Salt Hydrolysis

Salt hydrolysis

Does neutralisation mean that the pH of the solution is \(\text{7}\)? No. At the equivalence point of a reaction, the pH of the solution need not be \(\text{7}\). This is because of the interaction of the salt (formed by the reaction) and water.

At the equivalence point of an acid-base neutralisation reaction there is salt and water. The ions of water interact with the salt present and form a small quantity of excess hydronium ions (\(\text{H}_{3}\text{O}^{+}\)) or hydroxide ions (\(\text{OH}^{-}\)). This leads to pH values that are not equal to \(\text{7}\).

A simple rule for determining the likely pH of a solution is as follows:

• A \(\color{red}{\textbf{strong }\text{acid}} + \color{blue}{\textbf{strong }\text{base}}\) form a \(\color{darkgreen}{\textbf{neutral}}\) salt and water \(\color{darkgreen}{\textbf{solution}}\):

  \(\to\) pH = \(\text{7}\).

  \(\color{red}{\text{H}_{2}{\text{SO}}_{4}{\text{(l)}}} + \color{blue}{\text{NaOH(s)}} \to \color{darkgreen}{\textbf{Na}_{2}{\textbf{SO}}_{4}{\textbf{(aq)}}} + \color{darkgreen}{\text{H}_{2}{\text{O(l)}}}\)

• A \(\color{red}{\textbf{weak }\text{acid}} + \color{blue}{\textbf{strong }\text{base}}\) form a \(\color{blue}{\textbf{weak basic}}\) salt and water \(\color{blue}{\textbf{solution}}\):

  \(\to\) pH = approximately \(\text{9}\).

  \(\color{red}{\text{HF(l)}} + \color{blue}{\text{NaOH(s)}} \to \color{blue}{\textbf{NaF(aq)}} + \color{blue}{\text{H}_{2}{\text{O(l)}}}\)

• A \(\color{red}{\textbf{strong }\text{acid}} + \color{blue}{\textbf{weak }\text{base}}\) form a \(\color{red}{\textbf{weak acidic}}\) salt and water \(\color{red}{\textbf{solution}}\):

  \(\to\) pH = approximately \(\text{5}\).

  \(\color{red}{\text{H}_{2}{\text{SO}}_{4}{\text{(l)}}} + \color{blue}{\text{NH}_{3}{\text{(l)}}} \to \color{red}{\textbf{(NH}_{4}{\textbf{)}}_{2}{\textbf{SO}}_{4}{\textbf{(aq)}}} + \color{red}{\text{H}_{2}{\text{O(l)}}}\)