Comparing acid strengths

The relative strengths of acids can be determined by measuring the pH of equimolar aqueous solutions at the same temperature.

Consider the following data for 0.100 mol dm^-3 aqueous solutions of various acids at 298 K:

Interpreting the data:

• HCl has the lowest pH, making it the strongest acid in this group.
• CH₃COOH has the highest pH, indicating it is the weakest acid.

Comparing base strengths

A similar approach can be used to compare the strength of bases. However, for bases, a higher pH indicates a stronger base.

Consider the following data for 0.100 mol dm^-3 aqueous solutions of various bases at 298 K:

Interpreting the data:

• NaOH has the highest pH, making it the strongest base in this group.
• NH₃ has the lowest pH, indicating it is the weakest base.

The pH of salt solutions

The pH of a salt solution depends on the strengths of the acid and base from which the salt is derived:

1. Strong acid + strong base - The resulting salt solution is neutral (pH = 7). For example, hydrochloric acid reacts with sodium hydroxide to form sodium chloride (NaCl), a neutral salt.

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

1. Weak acid + strong base - The resulting salt solution is alkaline (pH > 7). For example, ethanoic acid reacts with sodium hydroxide to form sodium ethanoate (CH₃COONa), an alkaline salt.

CH₃COOH_(aq) + NaOH_(aq) ➔ CH₃COONa_(aq) + H₂O_(l)

1. Strong acid + weak base - The resulting salt solution is acidic (pH < 7). For example, hydrochloric acid reacts with ammonia to form ammonium chloride (NH₄Cl), an acidic salt.

HCl_(aq) + NH_3(aq) ➔ NH₄Cl_(aq)

In summary, when either the parent acid or base is weak, the resulting salt solution will not be neutral. The stronger of the two will determine whether the solution is acidic or alkaline.

Effect of dilution on pH

Diluting an acid decreases the concentration of H⁺ ions in the solution, resulting in an increase in pH. The extent of the pH change depends on whether the acid is strong or weak.

For a strong acid like hydrochloric acid (HCl):

• Diluting the acid by a factor of 10 increases the pH by 1 unit.
• [H⁺] = [acid]
• pH = $-$log₁₀[acid]

For a weak acid like ethanoic acid (CH₃COOH):

• Diluting the acid by a factor of 10 increases the pH by 0.5 units.
• [H⁺] = $\sqrt{(K_a\times [acid])}$
• pH = $-$log₁₀$\sqrt{(K_a\times [acid])}$

The table below shows the pH values of HCl and CH₃COOH at different concentrations at 298 K:

Calculating the Ka of a weak acid

The K_a value of a weak acid can be determined if the concentration and pH of a sample of the weak acid are known.

Use the simplified formula for K_a:

${\text{K}}_{\text{a }}=\frac{{[{\text{H}}^{+}]}^2}{[\text{HA}]}$

Where:

• [H⁺] is calculated from the pH, using the equation [H⁺] = 10^-pH.
• [HA] is the initial concentration of the weak acid.

Worked example 1 - Calculating the Ka of hydrofluoric acid

A solution is made by dissolving 0.200 g of hydrofluoric acid (HF) in 100 cm³ of water. The resulting solution has a pH of 3.14. Calculate the K_a for hydrofluoric acid. Give your answer to 3 significant figures.

Step 1: Conversion of cm³ into dm³

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

100 cm³ = 0.100 dm³

Step 2: Calculate number of moles of HF

n = $\frac{m}{M_r}=\frac{0.200}{20.0}=0.0100$ mol

Step 3: Calculate [HF]

c = $\frac{n}{V}=\frac{0.0100}{0.100}=0.100$ mol dm^-3

Step 4: Calculate [H⁺]

[H⁺] = 10^-pH = 10^-3.14 = 7.24 x 10^-4 mol dm^-3

Step 5: K_a equation

${\text{K}}_{\text{a }}=\frac{{[{\text{H}}^{+}]}^2}{[\text{HF}]}$

Step 6: Substition and correct evaluation

${\text{K}}_{\text{a }}=\frac{{(7.24\times {10}^{-4})}^2}{0.100}=5.25\times {10}^{-6}{\text{ mol dm}}^{-3}$

Therefore, the K_a of hydrofluoric acid in this solution is 5.25 × 10^-6 mol dm^-3.