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Chapter 14: Problem 113

The solubility of silver acetate in water at \(20{ }^{\circ} \mathrm{C}\) is \(10.5 \mathrm{~g} / \mathrm{L}\) of solution. Calculate \(\mathrm{K}_{\mathrm{sp}}\) for silver acetate.

Short Answer

Expert verified
The solubility product constant (K_sp) for silver acetate at 20°C is 0.00396.

Step by step solution

01

Calculate the number of moles of silver acetate dissolving in water

First, we'll convert the solubility of silver acetate given in g/L into moles/L (M). The molar mass of silver acetate (AgC₂H₃O₂) is: Molar mass of AgC₂H₃O₂ = 107.87 g/mol (Ag) + 12.01 g/mol (C) × 2 + 1.01 g/mol (H) × 3 + 16.00 g/mol (O) × 2 = 166.92 g/mol Now, we'll convert the solubility of silver acetate (10.5 g/L) into moles/L (M): \( \textrm{Molarity (M)} = \frac{10.5 \textrm{ g/L}}{166.92 \textrm{ g/mol}} = 0.0629 \textrm{ M} \) The concentration of silver acetate in water is 0.0629 M.
02

Write the balanced chemical equation for the dissolution of silver acetate

The dissolution of silver acetate in water can be represented by the following balanced chemical equation: \[ AgC_{2}H_{3}O_{2}(s) \rightleftharpoons Ag^{+}(aq) + C_{2}H_{3}O_{2}^{-}(aq) \]
03

Find the concentrations of Ag⁺ and C₂H₃O₂⁻ ions in the water

For every 1 mol of silver acetate dissolved, it gets separated into 1 mol of Ag⁺ ions and 1 mol of C₂H₃O₂⁻ ions. Therefore, the concentrations of Ag⁺ and C₂H₃O₂⁻ ions are equal to the concentration of silver acetate: Concentration of Ag⁺ ions: [Ag⁺] = 0.0629 M Concentration of C₂H₃O₂⁻ ions: [C₂H₃O₂⁻] = 0.0629 M
04

Calculate the K_sp for silver acetate

The solubility product constant (K_sp) for silver acetate can be calculated using the following equation: \[ K_{sp} = [Ag^{+}][C_{2}H_{3}O_{2}^{-}] \] Plugging in the concentrations of Ag⁺ and C₂H₃O₂⁻ ions, we get: K_sp = (0.0629)(0.0629) K_sp = 0.00396 So, the solubility product constant (K_sp) for silver acetate at 20°C is 0.00396.

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What do we mean by the position of a reaction's equilibrium, and what practical consequence can it have?

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