Define pH. pH cannot be calculated directly from the molar concentration of a weak acid or weak base. Why? Derive an equation for the pH of a weak acid.

Explanation

The pH of a solution is defined as the negative logarithim to base 20 of Hydrogen ion concentration (or)`p^(H)="log"_(10)[H^(+)]`
Strong acid and strong bases ionise completely in their aquous solutiosn. So the molar concentrations of strong acids or strong bases are equal to the molar concentrations of `H^(+)` and `OH^(-)` ions produced by them.
Weak acids and weak bases do not ionise completely. So the molar concentrations of `H^(+)` ions or `OH^(-)` ions produced by them are not equal to the molar concentrations of weak acids or bases from whcih they are produced. Hence pH of the aqueous solutions of weak acid and weak bases cannot be calculated from their molar concentrations.
`pH` of a weak acid:
Ionisation of weak acid
`underset(C(1-alpha))(HX(aq))+H_(2)O(l)hArr underset(C alpha)(H_(3)O^(+))(eq)+underset(Calpha)(X)(aq)`
Applying law of mass action
`K_(a)=([H_(3)O^(+)][X^(-)])/([HX]) K_(a)` is acid dissociationn constant.
`=(C alpha xxC alpha)/(C(1-alpha))`
Since the weak acid `alpha` is very small as compared to 1, `alpha` in the denominator can be neglected. Then
`K_(a)=C alpha^(2).` or `alpha=sqrt((K_(a))/C)`
`[H^(+)]=Cxxalpha=Csqrt((K_(a))/C)=sqrt(K_(a)xxC)`
`p^(H)=-log[H^(+)]=-logsqrt(K_(a)xxC)`
`:.p^(H)=1//2P_(K_(a))-1//2logC`