Chapter 1: Moles and equations
Calculating solution concentration by titration
A titration is often used to find the exact concentration of a solution. Worked example 20 shows the steps used to calculate the concentration of a solution of sodium hydroxide when it is neutralised by aqueous sulfuric acid of known concentration and volume. WORKED EXAMPLE 20 25.0 cm3 of a solution of sodium hydroxide is exactly neutralised by 15.10 cm3 of sulfuric acid of concentration 0.200 mol dm–3. 2NaOH + H2SO4
Na2SO4 + 2H2O
Calculate the concentration, in mol dm–3, of the sodium hydroxide solution. Step 1 Calculate the moles of acid. moles = concentration (mol dm–3) × volume of solution (dm3) 15.10 _____ = 0.003 02 mol H2SO4 0.200 × 1000 Step 2 Use the stoichiometry of the balanced equation to calculate the moles of NaOH. moles of NaOH = moles of acid (from step 1) × 2 Step 3 Calculate the concentration of NaOH. concentration (mol dm–3) number of moles of solute (mol) = ___________________________ volume of solution (dm3) 0.00604 = _______ 0.0250 = 0.242 mol dm–3 Note 1 In the first step we use the reagent for which the concentration and volume are both known. Note 2 In step 2, we multiply by 2 because the balanced equation shows that 2 mol of NaOH react with every 1 mol of H2SO4. Note 3 In step 3, we divide by 0.0250 because we have 25.0 changed cm3 to dm3 0.0250 = _____ . 1000
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Note 4 The answer is given to 3 significant figures because the smallest number of significant figures in the data is 3.
QUESTION 17 a The equation for the reaction of strontium hydroxide with hydrochloric acid is shown below. Sr(OH)2 + 2HCl
SrCl2 + 2H2O
25.0 cm3 of a solution of strontium hydroxide was exactly neutralised by 15.00 cm3 of 0.100 mol dm–3 hydrochloric acid. Calculate the concentration, in mol dm–3, of the strontium hydroxide solution. b 20.0 cm3 of a 0.400 mol dm–3 solution of sodium hydroxide was exactly neutralised by 25.25 cm3 of sulfuric acid. Calculate the concentration, in mol dm–3, of the sulfuric acid. The equation for the reaction is: H2SO4 + 2NaOH
Na2SO4 + 2H2O
Deducing stoichiometry by titration
We can use titration results to find the stoichiometry of a reaction. In order to do this, we need to know the concentrations and the volumes of both the reactants. The example below shows how to determine the stoichiometry of the reaction between a metal hydroxide and an acid. WORKED EXAMPLE 21 25.0 cm3 of a 0.0500 mol dm–3 solution of a metal hydroxide was titrated against a solution of 0.200 mol dm–3 hydrochloric acid. It required 12.50 cm3 of hydrochloric acid to exactly neutralise the metal hydroxide. Deduce the stoichiometry of this reaction. Step 1 Calculate the number of moles of each reagent. moles of metal hydroxide = concentration (mol dm–3) × volume of solution (dm3) 25.0 = 0.0500 × _____ = 1.25 × 10–3 mol 1000 moles of hydrochloric acid = concentration (mol dm–3) × volume of solution (dm3) 12.50 = 0.200 × _____ = 2.50 × 10–3 mol 1000 Step 2 Deduce the simplest mole ratio of metal hydroxide to hydrochloric acid. 1.25 × 10–3 moles of hydroxide : 2.50 × 10–3 moles of acid = 1 hydroxide : 2 acid
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