Strong acids

A strong acid completely dissociates in water to form H⁺ ions. Common strong acids include hydrochloric acid (HCl), sulphuric acid (H₂SO₄) and nitric acid (HNO₃).

\(HCl_{(aq)} ⟶ H^+_{(aq)} + Cl^-_{(aq)} \)


Calculating pH of strong acids

Because a strong acid completely dissociates, calculating the pH if you know its concentration is straightforward, since the molar ratio can be used to determine H⁺ concentration.

For example: Calculate the pH of 0.1 mol dm^-3 hydrochloric acid.

The molar ratio of HCl to H⁺ is 1:1, so 0.1 mol dm^-3 HCl also contains 0.1 mol dm^-3 H⁺.

\(pH = - \log_{10}{[H^+]} \)

\(pH = - \log_{10}{0.1} \)

\(pH = 1 \)


Weak acids

A weak acid partially dissociates in water. Common weak acids include ethanoic acid (CH₃COOH) and methanoic acid (HCOOH).

The dissociation of a weak acid (HA) can be expressed as an equilibrium. Only a minority of HA have dissociated into H⁺ and A^-, with the majority remaining as HA molecules.

\(HA_{(aq)} ⇌ H^+_{(aq)} + A^-_{(aq)} \)

The position of equilibrium varies from one weak acid to another. The further to the left (HA) it lies, the weaker the acid is. This is because fewer H⁺ is dissociated.

The position of equilibrium (and therefore the strength of the weak acid) can be expressed by writing an equilibrium constant for the reaction. The equilibrium constant for the dissociation of a weak acid is known as the acid dissociation constant, K_a.

\(K_a = \dfrac{[H^+][A^-]}{[HA]} \)

The smaller the value of K_a, the weaker the acid. For example, the K_a of ethanoic acid is 1.7 x 10^-5 mol dm^-3.

Because K_a values are typically very small, the values are often converted into pK_a. The relationship between K_a and pK_a is the same as that for [H⁺] and pH.

\( pK_a = - \log_{10}{K_a} \)

The larger the value of pK_a, the weaker the acid. For example, the pK_a of ethanoic acid is 4.8. pKa does not have any units.


Calculating pH of weak acids

Because a weak acid partially dissociates, calculating pH is more complicated.

Consider the K_a for a weak acid (HA):

\(K_a = \dfrac{[H^+][A^-]}{[HA]} \)

Since the molar ratio of H⁺ and A^- is 1:1, the concentration of H⁺ is identical to that of A^-, so the equation can be written as:

\(K_a = \dfrac{[H^+]^2}{[HA]} \)

Note: In reality, the concentration of H⁺ is not identical to that of A^-, because water also dissociates to produce some H⁺. To simplify the calculation, you can assume that this is negligible and use [H⁺] = [A^-]. In addition, the value of [HA] will have decreased slightly, because a tiny proportion of the HA will have dissociated into H⁺ and A^-. To simplify the calculation, you can assume that none of the HA have dissociated, and use the value of [HA] provided.

In order to calculate pH, we must find the value of [H⁺]. Rearranging the equation gives:

\([H^+]^2 = K_a[HA] \)

\([H^+] = \sqrt{K_a[HA]} \)

This allows us to calculate the pH:

\(pH = - \log_{10}{[H^+]} \)

For example: Calculate the pH of 0.1 mol dm^-3 ethanoic acid (K_a = 1.7 x 10^-5 mol dm^-3).

\(CH_3COOH ⇌ H^+ + CH_3COO^- \)

\(K_a = \dfrac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \dfrac{[H^+]^2}{[CH_3COOH]} \)

\(1.7 × 10^{-5} = \dfrac{[H^+]^2}{[0.1]} \)

\([H^+]^2 = 1.7 × 10^{-5} × 0.1 \)

\([H^+] = \sqrt{1.7 × 10^{-5} × 0.1} = 1.30 × 10^{-3} \)

\(pH = - \log_{10}{[1.30 × 10^{-3}]} = 2.88 \)


Calculating K_a for a weak acid given the pH of a solution containing a known mass of acid

If the pH is known, then [H⁺] can be calculated. If the mass of acid is known, then you can calculate the moles of acid present, and therefore [HA].

You can then calculate the value of K_a by inserting the values of [H⁺] and [HA] into the equation below:

\(K_a = \dfrac{[H^+]^2}{[HA]} \)


Diluting strong and weak acids

Diluting a strong acid 10 times will increase its pH by 1, because pH is a logarithmic scale. Diluting it 100 and 1000 times will increase its pH by 2 and 3 respectively.

Because weak acids are not fully dissociated, diluting them causes the equilibrium to shift to oppose the change (Le Chatelier's Principle). Therefore diluting a weak acid 10 times will increase its pH by less than 1.


Difference in enthalpy changes of neutralisation values for strong and weak acids

The enthalpy of neutralisation of a weak acid with a strong base is less than the enthalpy of neutralisation of a strong acid with a strong base as some energy is used in dissociating the weak acid fully.

For example, the enthalpy of neutralisation for sodium hydroxide and ethanoic acid is -56.1 kJ mol^-1, while for sodium hydroxide and hydrochloric acid is -57.9 kJ mol^-1.