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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

Expert verified

(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

01

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.

02

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)

03

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

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Most popular questions from this chapter

Convert the values of Kc to values of KP or the values of KP to values of Kc .

\((a)\,C{l_2}\left( g \right) + B{r_2}\left( g \right)\rightleftharpoons 2BrCl(g)\,\,\,{K_C} = 4.7 \times {10^{ - 2}}\,at\,25^\circ C\)

\((b)2{\rm{S}}{{\rm{O}}_2}(g) + {{\rm{O}}_2}(g)\rightleftharpoons 2{\rm{S}}{{\rm{O}}_3}(g){K_P} = 48.2 at {500^\circ} {\rm{C}}\)

\((c){\rm{CaC}}{{\rm{l}}_2} \cdot 6{{\rm{H}}_2}{\rm{O}}(s)\rightleftharpoons {\rm{CaC}}{{\rm{l}}_2}(s) + 6{{\rm{H}}_2}{\rm{O}}(g){K_P} = 5.09 \times {10^{ - 44}}at{25^\circ }{\rm{C}}\)

\((d){{\rm{H}}_2}{\rm{O}}(l)\rightleftharpoons {{\rm{H}}_2}{\rm{O}}(g)\,{K_P} = 0.196\,at\,{60^\circ }{\rm{C}}\)

Question: Antimony pentachloride decomposes according to this equation:

An equilibrium mixture in a 5.00-L flask at 4480C contains 3.85 g of \({\rm{SbC}}{{\rm{l}}_5}\),9.14 g of \({\rm{SbC}}{{\rm{l}}_3}\)and 2.84 g of \({\rm{C}}{{\rm{l}}_2}\).How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature?

Question:Calculate the value of the equilibrium constant \({K_P}\) for the reaction \(2NO(g) + C{l_2}(g) \rightleftharpoons 2NOCl(g)\) from these equilibrium pressures: NO, \(0.050atm;C{l_2},0.30atm;NOCl,1.2atm\)

Question:Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures. \(C{H_4}(g) + {H_2}O(g) \rightleftharpoons 3{H_2}(g) + CO(g)\)

Assume that the change in pressure of \({H_2}S\) is small enough to be neglected in the following problem.(a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H2S with an initial pressure of 0.824 atm.

\(2{H_2}S(g) \rightleftharpoons 2{H_2}(g) + {S_2}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_p} = 2.2 \times {10^{( - 6)}}\)

(b) Show that the change is small enough to be neglected.

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