Reasons for errors using formulas

1. The formula is remembered wrongly.

A simple formula often used to find the concentration of a solution is:

c = n / V (where c is the concentration, n is ‘number of moles’ (actually amount in mol) and V is the volume).

I’ve seen several variations of this in exams, e.g. c = V / n or n = c / V. The key to stating formulas correctly is to check the units. Concentration is normally expressed as mol dm⁻³. Once a student realises this the formula is redundant as mol dm⁻³ literally means the amount in mol divided by the volume in litres (dm³). Another example where the formula is often quoted wrongly is the relationship between the velocity of light (c),  the frequency (ν) and the wavelength (λ).

Looking at the units it is obvious that velocity (m s⁻¹) = wavelength (m) x frequency (s⁻¹), i.e. c = λν.

2. The formula is applied wrongly

One of the most common examples of this is the formula M₁V₁ = M₂V₂ which students often use to solve titration problems. It does work in the simplest case when one mole of an acid is neutralised by one mole of a base,

e.g. HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

but certainly does not work when it comes to the neutralisation of diprotic acids,

e.g. H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l).

My advice is to never use it! Instead always work from first principles.

Worked example:

26.80 cm³ of a solution of sulfuric acid was required to neutralise 25.00 cm³ of 0.150 mol dm⁻³ potassium hydroxide solution. Calculate the concentration of the sulfuric acid solution in mol dm^-3.

Step 1: Write the stoichiometric equation

H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)

Step 2: Calculate the amount (in mol) of potassium hydroxide present.

1000 cm³ of 0.150 mol dm⁻³ KOH contains 0.150 mol

so 25.00 cm³ contains (25.00/1000) x 0.150 = 3.75 x 10⁻³ mol.

Step 3: Calculate the amount (in mol) of sulfuric acid required to neutralise the potassium hydroxide solution.

From the equation 1 mol of H₂SO₄ neutralises 2 mol of KOH so amount of H₂SO₄ required = ½ x 3.75 x 10⁻³ = 1.875 x 10⁻³ mol.

Step 4: Calculate the concentration of the sulfuric acid in mol dm⁻³.

26.80 cm³ of H₂SO₄ contains 1.875 x 10⁻³ mol

1000 cm³ of H₂SO₄ contains (1000/26.80) x 1.875 x 10⁻³ = 0.0700 mol

so the concentration of the sulfuric acid = 0.0700 mol dm⁻³ (to 3 significant figures).

3. A formula cannot be used at all.

A back to basics stepwise approach is the only way to answer titration questions where a solid, rather than a solution, is being titrated as the formula M₁V₁ = M₂V₂ cannot be applied.

Example:

What mass of calcium carbonate will react exactly with 20.00 cm³ of 0.200 mol dm⁻³ hydrochloric acid?

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

Amount of acid reacting = 20.00/1000 x 0.200 = 4.00 x 10⁻³ mol

Since two mol of HCl react with one mol of CaCO₃

Amount of CaCO₃ that reacts = ½ x 4.00 x 10^-3 = 2.00 x 10⁻³ mol

M(CaCO₃) = 40.08 + 12.01 + (3 x 16.00) = 100.09 g mol⁻¹

Mass of CaCO₃ that reacts with the acid = 100.09 x 2.00 x 10⁻³ = 0.200 g ( to 3 significant figures).

4.The wrong units are used.

The classic example of where this often occurs is when students use the equation

ΔG = ΔH – TΔS

ΔG and ΔH values are usually quoted in kJ mol⁻¹ but entropy values are normally quoted in J K⁻¹ mol^−1. Students constantly need reminding that they must divide the entropy value by 1000 to convert it into kJ K⁻¹ mol⁻¹ when using it in the equation. Of course they also need to ensure the temperature is given in Kelvin not ^oC.

Other examples of formulas where units must be consistent are the gas equation pV = nRT and rate equations as the units of the rate constant depend upon the overall order of the reaction.

5. Problems with the Henderson–Hasselbalch equation.

This formula causes particular problems although strictly speaking calculations involving buffer solutions should only appear in Paper 3.

The theory of buffer solutions is covered at Higher Level in the AHL Topic 18 but buffer calculations only appear in two of the options in Paper 3. However even though buffers are not coved at Standard Level in the core, Option D (see D.4 pH regulation of the stomach) includes the application of the Henderson–Hasselbalch equation at both SL and HL whereas in option B (B.7 Proteins and enzymes) it is only on the HL syllabus. The use of the Henderson–Hasselbalch equation in Option D can cause real problems for Standard Level students who have not covered either equilibrium calculations or the definitions of K_a, K_b, pK_a and pK_b.

The Henderson–Hasselbalch equation is given in Section 1 of the IB data booklet but it seems unwise for students to use it if they do not fully understand it. It is easy to use wrongly and and I have never taught it to my students as I prefer them to be able to work out all buffer calculations from first principles.

Example using first principles

Calculate the pH of a buffer solution made by adding 20.0 cm³ of 1.00 mol dm⁻³ sodium hydroxide solution to 40.0 cm³ of 1.00 mol dm^-3 ethanoic acid solution. Section 21 of the IB data booklet gives the pK_a of ethanoic acid = 4.76

(Standard level students need to understand that K_a is the equilibrium constant (acid dissociation constant) for a weak acid and that pK_a = − log₁₀K_a (this is rather similar to pH = − log₁₀[H⁺(aq)]). The relationship K_a = 10^−pKa is also helpful as the data booklet only lists pK_a values, not K_a values.)

Worked solution:

Step 1: Write the equation for the reaction.

CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)

After the reaction all the sodium hydroxide solution will have been neutralised to form sodium ethanoate and the excess ethanoic acid remains.

Step 2: Calculate the amounts of the sodium ethanoate and the excess ethanoic acid in the resulting solution.

Initial amount of NaOH = 20.0/1000 x 1.00 = 2.00 x 10⁻² mol

Initial amount of CH₃COOH = 40.0/1000 x 1.00 = 4.00 x 10⁻² mol

Since all the NaOH reacts to form CH₃COONa

Final amount of CH₃COONa = 2.00 x 10⁻² mol

Final amount of CH₃COOH = 4.00 x 10⁻² − 2.00 x 10⁻² = 2.00 x 10⁻² mol.

Step 3: Calculate the concentrations of the sodium ethanoate and the excess ethanoic acid in the resulting solution.

Final volume = 20.0 + 40.0 = 60.0 cm³ (assuming the volumes are exactly additive).

[CH₃COONa(aq)] = 1000 / 60.0 x 2.00 x 10⁻² = 0.333  mol dm⁻³

[CH₃COOH(aq)] = 1000 / 60.0 x 2.00 x 10⁻² = 0.333  mol dm⁻³

Step 4: Write the equilibrium expression for the dissociation of ethanoic acid.

CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq)

K_a = [H⁺(aq)] x [CH₃COO⁻(aq)] / [CH₃COOH(aq)]

Step 5: Substitute the values for the concentration of the acid and the salt into the equilibrium expression to find [H⁺(aq)].

K_a = [H⁺(aq)] x [CH₃COO⁻(aq)] / [CH₃COOH(aq)] = [H⁺(aq)] x 0.333 / 0.333 = [H⁺(aq)]

K_a = 10^−4.7 so

[H⁺(aq)] = 10^−4.76 mol dm⁻³

Step 6: Convert the hydrogen ion concentration into its pH value

pH = − log₁₀[H⁺(aq)]

so pH = − log₁₀10^−4.76 = 4.76

Note that this example shows that at the half-equivalence point (i.e. when exactly half of the excess weak acid has been neutralised by the base, so that [acid] = [salt]) the pH = pK_a.