Conjugate Acid-Base Pairs Continued

Conjugate acid-base pairs

Look at the reaction between hydrochloric acid and ammonia to form ammonium and chloride ions:

\(\text{HCl}(\text{aq}) + \text{NH}_{3}(\text{g})\) \(\leftrightharpoons\) \(\text{NH}_{4}^{+}(\text{aq}) + \text{Cl}^{-}(\text{aq})\)

Looking at the forward reaction (i.e. the reaction that proceeds from left to right), the changes that take place can be shown as follows:

• \({\color{red}{\text{HCl(aq)}}} \to {\text{Cl}}^{-}{\text{(aq)}} + {\color{darkgreen}{{\text{H}}^{+}{\text{(aq)}}}}\)

• \({\color{blue}{\text{NH}_{3}{\text{(g)}}}} + {\color{darkgreen}{\text{H}^{+}{\text{(aq)}}}} \to {\text{NH}_{4}^{+}{\text{(aq)}}}\)

• In the forward reaction, HCl is a proton donor (\({\color{red}{\textbf{acid}}}\)) and \(\text{NH}_{3}\) is a proton acceptor (\({\color{blue}{\textbf{base}}}\)).

Looking at the reverse reaction (i.e. the reaction that proceeds from right to left), the changes that take place are as follows:

• \({\color{blue}{\text{Cl}^{-}{\text{(aq)}}}} + {\color{darkgreen}{{\text{H}}^{+}{\text{(aq)}}}} \to {\text{HCl(aq)}}\)

• \({\color{red}{\text{NH}_{4}^{+}{\text{(aq)}}}} \to {\text{NH}_{3}{\text{(g)}}} + {\color{darkgreen}{{\text{H}}^{+}{\text{(aq)}}}}\)

• In the reverse reaction, the chloride ion (\(\text{Cl}^{-}\)) is the proton acceptor (\({\color{blue}{\textbf{base}}}\)) and the ammonium ion (\(\text{NH}_{4}^{+}\)) is the proton donor (\({\color{red}{\textbf{acid}}}\)).

In the forward reaction \(\color{red}{\text{HCl}}\) donates a proton (\(\text{H}^{+}\)) to form \(\color{blue}{\text{Cl}^{-}}\). In the reverse reaction \(\color{blue}{\text{Cl}^{-}}\) accepts a proton to form \(\color{red}{\text{HCl}}\). \(\color{blue}{\text{Cl}^{-}}\) is the \(\color{blue}{\text{conjugate base}}\) of the \(\color{red}{\text{acid HCl}}\). So \(\text{HCl}\) and \(\text{Cl}^{-}\) are a \({\color{orange}{\textbf{conjugate acid-base pair}}}\).

Similarly, in the forward reaction \(\color{blue}{\text{NH}_{3}}\) accepts a proton (\(\text{H}^{+}\)) to form \(\color{red}{{\text{NH}}_{4}^{+}}\). In the reverse reaction \(\color{red}{{\text{NH}}_{4}^{+}}\) donates a proton to form \(\color{blue}{\text{NH}_{3}}\). \(\color{red}{\text{NH}_{4}^{+}}\) is the \(\color{red}{\text{conjugate acid}}\) of the \(\color{blue}{\text{base NH}_{3}}\). So \(\text{NH}_{3}\) and \(\text{NH}_{4}^{+}\) are a \({\color{purple}{\textbf{conjugate acid-base pair}}}\).

The reaction between ammonia and water is another example:

Example: Conjugate Acid-Base Pairs

Question

Determine the conjugate acid-base pairs for the following reaction:

\(\text{HNO}_{3}(\text{aq}) + \text{OH}^{-}(\text{aq})\) \(\rightleftharpoons\) \(\text{NO}_{3}^{-}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})\)

Step 1: Which reactant is an acid and which is a base?

\(\text{HNO}_{3}\) is nitric acid. It donates a proton in the forward reaction:

\(\text{HNO}_{3}(\text{aq})\) \(\to\) \(\text{NO}_{3}^{-}(\text{aq}) + \text{H}^{+}(\text{aq})\)

\(\text{OH}^{-}\) accepts a proton in the forward reaction and is therefore the base:

\(\text{OH}^{-}(\text{aq}) + \text{H}^{+}(\text{aq})\) \(\to\) \(\text{H}_{2}\text{O}(\text{l})\)

Step 2: Which product is the conjugate base of the acid?

Nitric acid (\(\text{HNO}_{3}\)) donates a proton to become \(\text{NO}_{3}^{-}\). In the reverse reaction \(\text{NO}_{3}^{-}\) accepts a proton to become \(\text{HNO}_{3}\). Therefore \(\text{NO}_{3}^{-}\) is the conjugate base of \(\text{HNO}_{3}\).

\(\text{NO}_{3}^{-}(\text{aq}) + \text{H}^{+}(\text{aq})\) \(\to\) \(\text{HNO}_{3}(\text{aq})\)

Step 3: Which product is the conjugate acid of the base?

\(\text{OH}^{-}\) accepts a proton to become \(\text{H}_{2}\text{O}\). In the reverse reaction \(\text{H}_{2}\text{O}\) donates a proton to become \(\text{OH}^{-}\). Therefore \(\text{H}_{2}\text{O}\) is the conjugate acid of \(\text{OH}^{-}\).

\(\text{H}_{2}\text{O}(\text{l})\) \(\to\) \(\text{OH}^{-}(\text{aq}) + \text{H}^{+}(\text{aq})\)

Step 4: Label the conjugate acid-base pairs in this reaction