Question

Acetic acid is a weak acid that reacts with water according to this equation:

\(C{H_3}C{O_2}H(aq) + {H_2}O(aq) \rightleftharpoons {H_3}{O^ + }(aq) + C{H_3}CO_2^ - (aq)\)

Will any of the following increase the percent of acetic acid that reacts and produces \(C{H_3}CO_2^ - \)ion?

(a) Addition of \(HCl\)

(b) Addition of \(NaOH\)

(c) Addition of \(NaC{H_3}C{O_2}\)

Short answer

(b) Addition of \({\rm{NaOH}}\) increases the percentage of acetic acid that reacts and produces \({\rm{C}}{{\rm{H}}_3}{\rm{CO}}_2^ - \).

Step-by-step solution

Increasing percentage of acetic acid reacting and producing ion

The reaction

\({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{O}}_2}{\text{H}}({\text{aq}}) + {{\text{H}}_2}{\text{O}}({\text{aq}}) \rightleftharpoons {{\text{H}}_3}{\text{O}}({\text{aq}}) + {\text{C}}{{\text{H}}_3}{\text{C}}{{\text{O}}_2}({\text{aq}})\)

Let us see what will increase the percent of acetic acid that reacts and produces \({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}^ - \)ion.

Increasing percentage of acetic acid after addition of HCl

\({\rm{HCl}}\) is a strong acid and dissociates completely into \({\rm{H}}{{\rm{\;}}^ + }{\rm{and\; C}}{{\rm{l}}^ - }\)ions. Therefore, addition of \({{\rm{H}}^ + }\)ions will move the equilibrium to the left, resulting increase in concentration of \({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}\) (decrease the percent of acetic acid that reacts)

Increasing percentage of acetic acid after addition of NaOH

\({\rm{NaOH}}\)is a strong base, and dissociates completely into \({\rm{Na}}{{\rm{\;}}^ + }{\rm{and\;}}O{H^ - }\)ions. The \(O{H^ - }\)ion will react with \({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}\) (by taking its \({{\rm{H}}^ + }\)ion) and produce \({\rm{C}}{{\rm{H}}_3}{\rm{CO}}_2^ - \). Therefore, the concentration of \({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}\)will decrease (increase the percent of acetic acid that reacts), and the concentration of \({\rm{C}}{{\rm{H}}_3}{\rm{CO}}_2^ - \) will increase