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Chapter 14: Problem 26

Vinegar-a solution of acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) in water \(-\) is used in varying concentrations for different household tasks. The following equilibrum exists in vinegar. $$\begin{array}{c}{\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows} \\ {\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)}\end{array}$$ If the concentration of the acetic acid solution at equilibrium is 3.00 \(\mathrm{M}\) and the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=7.22 \times 10^{-3},\) what is the \(K_{e q}\) value for acetic acid?

Short Answer

Expert verified
The equilibrium constant (\(K_{eq}\)) for the reaction of acetic acid in water is \(1.76 \times 10^{-5}\). This small value suggests that acetic acid remains largely un-ionized at equilibrium in an aqueous solution.

Step by step solution

01

Understand the Equilibrium Expression

The equilibrium expression for a chemical reaction signifies the ratio between product concentrations and reactant concentrations each raised to the power of their respective stoichiometric coefficients. In this case, the equilibrium expression for the reaction is \(K_{eq} = [\mathrm{H}_3 \mathrm{O}^{+}][\mathrm{CH}_3 \mathrm{COO}^{-}]/[\mathrm{CH}_3 \mathrm{COOH}] \cdot [\mathrm{H}_2 \mathrm{O}]\)
02

Simplify the Equilibrium Expression

Since water is a solvent in this case and remains in excess throughout the reaction, its concentration, [\mathrm{H}_2\mathrm{O}], is considered approximately constant and is encompassed in the equilibrium constant itself. It need not explicitly feature in our equilibrium expression. So, our simplified equilibrium expression becomes: \(K_{eq} = [\mathrm{H}_3 \mathrm{O}^{+}][\mathrm{CH}_3 \mathrm{COO}^{-}]/[\mathrm{CH}_3 \mathrm{COOH}]\)
03

Substitute the Given Values

The problem provides the concentrations of the chemicals at equilibrium. Substituting them into the simplified equilibrium expression, we get: \(K_{eq} = [(7.22 \times 10^{-3})(7.22 \times 10^{-3})]/[3]\)
04

Calculation of \(K_{eq}\)

The equation \(K_{eq} = [(7.22 \times 10^{-3})(7.22 \times 10^{-3})]/[3]\) simplifies to \(K_{eq} = 1.76 \times 10^{-5}\).
05

Interpret the Results

The obtained equilibrium constant, \(K_{eq}\), is a small value, indicating that the reaction tends to favor the reactant side at equilibrium under the given conditions. In other words, acetic acid doesn't ionize significantly in the solution.

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Most popular questions from this chapter

Why must a balanced chemical equation be used when determining \(K_{e q} ?\)

Determine the value of the equilibrium constant for each reaction below assuming that the equilibrium concentrations are as specified. $$\text {(a.)}A+B \rightleftarrows C ;[A]=2.0 ;[B]=3.0 ;[C]=4.0$$ $$\begin{array}{l}{\text { b. } \mathrm{D}+2 \mathrm{E} \rightleftarrows \mathrm{F}+3 \mathrm{G} ;[\mathrm{D}]=1.5 ;[\mathrm{E}]=2.0} \\\ {[\mathrm{F}]=1.8 ;[\mathrm{G}]=1.2}\end{array}$$ $$\mathrm{c} \cdot \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NH}_{3}(\mathrm{g}) ;\left[\mathrm{N}_{2}\right]=0.45\( \)\left[\mathrm{H}_{2}\right]=0.14 ;\left[\mathrm{NH}_{3}\right]=0.62$$

Describe and explain how the concentrations of \(A, B, C,\) and \(D\) change from the time when A and \(B\) first combine to the point at which equilibrium is established for the reaction $$A+B \rightleftarrows C+D$$

What is the change in the rate of the forward reaction after the reaction is at equilibrium?

The reaction below has an equilibrium constant of \(4.9 \times 10^{11}\) . $$\mathrm{Fe}(\mathrm{OH})_{2}(s)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftarrows \mathrm{Fe}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)$$ Write the equilibrium constant expression, and determine the concentration of \(\mathrm{Fe}^{2+}\) ions in equilibrium when the hydronium ion concentration is \(1.0 \times 10^{-7} \mathrm{mol} / \mathrm{L}\)
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