Concentration Calculations

Titrations

Volumetric analysis is a process that uses the volume and concentration of one chemical reactant (a standard solution) to determine the concentration of another unknown solution
The technique most commonly used is a titration
The volumes are measured using two precise pieces of equipment, a volumetric or graduated pipette and a burette

The steps in a titration are:
- Measuring a known volume (usually 20 or 25 cm³) of one of the solutions with a volumetric or graduated pipette and placing it into a conical flask
- The other solution is placed in the burette
- A few drops of the indicator are added
- The tap on the burette is carefully opened and the solution added, portion by portion, to the conical flask until the indicator just changes colour
- Multiple trials are carried out until concordant results are obtained

Calculating concentration

Concentration calculations involve bringing together the skills and knowledge you have acquired in molar concentration and applying them to problem solving
You should be able to easily convert between moles, mass, concentrations and volumes ( of solutions and gases)
The four steps involved in problem solving are:
- write the balanced equation for the reaction
- determine the mass/ moles/ concentration/ volume of the of the substance(s) you know about
- use the balanced equation to deduce the mole ratios of the substances present
- calculate the mass/ moles/ concentration/ volume of the of the unknown substance(s)

Worked example

25.0 cm³ of 0.050 mol dm^-3 sodium carbonate was completely neutralised by 20.0 cm³ of dilute hydrochloric acid.Calculate the concentration in mol dm^-3 of the hydrochloric acid.

Answer:

Step 1: Write the balanced equation for the reaction

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Step 2: Determine the moles of the known substance, in this case sodium carbonate. Don't forget to divide the volume by 1000 to convert cm³ to dm³

moles = volume x concentration

amount (Na₂CO₃) = 0.0250 dm³ x 0.050 mol dm^-3 = 0.00125 mol

Step 3: Use the balanced equation to deduce the mole ratio of sodium carbonate to hydrochloric acid:

1 mol of Na₂CO₃ reacts with 2 mol of HCl, so the mole ratio is 1 : 2

Therefore 0.00125 moles of Na₂CO₃ react with 0.00250 moles of HCl

Step 4: Calculate the concentration of the unknown substance, hydrochloric acid

$concentration = \frac{moles}{volume} concentration (HCl) = \frac{0.00250 mol}{0.0200 {dm}^{3}} = 0.125 mol {dm}^{- 3}$

Worked example

Calculate the volume of hydrochloric acid of concentration 1.0 mol dm^-3 that is required to react completely with 2.5 g of calcium carbonate.

Answer:

Step 1: Write the balanced equation for the reaction

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Step 2: Determine the moles of the known substance, calcium carbonate

Amount of CaCO₃ = $\frac{2.5 g}{100.09 g {mol}^{- 1}} = 0.025 mol$

Step 3: Use the balanced equation to deduce the mole ratio of calcium carbonate to hydrochloric acid:

1 mol of CaCO₃ requires 2 mol of HCl

So 0.025 mol of CaCO₃ requires 0.050 mol of HCl

Step 4: Calculate the volume of HCl required

Volume of HCl = $\frac{moles}{concentration} = \frac{0.050 mol}{1.0 mol {dm}^{- 3}} = 0.050 {dm}^{3}$

Examiner Tip

When performing titration calculations using monoprotic acids (meaning one H⁺) such as HCl, the number of moles of the acid and alkali will be the same. This allows you to use the relationship

C₁V₁ =C₂V₂

where C₁ and V₁ are the concentration and volume of the acid and C₂ and V₂ are the concentration and volume of the alkali. There is no need to convert the units of volume to dm³ as this is a ratio.Simply re-arrange the formula to solve for the unknown quantity.

Worked example

A 0.675 g sample of a solid acid, HA, was dissolved in distilled water and made up to 100.0 cm³ in a volumetric flask. 25.0 cm³ of this solution was titrated against 0.100 mol dm^-3 NaOH solution and 12.1 cm³were required for complete reaction. Determine the molar mass of the acid.

Answer:

Step 1: Write the equation for the reaction

HA (aq) + NaOH (aq) → NaA (aq) + H₂O (l)

Step 2: Calculate the number of moles of the NaOH

n(NaOH)_sample = $(\frac{12.1 {cm}^{3}}{1000}) {dm}^{3} \times 0.100 mol {dm}^{- 3} = 1.21 \times 10^{- 3} mol$

Step 3: Deduce the number of moles of the acid

Since the acid is monoprotic the number of moles of HA is also 1.21 x 10^-3 mol

This is present in 25.0 cm³ of the solution

Step 4: Scale up to find the amount in the original solution

$n {(NaOH)}_{original} = \frac{1.21 \times 10^{- 3} mol \times 100.0 {cm}^{3}}{25.0 {cm}^{3}} = 4.84 \times 10^{- 3} mol$

Step 5: Calculate the molar mass

$moles = \frac{mass}{molar mass}$

$molar mass = \frac{mass}{moles} = \frac{0.675 g}{4.84 \times 10^{- 3}} = 139 g {mol}^{- 1}$

Back titration

A back titration is a common technique used to find the concentration or amount of an unknown substance indirectly
The principle is to carry out a reaction with the unknown substance and an excess of a further reactant such as an acid or an alkali
The excess reactant, after reaction, is then analysed by titration and the mole ratios are used to deduce the moles or concentration of the original substance being analysed

Worked example

The percentage by mass of calcium carbonate, CaCO₃, in a sample of marble was determined by adding excess hydrochloric acid to ensure that all the calcium carbonate had reacted. The excess acid left was then titrated with aqueous sodium hydroxide. A student added 27.20 cm³ of 0.200 mol dm^-3 HCl to 0.188 g of marble. The excess acid required 23.80 cm³ of 0.100 mol dm^-3 NaOH for neutralisation. Calculate the percentage of calcium carbonate in the marble.

Answer:

Step 1: Write the equation for the titration reaction:

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

Step 2: Calculate the number of moles of the NaOH

n(NaOH) = 0.02380 dm³ x 0.100 mol dm^-3= 2.380 x 10^-3 mol

Step 3: Deduce the number of moles of the excess acid

Since the reacting ratio is 1:1 the number of moles of HCl is also 2.380 x 10^-3 mol

Step 4: Find the amount of HCl in the original solution and then the amount reacted

n(HCl)_original= 0.02720 dm³ x 0.200 mol dm^-3= 5.440 x 10^-3 mol

n(HCl)_reacted= 5.440 x 10^-3 mol – 2.380 x 10^-3 mol = 3.060 x 10^-3 mol

Step 5: Write the equation for the reaction with the calcium carbonate

2HCl (aq) + CaCO₃ (s) → CaCl₂ (aq) + CO₂ (g) + H₂O (l)

Step 6: Deduce the number of moles of the calcium carbonate that reacted

Since the reacting ratio is 2:1 the number of moles of CaCO₃ is (3.060 x 10^-3 mol) ÷ 2

n(CaCO₃) = 1.530 x 10^-3 mol

Step 7: Calculate the mass of calcium carbonate in the sample of marble

mass = moles x molar mass = 1.530 x 10^-3 mol x 100.09 g mol^-1 = 0.1531g

Step 8: Calculate the percentage of calcium carbonate in the marble

Percentage of CaCO₃ in marble = $\frac{0.1531 \times 100}{0.188} = 81.5 %$

Examiner Tip

Rounding off when you take averages. When you have an average of burette readings that comes to three decimal places, e.g.(23.20 cm³ + 23.25 cm³) ÷ 2 = 23.225 cm³

You CANNOT show more than two decimal places because that would make the average more precise than the readings.

To manage this situation you need to follow a simple rule. If the last digit is between a 5 and 9 then you round up; if the digit is between 0 and 4 you round down. So in this case the value recorded would be 23.23 cm³

Concentration Calculations (SL IB Chemistry)

Revision Note