bStrong acids/b
A strong acid ucompletely/u dissociates in water to form Hsup+/sup ions. Common strong acids include hydrochloric acid (HCl), sulphuric acid (Hsub2/subSOsub4/sub) and nitric acid (HNOsub3/sub).
(HCl(aq) ⟶ H^+{(aq)} + Cl-{(aq)} )
bCalculating pH of strong acids/b
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 Hsup+/sup concentration.
For example: Calculate the pH of 0.1 mol dmsup-3/sup hydrochloric acid.
The molar ratio of HCl to Hsup+/sup is 1:1, so 0.1 mol dmsup-3/sup HCl also contains 0.1 mol dmsup-3/sup Hsup+/sup.
(pH = - \log10{H+} )
(pH = - \log10{0.1} )
(pH = 1 )
bWeak acids/b
A weak acid upartially/u dissociates in water. Common weak acids include ethanoic acid (CHsub3/subCOOH) 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 Hsup+/sup and Asup-/sup, 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 Hsup+/sup 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, Ksuba/sub.
(Ka = \dfrac{[H^+]A-}{HA} )
The smaller the value of Ksuba/sub, the weaker the acid. For example, the Ksuba/sub of ethanoic acid is 1.7 x 10sup-5/sup mol dmsup-3/sup.
Because Ksuba/sub values are typically very small, the values are often converted into pKsuba/sub. The relationship between Ksuba/sub and pKsuba/sub is the same as that for Hsup+/sup and pH.
( pKa = - \log_{10}{K_a} )
The larger the value of pKsuba/sub, the weaker the acid. For example, the pKsuba/sub of ethanoic acid is 4.8. pKa does not have any units.
bCalculating pH of weak acids/b
Because a weak acid partially dissociates, calculating pH is more complicated.
Consider the Ksuba/sub for a weak acid (HA):
(Ka = \dfrac{[H^+]A-}{HA} )
Since the molar ratio of Hsup+/sup and Asup-/sup is 1:1, the concentration of Hsup+/sup is identical to that of Asup-/sup, so the equation can be written as:
(Ka = \dfrac{H+^2}{HA} )
Note: In reality, the concentration of Hsup+/sup is not identical to that of Asup-/sup, because water also dissociates to produce some Hsup+/sup. To simplify the calculation, you can assume that this is negligible and use Hsup+/sup = Asup-/sup. In addition, the value of HA will have decreased slightly, because a tiny proportion of the HA will have dissociated into Hsup+/sup and Asup-/sup. 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 Hsup+/sup. Rearranging the equation gives:
(H+^2 = K_aHA )
(H+ = \sqrt{KaHA} )
This allows us to calculate the pH:
(pH = - \log10{H+} )
For example: Calculate the pH of 0.1 mol dmsup-3/sup ethanoic acid (Ksuba/sub = 1.7 x 10sup-5/sup mol dmsup-3/sup).
(CH3COOH ⇌ H^+ + CH_3COO^- )
(Ka = \dfrac{[H^+]CH3COO-}{CH3COOH} = \dfrac{H+^2}{CH3COOH} )
(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 = - \log10{1.30 × 10-3} = 2.88 )
bCalculating Ksuba/sub for a weak acid given the pH of a solution containing a known mass of acid/b
If the pH is known, then Hsup+/sup 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 Ksuba/sub by inserting the values of Hsup+/sup and HA into the equation below:
(Ka = \dfrac{H+^2}{HA} )
bDiluting strong and weak acids/b
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.
bDifference in enthalpy changes of neutralisation values for strong and weak acids/b
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 molsup-1/sup, while for sodium hydroxide and hydrochloric acid is -57.9 kJ molsup-1/sup.