/dm3

Question

In a titration, 25.00 cm3 of 0.200 mol/dm3 sodium hydroxide solution is exactly neutralised by 22.70 cm3 of a dilute solution of hydrochloric acid.

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Calculate the concentration of the hydrochloric acid.

Volume of sodium hydroxide solution = 25.00 ÷ 1000 = 0.0250 dm3

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 dm3

Concentration of hydrochloric acid = 0.005 mol ÷ 0.0227

= 0.220 mol/dm3

Calculating a volume

Worked example

25.00 cm3 of 0.300 mol/dm3 sodium hydroxide solution is exactly neutralised by 0.100 mol/dm3 sulfuric acid. Calculate the volume of sulfuric acid needed.

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 25.0 ÷ 1000 = 0.0250 dm3

Amount of sodium hydroxide = concentration × volume

Amount of sodium hydroxide = 0.300 mol/dm3 × 0.0250 dm3

= 0.00750 mol

Step 2: Find the amount of sulfuric acid in moles

The balanced equation is:

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

So the mole ratio NaOH:H2SO4 is 2:1.

Therefore 0.00750 mol of NaOH reacts with (0.00750 ÷ 2) = 0.00375 mol of H2SO4

Step 3: Calculate the volume of sulfuric acid

Rearrange:

Concentration in mol/dm3 = \frac{\textup{amount~of~solute~in~mol}}{\textup{volume~in~dm}^3}

Volume in dm3 = \frac{\textup{amount~of~solute~in~mol}}{\textup{concentration~in~mol/dm}^3}

Volume in dm3 = \frac{\textup{0.00375}}{\textup{0.100}}

= 0.0375 dm3 (37.5 cm3)

Question

25.00 cm3 of 0.100 mol/dm3 sodium hydroxide solution is exactly neutralised by 0.125 mol/dm3 hydrochloric acid.

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Calculate the volume of hydrochloric acid needed.

Volume of sodium hydroxide solution = 25.00 ÷ 1000 = 0.0250 dm3

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 dm3 (20 cm3)

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