Problem 40 - Chemistry and Chemical Reactivity

Question

At \(20^{\circ} \mathrm{C},\) a saturated aqueous solution of silver acetate, \(\mathrm{AgCH}_{3} \mathrm{CO}_{2},\) contains \(1.0 \mathrm{g}\) of the silver compound dissolved in \(100.0 \mathrm{mL}\) of solution. Calculate \(K_{\mathrm{sp}}\) for silver acetate. \(\mathrm{AgCH}_{3} \mathrm{CO}_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq})\)

Step-by-Step Solution

Answer

The solubility product \( K_{sp} \) for silver acetate is 0.00359.

Step 1: Calculate Molar Mass of Silver Acetate

Silver acetate, \( \mathrm{AgCH}_3\mathrm{CO}_2 \), consists of the elements silver (Ag), carbon (C), hydrogen (H), and oxygen (O). To find its molar mass, sum the atomic masses: \( \text{Ag} = 107.87 \text{g/mol}, \ \text{C} = 12.01 \text{g/mol}, \ \text{H} = 1.01 \text{g/mol}, \text{and} \ \text{O} = 16.00 \text{g/mol}.\)\[ \text{Molar mass of } \mathrm{AgCH}_3\mathrm{CO}_2 = 107.87 + (2 \times 12.01) + (3 \times 1.01) + (2 \times 16.00) = 166.91 \text{ g/mol}. \]

Step 2: Calculate Moles of Silver Acetate

Given that 1.0 g of silver acetate is dissolved in 100.0 mL of solution, calculate the number of moles:\[ \text{Moles of } \mathrm{AgCH}_3\mathrm{CO}_2 = \frac{1.0 \text{ g}}{166.91 \text{ g/mol}} = 0.005992 \text{ moles}. \]

Step 3: Find Molarity of Silver Ions

Since the solution is 100.0 mL (or 0.1 L), the concentration of silver ions, \( \mathrm{Ag}^+ \), in the solution is:\[ \text{Molarity of } \mathrm{Ag}^+ = \frac{0.005992 \text{ moles}}{0.1 \text{ L}} = 0.05992 \text{ M}. \]

Step 4: Apply Solubility Product Expression

For the equilibrium \( \mathrm{AgCH}_3\mathrm{CO}_2 (s) \rightleftharpoons \mathrm{Ag}^+ (aq) + \mathrm{CH}_3\mathrm{CO}_2^- (aq) \), where silver acetate dissolves to form \( \mathrm{Ag}^+ \) and \( \mathrm{CH}_3\mathrm{CO}_2^- \), the solubility product, \( K_{sp} \), is given by:\[ K_{sp} = [\mathrm{Ag}^+][\mathrm{CH}_3\mathrm{CO}_2^-]. \]Since each mole of silver acetate produces one mole of \( \mathrm{Ag}^+ \) and one mole of \( \mathrm{CH}_3\mathrm{CO}_2^- \), we have:\[ K_{sp} = (0.05992) \times (0.05992) = 0.00359. \]

Key Concepts

Equilibrium Expression

In the context of solubility, the equilibrium expression is vital as it helps us understand how a solid dissolves into its ionic components in a solution. Here, we discuss the equilibrium involving silver acetate. The dissociation of silver acetate (\(\mathrm{AgCH}_3\mathrm{CO}_2\)) can be described by the following chemical equation:
\[\mathrm{AgCH}_3\mathrm{CO}_2(s) \rightleftharpoons \mathrm{Ag}^+(aq) + \mathrm{CH}_3\mathrm{CO}_2^-(aq)\]
Here, an equilibrium is established between the solid silver acetate and its ions in solution. The equilibrium expression, known as the solubility product constant (\(K_{sp}\)), reflects this balance.

• The equilibrium expression for this reaction is represented as: \[K_{sp} = [\mathrm{Ag}^+][\mathrm{CH}_3\mathrm{CO}_2^-]\]
• Each ion's concentration is raised to the power of its coefficient in the balanced equation (in this case, both coefficients are 1).
• The product of these concentrations gives us the magnitude of \(K_{sp}\), indicating the solubility of the compound.

Molar Mass Calculation

Calculating the molar mass of a compound is a crucial step in finding the number of moles in a given mass. Let's look at silver acetate, \(\mathrm{AgCH}_3\mathrm{CO}_2\), which comprises silver (Ag), carbon (C), hydrogen (H), and oxygen (O).

• Silver (\(\text{Ag}\)) has an atomic mass of \(107.87 \text{ g/mol}\).
• Carbon (\(\text{C}\)) is \(12.01 \text{ g/mol}\), and considering there are two carbon atoms, it contributes \(2 \times 12.01\).
• Hydrogen (\(\text{H}\)) has \(1.01 \text{ g/mol}\), so for three hydrogen atoms, it adds \(3 \times 1.01\).
• Oxygen (\(\text{O}\)) is \(16.00 \text{ g/mol}\) for two atoms giving \(2 \times 16.00\).

Summing these up, the molar mass becomes:\[107.87 + (2 \times 12.01) + (3 \times 1.01) + (2 \times 16.00) = 166.91 \text{ g/mol}\]This molar mass is critical for converting the mass of silver acetate in a solution to moles, which is essential for further calculations.

Silver Acetate Dissolution

Dissolution refers to the process where a solid compound breaks down into ions in a solution. For silver acetate (\(\mathrm{AgCH}_3\mathrm{CO}_2\)), it dissolves into silver ions (\(\mathrm{Ag}^+\)) and acetate ions (\(\mathrm{CH}_3\mathrm{CO}_2^-\)).

• When one mole of silver acetate dissolves, it produces one mole of \(\mathrm{Ag}^+\) and one mole of \(\mathrm{CH}_3\mathrm{CO}_2^-\), maintaining a 1:1 ratio.
• The molarity of these ions in a solution helps in determining the solubility product, \(K_{sp}\).
• For example, in the solution containing 0.005992 moles of silver acetate in 100.0 mL, the concentrations of both ions would equal 0.05992 \text{ M}.

This dissolution process is key for calculating \(K_{sp}\), as it allows us to find the equilibrium between the ions and the solid phase. By understanding each step thoroughly, one can easily determine the solubility equilibrium of silver acetate.