# Titration for SQA National 5 Chemistry

1. Titration experimental setup
2. Concentration
3. Titration calculations

Titration is a method used to determine the concentration of solutions by measuring the volumes of solutions required for them to react fully. Before the titration has started, a volumetric solution is obtained by carefully measuring the mass or volume of a material, dissolving it in a solvent, pouring it into a volumetric flask using a funnel and filling the flask with distilled water or other solvent up to the mark of the volumetric flask. The volumetric flask is then usually covered with a stopper and shaken to mix the contents inside.

Before the titration, one of the solutions is placed into a burette by using a funnel which is removed afterwards to prevent potential drops from affecting the volume measured. The solution within the burette, the titrant, is slowly poured into a beaker to set the meniscus of the titrant to 0 and to ensure that there are no air bubbles present at a stopcock.

Another solution is placed into a conical flask using a pipette and a pipette filler. A few drops of an indicator are added to the conical flask. The flask is placed directly below the burette to avoid spillages and the titrant is slowly poured to the conical flask while stirring it. When the indicator changes the colour to signify the end of the reaction, the volume of the titrant used is recorded and the titration is repeated until several concordant results (within 0.2 cm³ of each other) are obtained. Usually the mean/average of the concordant results is used for further calculations.

In the image from left to right: volumetric flask, burette, funnel, pipette, pipette filler, conical flask.

Concentration is commonly given in moles per litre (mol/l or mol/dm³). The mole concentration is obtained by dividing the quantity of a material in moles by the volume of a solution.

1 dm³ = 1 l = 1000 cm³ = 1000 ml

To find an unknown concentration of a solution by using titration, find the mole ratio from a balanced chemical equation for the compound of unknown concentration and a compound with a known concentration. Then find the moles of known concentration compound, use the mole ratio to find the missing moles of the other compound and find the concentration of the other solution. A similar method can be used to find the volume or the number of moles that are not provided for one of the compounds.

# 1

Mary is performing a NaOH with HCl titration and she has chosen to use an indicator that is yellow in the base but red in the acid. Mary's setup just before she starts the titration is shown in the image. What is wrong with the titration setup provided?

One of the solutions during a titration is placed into a conical flask with a few drops of an indicator. The flask is placed directly below the burette to avoid spillages and the titrant is slowly poured to the conical flask while stirring it. When the indicator changes the colour to signify the end of the reaction, the volume of the titrant used is recorded.

The indicator has to be in the conical flask, not the burette.

# 2

What is the volume of 1.73 mol/dm³ aluminium hydroxide solution needed to titrate 5.00 cm³ of 0.300 M sulphuric acid solution?

2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

n(H₂SO₄) = c(H₂SO₄) × V(H₂SO₄) = 0.300 × 0.00500 = 1.50 × 10⁻³ mol

N(Al(OH)₃) : N(H₂SO₄) = 2 : 3
N₁n₂ = N₂n₁
N(Al(OH)₃) × n(H₂SO₄) = N(H₂SO₄) × n(Al(OH)₃)
2 × 1.50 × 10⁻³ = 3 × n(Al(OH)₃)
n(Al(OH)₃) = 2 × 1.50 × 10⁻³ ÷ 3 = 1.00 × 10⁻³ mol

c(Al(OH)₃) = n(Al(OH)₃) ÷ V(Al(OH)₃)
1.73 = 1.00 × 10⁻³ ÷ V(Al(OH)₃)
V(Al(OH)₃) = 1.00 × 10⁻³ ÷ 1.73 = 5.78 × 10⁻⁴ dm³

5.78 × 10⁻⁴ dm³

# 3

Name the chemical equipment presented in the image.

Volumetric flask is usually taller than the conical flask. Burette is long and has a stopcock at the end of it. Conical flask has a cone shape.

# 4

Sophie has 5.37 mol of AgNO₃ available in the lab. What is the maximum volume of 3.0 mol/dm³ silver nitrate solution that she can make?

c = m ÷ V
3.0 = 5.37 ÷ V
V = 5.37 ÷ 3.0 = 1.8 dm³

1.8 dm³

# 5

A 250 cm³ sulphuric acid solution has been made by using 0.055 mol of sulphuric acid. Find the mole concentration of the solution produced.

250 cm³ = 0.25 dm³
c = n ÷ V
c = 0.055 ÷ 0.25 = 0.22 mol/dm³

0.22 mol/dm³

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