Titrations determine neutralisation volumes

Titration is an analytical method used to determine the concentration of a solution of known volume by gradually adding a solution of known concentration to it until neutralisation occurs.

• A solution with an unknown volume and concentration is referred to as the analyte. It is measured using a pipette and placed into a conical flask.
• The standard solution, which has a known concentration, is filled into a burette and slowly added to the analyte.
• To visually determine when neutralisation has occurred, an indicator is added to the flask, causing a colour change at the end point - the point at which the reaction has just reached completion and neither reactant is in excess.

By calculating the volume of the standard solution required to reach the end point, you can determine the concentration of the analyte.

Steps in carrying out a titration

To perform a titration accurately, follow these steps:

1. Measure the analyte into a conical flask using a pipette, and add an indicator.
2. Fill the burette with the standard solution and note the initial reading.
3. Gradually add the standard solution to the flask, swirling to mix, until you approximately reach the point of neutralisation.
4. Refill the burette and repeat steps 2 and 3 to get an approximate idea of where neutralisation occurs.
5. Conduct a precise titration by adding the standard solution slowly to identify the exact end point. Record the final burette reading, also known as the titre volume.
6. Perform the titration multiple times to obtain concordant results, which are titre volumes that differ by no more than 0.10 cm³.
7. Calculate the mean titre volume using the concordant results.

Measuring burette readings accurately

• When reading a 50 cm³ burette, the volume should be recorded to the nearest half division (±0.05 cm³), with the bottom of the meniscus either on a mark or between two marks.
• The meniscus is the curved surface of the liquid in the burette, which is caused by the adhesive forces between the liquid and the glass.
• To ensure accurate readings, always read the volume at the bottom of the meniscus, as shown in the image below.

Indicators mark reaction completion

Indicators are crucial for visually identifying the end point of a titration through a colour change, indicating neutralisation.

Common indicators include:

• Methyl orange, which changes from yellow to red when adding acid to alkali.
• Phenolphthalein, which changes from pink to colourless when adding acid to alkali. Conduct titrations against a white background to easily observe the colour change.

Calculating concentration in titrations

Titration experiments can be used to determine the concentration of a solution.

Here are the steps:

1. Write a balanced equation for the reaction.
2. Use the titration volumes and a known concentration to calculate the moles of one reactant.
3. Use stoichiometry to relate moles of this reactant to the moles of the reactant whose concentration is unknown.
4. Divide the moles by the volume to calculate the unknown concentration.

Worked example 1 - Calculating concentration a from titration

30.0 cm³ of 0.100 mol dm^-3 NaOH neutralises 20.0 cm³ of HCl.

Calculate the concentration of the HCl solution.

Step 1: Write the balanced equation

NaOH_(aq) + HCl_(aq) ➔ NaCl_(aq) + H₂O_(l)

Step 2: Conversion of cm³ to dm³

To convert from cm³ into dm³, divide by 1,000

30.0 cm³ = 0.0300 dm³

20.0 cm³ = 0.0200 dm³

Step 3: Calculate number of moles of NaOH

$\text{n }=\text{ c }\times \text{ V }=0.100\times 0.0300=3.00\times {10}^{-3}\text{ mol}$

Step 4: Calculate number of moles moles of HCl

HCl : NaOH mole ratio = 1:1

Moles of HCl = 3.00 x 10^-3 mol

Step 5: Calculate concentration of HCl

$\text{c }=\frac{\text{n}}{\text{V}}=\frac{3.00\times {10}^{-3}}{0.0200}=0.150{\text{ mol dm}}^{-3}$

The concentration of the HCl solution is 0.150 mol dm^-3.

The next example covers a more complex reaction stoichiometry.

Worked example 2 - Calculating concentration from a titration

25.0 cm³ of 0.200 mol dm^-3 H₂SO₄ neutralises 50.0 cm³ of KOH.

Calculate the concentration of the KOH solution.

Step 1: Write the balanced equation

H₂SO_4(aq) + 2KOH_(aq) ➔ K₂SO_4(aq) + 2H₂O_(l)

Step 2: Conversion of cm³ to dm³

To convert from cm³ into dm³, divide by 1,000

250 cm³ = 0.0250 dm³

50.0 cm³ = 0.0500 dm³

Step 3: Calculate number of moles of H₂SO₄

$\text{n }=\text{ c }\times \text{ V }=0.200\times 0.0250=5.00\times {10}^{-3}\text{ mol}$

Step 4: Calculate number of moles of KOH

KOH : H₂SO₄ mole ratio = 2:1

Moles of KOH = 2 x 5.00 x 10^-3 = 0.0100 mol

Step 5: Calculate concentration of KOH

$\text{c }=\frac{\text{n}}{\text{V}}=\frac{0.0100}{0.0500}=0.200{\text{ mol dm}}^{-3}$

The concentration of the KOH solution is 0.200 mol dm^-3.

A similar method can also calculate the volume of one reagent needed to completely react with a certain amount of another reagent.

Worked example 3 - Calculating volume from a titration

Calculate the volume (in dm³) of 0.250 mol dm^-3^ Na₂CO₃ solution required to completely neutralise 40.0 cm³ of 0.100 mol dm^-3 HCl.

Step 1: Write the balanced equation

Na₂CO_3(aq) + 2HCl_(aq) ➔ 2NaCl_(aq) + H₂O_(l) + CO_2(g)

Step 2: Conversion of cm³ to dm³

To convert from cm³ into dm³, divide by 1,000

40.0 cm³ = 0.0400 dm³

Step 3: Calculate number of moles of HCl

$\text{n }=\text{ c }\times \text{ V }=0.100\times 0.0400=4.00\times {10}^{-3}\text{ mol}$

Step 4: Calculate number of moles of Na₂CO₃

Na₂CO₃ : HCl mole ratio = 1:2

${\text{Moles of Na}}_2{\text{CO}}_3=\frac{4.00\times {10}^{-3}}{2}=2.00\times {10}^{-3}\text{ mol}$

Step 5: Calculate volume of Na₂CO₃

$\text{V }=\frac{\text{n}}{\text{c}}=\frac{2.00\times {10}^{-3}}{0.250}=8.00\times {10}^{-3}{\text{ dm}}^{-3}$

Standard solutions enable calculation of unknown concentrations

To determine an unknown concentration via titration, it's essential to use a standard solution with a precisely known concentration.

These solutions are prepared by:

1. Weighing a precise mass of solute using a balance.
2. Dissolving the weighed solute in distilled water in a beaker.
3. Transferring the solution to a volumetric flask of a specific volume.
4. Rinsing the beaker several times with distilled water and adding the rinsings to the volumetric flask.
5. Diluting the solution to the mark on the volumetric flask with a pipette.
6. Inverting the flask several times to ensure thorough mixing of the solution.

The formula for the concentration of the stock solution is:

${\text{concentration in g dm}}^{\text{-3 }}=\frac{\text{mass of solute in g}}{{\text{volume in dm}}^3}$

Worked example 4 - Determining the mass of solid Na_2_CO_3_ needed

What mass in g of solid sodium carbonate (Na₂CO₃) is needed to make 250 cm³ of a 0.400 mol dm^-3 solution?

Step 1: Conversion of cm³ into dm³

To convert from cm³ into dm³, divide by 1,000

250 cm³ = 0.250 dm³

Step 2: Calculate number of moles of Na_2_CO_3_ required

n = c x V = 0.400 x 0.250 = 0.100 mol

Step 3: Calculate mass of Na_2_CO_3_ required

m = n x M_r = 0.100 x 106.0 = 10.6 g

Uncertainty in measurements

All measurements have some degree of uncertainty due to the sensitivity limits of the equipment used.

The uncertainty varies depending on the equipment. For instance:

• A 50 cm³ burette has scale markings every 0.1 cm³.
• Readings should be within 0.05 cm³ of the true value (assuming no gross errors).
• So the uncertainty is ±0.05 cm³. The ± sign indicates the range containing the true value (also called the margin of error).

In general, for any measuring device, the uncertainty is half the smallest increment the equipment can measure, in either direction.

Manufacturers typically provide uncertainty values for equipment based on its construction accuracy. These are often printed directly on the equipment.

When combining measurements with the same units, their uncertainties must also be combined.

Worked example 5 - Calculating the total uncertainty in mass measurement

A student uses electronic scales (measures to nearest 0.05 g) to weigh out 1.65 g of a solid after zeroing the scales.

Calculate the total uncertainty.

Step 1: Determine number of readings

There are two readings:

• Initial: 0.00 g
• Final: 1.65 g

Step 2: Calculate total uncertainty

Each has an uncertainty of $\frac{0.05}{2}$ = 0.025 g.

Total uncertainty = 0.025 + 0.025 = 0.05 g

Calculating percentage uncertainty

The percentage uncertainty of a measurement can be calculated using:

Percentage uncertainty = $\frac{uncertainty}{reading}\times 100$

Worked example 6 - Calculating percentage uncertainty

A 500 cm³ volumetric flask has a manufacturer's stated uncertainty of ±0.50 cm³.

Calculate the flask's percentage uncertainty to 2 decimal places.

Step 1: Equation

Percentage uncertainty = $\frac{uncertainty}{reading}\times 100$

Step 2: Substitution and correct evaluation

Percentage uncertainty = $\frac{0.50}{500}\times 100$ = 0.10%

Minimising percentage uncertainty

Percentage uncertainty can be reduced in two key ways:

1. Use the most precise equipment available
2. Plan experiments strategically

For example, when using a measuring cylinder with 0.1 cm³ increments:

• Measuring 5 cm³ of liquid: % uncertainty = $\frac{0.05}{5}\times 100$ = 1%
• Measuring 10 cm³ of liquid: % uncertainty = $\frac{0.05}{10}\times 100$ = 0.5% So using a larger volume reduces the percentage uncertainty.

Systematic vs. random errors

Systematic errors are consistent each time an experiment is repeated. They often stem from issues with the experimental setup or equipment.

For example, using a 10.00 cm³ pipette that actually only delivers 9.95 cm³ - the sample will be ~0.05 cm³ too small every time.

Random errors vary between repeated experiments. They cause results to differ slightly each time.

For eample, reading errors from a burette - readings may be slightly above or below the true value.

Calculating total uncertainty in titrations

To find the total uncertainty in a titration result:

1. Calculate the percentage uncertainty for each piece of equipment.
2. Add these individual percentage uncertainties to get the overall percentage uncertainty.
3. Use this to determine the absolute uncertainty in the final result.

Worked example 7 - Calculating total uncertainty in titration

20.00 cm³ of KOH solution (uncertainty ±0.080 cm³) is neutralised by 35.40 cm³ of HCl (uncertainty ±0.15 cm³). The KOH concentration is calculated as 1.250 mol dm^-3.

Calculate the absolute uncertainty in this concentration to 2 significant figures.

Step 1: Individual percentage uncertainties

KOH: $\frac{0.080}{20.00}\times$ = 0.40%

HCl: $\frac{0.15}{35.40}\times 100$ = 0.42%

Step 2: Calculate total percentage uncertainty

0.40 + 0.42 = 0.82%

Step 3: Calculate absolute uncertainty in final result

0.82% of 1.250 mol dm^-3 = ±0.010 mol dm^-3