Titration calculations - Higher
The results of a titrationA quantitative procedure in which two solutions react in a known ratio, so if the concentration of one solution is known and the volumes of both are measured, the concentration of the other solution can be determined. can be used to calculate the concentrationThe concentration of a solution tells us how much of a substance is dissolved in water. The higher the concentration, the more particles of the substance are present. of a solutionMixture formed by a solute and a solvent., or the volume of solution needed.
Alisha Kakar explains how the results of a titration can be used to calculate the concentration of a solution
Calculating a concentration
Worked example
In a titration, 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution is exactly neutralisationThe reaction between an acid and a base to form a salt plus water. 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 = \(\frac{\textup{amount~of~solute~in~mol}}{\textup{volume~in~dm}^3}\)
amountIn chemistry, a measure of the number of particles in a substance. of soluteThe dissolved substance in a solution.in 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 chemical equationA chemical equation written using the symbols and formulae of the reactants and products, so that the number of units of each element present is the same on both sides of the arrow. is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
So the mole ratioThe ratio of the amounts of two substances as shown in a balanced equation. 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)
