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Chapter 14: Problem 97

At \(25^{\circ} \mathrm{C}\), the solubility in water of the moderately soluble salt silver acetate, \(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) is \(10.6 \mathrm{~g} / \mathrm{L}\). (a) Write the chemical equation for the dissolving of silver acetate in water. (b) Write the \(K_{\mathrm{sp}}\) expression for silver acetate. (c) Calculate the value of \(K_{\mathrm{sp}}\) (show your work).

Short Answer

Expert verified
(a) The balanced chemical equation for the dissolving of silver acetate in water is: \( \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2(s)} \rightleftharpoons \mathrm{Ag}^{+}_{(aq)} + \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2^{-}}_{(aq)} \) (b) The Ksp expression for silver acetate is: \( K_{\mathrm{sp}} = [\mathrm{Ag}^{+}][\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2^{-}}] \) (c) The value of Ksp for silver acetate is 4.02 x 10鈦宦�.

Step by step solution

01

(a) Write the chemical equation for the dissolving of silver acetate in water.

The balanced chemical equation for the dissolving of silver acetate in water can be written as: \( \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2(s)} \rightleftharpoons \mathrm{Ag}^{+}_{(aq)} + \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2^{-}}_{(aq)} \) The given reaction shows that silver acetate dissociates into silver ions (Ag+) and acetate ions (C2H3O2-) in water.
02

(b) Write the Ksp expression for silver acetate.

The solubility product constant (Ksp) expression for silver acetate can be written as: \( K_{\mathrm{sp}} = [\mathrm{Ag}^{+}][\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2^{-}}] \) Ksp represents the equilibrium constant for the solubility of a slightly soluble ionic compound in water, and it depends only on the concentrations of the ions at equilibrium.
03

(c) Calculate the value of Ksp (show your work).

To calculate the value of Ksp, we need the equilibrium concentrations of Ag+ and C2H3O2-. We are given the solubility of silver acetate in water, which is equal to 10.6 g/L. We can use this information to calculate the molar solubility and then the equilibrium concentrations of the ions. 1) Find the molar mass of silver acetate: \( \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \) = 107.87 (Ag) + 2(12.01) (C) + 3(1.01) (H) + 2(16.00) (O) = 166.92 g/mol 2) Convert the solubility from g/L to mol/L (molar solubility): \( 10.6 \mathrm{g/L} \cdot \frac{1 \mathrm{mol}}{166.92\mathrm{g}} = 0.0634\mathrm{mol/L}\) 3) Since the stoichiometric coefficients are 1:1 in the balanced chemical equation, the equilibrium concentrations of Ag+ and C2H3O2- are equal to the molar solubility: [\(\mathrm{Ag}^+\)] = [\(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2^{-}}\)] = 0.0634 mol/L 4) Calculate the Ksp value using the Ksp expression: \( K_{\mathrm{sp}} = (0.0634) (0.0634) = 4.02 \times 10^{-3}\) The value of Ksp for silver acetate is 4.02 x 10鈦宦�.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equation for Dissolution
When a salt like silver acetate (\(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)) is added to water, it dissolves and dissociates into its ions. This process can be expressed with a chemical equation that specifies how the solid compound breaks down into ions in its dissolved or aqueous form.
The balanced chemical equation for the dissolution of silver acetate in water is:
\( \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2(s)} \rightleftharpoons \mathrm{Ag}^{+}_{(aq)} + \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2^{-}}_{(aq)} \)
  • The arrow pointing in both directions (\(\rightleftharpoons\)) indicates equilibrium, meaning both the forward and reverse reactions occur simultaneously.
  • \(\mathrm{Ag}^{+}\) is the silver ion, and \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2^{-}}\) is the acetate ion.
This equation shows that for every mole of silver acetate that dissolves, one mole of silver ions and one mole of acetate ions are produced. This reaction reaches a point where the rate of dissolution equals the rate of precipitation, establishing equilibrium.
Molar Solubility Calculation
Molar solubility is a measure of how many moles of a substance can dissolve in a liter of solution to reach saturation. This is closely linked to how easily a compound dissociates into ions.
The solubility of silver acetate is given as 10.6 g/L. To convert this solubility into molar solubility (mol/L), we need its molar mass.
First, calculate the molar mass of silver acetate:
\[ \text{Molar Mass of } \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} = 107.87 + 2(12.01) + 3(1.01) + 2(16.00) = 166.92 \, \text{g/mol} \]
Then, we convert the solubility from grams per liter to moles per liter:
\[ 10.6 \, \text{g/L} \times \frac{1 \, \text{mol}}{166.92 \, \text{g}} = 0.0634 \, \text{mol/L} \]
The result shows the molar solubility is 0.0634 mol/L. This tells us how many moles of silver acetate will dissolve in a liter of water to reach equilibrium.
Silver Acetate Equilibrium
At equilibrium, the concentration of ions in the solution remains constant. Silver acetate establishes a dynamic balance between the dissolved ions and the undissolved solid. The extent of its dissolution is quantified by the solubility product constant, or \(K_{\text{sp}}\).
The expression for \(K_{\text{sp}}\) is based on the concentration of the ions that result from dissociation:
\[ K_{\text{sp}} = [\mathrm{Ag}^{+}][\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2^{-}}] \]
For silver acetate, since the stoichiometry of the dissolution involves a 1:1 ratio, both the silver ion and the acetate ion have concentrations equal to the molar solubility. If the molar solubility is \(0.0634 \, \text{mol/L}\), then:
  • \([\mathrm{Ag}^{+}] = 0.0634 \, \text{mol/L}\)
  • \([\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2^{-}}] = 0.0634 \, \text{mol/L}\)
The calculated \(K_{\text{sp}}\) for silver acetate is then:\[ K_{\text{sp}} = (0.0634)(0.0634) = 4.02 \times 10^{-3} \]
This constant indicates the position of equilibrium鈥攈ow far the dissolution of silver acetate proceeds before equilibrium is reached, reflecting its limited solubility in water.

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Most popular questions from this chapter

\(K_{\text {eq }}=3.9 \times 10^{-11}\) for the dissolution of calcium fluoride in water: \(\mathrm{CaF}_{2}(s) \rightleftarrows \mathrm{Ca}^{2+}(a q)+2 \mathrm{~F}^{-}(a q)\) (a) What is another name for \(K_{\mathrm{eq}}\) for this reaction? (b) If the equilibrium calcium ion concentration in a saturated aqueous solution of calcium fluoride is \(3.3 \times 10^{-4} \mathrm{M}\), what is the equilibrium fluoride ion concentration? (c) Which is larger, the rate constant for the forward reaction or the rate constant for the reverse reaction? (d) Which is larger, \(E_{a}\) for the forward reaction or \(E_{\mathrm{a}}\) for the reverse reaction? (e) Which is larger, the rate of the forward reaction or the rate of the reverse reaction? (f) For lithium carbonate, \(K_{\mathrm{sp}}=0.0011\). Write the balanced chemical equation and the equilibrium expression for the dissolution of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) in water. (g) Which is more soluble in water, calcium fluoride or lithium carbonate?

On the basis of \(K_{\text {eq }}\) values, which reaction goes essentially to completion? How would you describe the other reaction? (a) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \leftrightarrows 2 \mathrm{H}_{2} \mathrm{O}(g)\); \(K_{\mathrm{eq}}=3 \times 10^{81}\) (b) \(2 \mathrm{HF}(g) \rightleftarrows \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) \(K_{\mathrm{eq}}=1 \times 10^{-95}\)

In theory, all reactions are reversible, but in practice, some are not. Explain why.

Write the equilibrium constant expression for (a) \(2 \mathrm{FeCl}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) \(+6 \mathrm{HCl}(g)\) (b) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \rightleftarrows 2 \mathrm{Fe}(l)+3 \mathrm{CO}_{2}(g)\) (c) \(\mathrm{PbSO}_{4}(s) \rightleftarrows \mathrm{Pb}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\) (d) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{2} \mathrm{CO}_{3}(a q)\)

At \(25^{\circ} \mathrm{C}\), the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in water is \(2.60 \times 10^{-6} \mathrm{M}\). What are the equilibrium concentrations of the cation and the anion in a saturated solution?
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