Textbook Question
(b) If 0.0490 g of AgIO3 dissolves per liter of solution, calculate the solubility-product constant.
Ch.17 - Additional Aspects of Aqueous Equilibria
Chapter 17, Problem 57
Using Appendix D, calculate the molar solubility of AgBr in (b) 3.0 × 10^-2 M AgNO3 solution and (c) 0.10 M NaBr solution.
Verified step by step guidance1
Step 1: Identify the relevant solubility product constant (K_sp) for AgBr from Appendix D. The dissolution of AgBr can be represented by the equation: \[ \text{AgBr(s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Br}^- (aq) \].
Step 2: For part (b), consider the common ion effect due to the presence of Ag^+ ions from AgNO_3. The initial concentration of Ag^+ is 3.0 \times 10^{-2} \text{ M}. Let the molar solubility of AgBr be 's'. The equilibrium concentration of Ag^+ will be \(3.0 \times 10^{-2} + s\).
Step 3: Write the expression for the solubility product constant: \(K_{sp} = [\text{Ag}^+][\text{Br}^-]\). Substitute the equilibrium concentrations into this expression: \(K_{sp} = (3.0 \times 10^{-2} + s)(s)\).
Step 4: For part (c), consider the common ion effect due to the presence of Br^- ions from NaBr. The initial concentration of Br^- is 0.10 \text{ M}. Let the molar solubility of AgBr be 's'. The equilibrium concentration of Br^- will be \(0.10 + s\).
Step 5: Write the expression for the solubility product constant for part (c): \(K_{sp} = [\text{Ag}^+][\text{Br}^-]\). Substitute the equilibrium concentrations into this expression: \(K_{sp} = (s)(0.10 + s)\).
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Molar Solubility
Molar solubility is the number of moles of a solute that can dissolve in a liter of solution at equilibrium. It is a measure of how much of a substance can be dissolved in a solvent, and it is often expressed in moles per liter (M). Understanding molar solubility is crucial for predicting how much of a compound will dissolve in a given solution, especially in the context of saturated solutions.
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Molar Solubility Example
Common Ion Effect
The common ion effect refers to the decrease in solubility of an ionic compound when a common ion is added to the solution. This phenomenon occurs because the addition of a common ion shifts the equilibrium position of the dissolution reaction, leading to a lower concentration of dissolved ions. In the case of AgBr, the presence of AgNO3 or NaBr introduces Ag+ or Br- ions, respectively, which can significantly affect the solubility of AgBr.
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Ksp (Solubility Product Constant)
The solubility product constant (Ksp) is an equilibrium constant that applies to the solubility of sparingly soluble ionic compounds. It is defined as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced equation. For AgBr, Ksp can be used to calculate its molar solubility in different solutions by setting up an equilibrium expression that incorporates the concentrations of Ag+ and Br- ions.
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Related Practice
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