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Determine the molar solubility for Ag:CrO. (Ksp = 1.2 x 10-1"). PREV 3 NEXT Based on your ICE table, set up the expression for Ksp in order to determine the unknown. Do not combine or simplify terms. Ksp = 1.2 x 10-12

Determine the molar solubility for Ag:CrO. (Ksp = 1.2 x 10-1"). PREV 3 NEXT Based on your ICE table, set up the expression for Ksp in order to determine the unknown. Do not combine or simplify terms. Ksp = 1.2 x 10-12

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The final portion asks for molar solubility.

**Question 5 of 30** **Determine the molar solubility for Ag₂CrO₄ (Ksp = 1.2 × 10⁻¹²).** Based on your ICE table, set up the expression for Ksp in order to determine the unknown. Do not combine or simplify terms. \[ K_{sp} = \_\_\_\_\_\_ = 1.2 \times 10^{-12} \] **Options:** \[ [0] \quad [1.2 \times 10^{-12}] \quad [2.4 \times 10^{-12}] \quad [1.2 \times 10^{-12}]^2 \quad [2.4 \times 10^{-12}]^2 \quad [x] \quad [x]^2 \quad [2x] \] \[ [2x]^2 \quad [1.2 \times 10^{-12} - x] \quad [2.4 \times 10^{-12} - 2x] \] \[ [1.2 \times 10^{-12} + x] \quad [2.4 \times 10^{-12} + 2x] \] \[ [2.4 \times 10^{-12} = 2x] \] **Reset Button** --- This interactive exercise asks students to use their knowledge of chemical equilibria to set up the solubility product expression for silver chromate \((\text{Ag}_2\text{CrO}_4)\). The goal is to solve for the molar solubility, given the solubility product constant (\(K_{sp}\)). The students should choose the correct mathematical expression from the options to calculate \(K_{sp}\), without simplifying or combining the terms, reflecting their understanding of solubility equilibrium principles.
Transcribed Image Text:**Question 5 of 30** **Determine the molar solubility for Ag₂CrO₄ (Ksp = 1.2 × 10⁻¹²).** Based on your ICE table, set up the expression for Ksp in order to determine the unknown. Do not combine or simplify terms. \[ K_{sp} = \_\_\_\_\_\_ = 1.2 \times 10^{-12} \] **Options:** \[ [0] \quad [1.2 \times 10^{-12}] \quad [2.4 \times 10^{-12}] \quad [1.2 \times 10^{-12}]^2 \quad [2.4 \times 10^{-12}]^2 \quad [x] \quad [x]^2 \quad [2x] \] \[ [2x]^2 \quad [1.2 \times 10^{-12} - x] \quad [2.4 \times 10^{-12} - 2x] \] \[ [1.2 \times 10^{-12} + x] \quad [2.4 \times 10^{-12} + 2x] \] \[ [2.4 \times 10^{-12} = 2x] \] **Reset Button** --- This interactive exercise asks students to use their knowledge of chemical equilibria to set up the solubility product expression for silver chromate \((\text{Ag}_2\text{CrO}_4)\). The goal is to solve for the molar solubility, given the solubility product constant (\(K_{sp}\)). The students should choose the correct mathematical expression from the options to calculate \(K_{sp}\), without simplifying or combining the terms, reflecting their understanding of solubility equilibrium principles.
**Title: Calculating Molar Solubility Using an ICE Table** **Problem:** Determine the molar solubility for Ag₂CrO₄ (Ksp = 1.2 × 10⁻¹²). **Instructions:** Based on the given values, set up an ICE table in order to determine the unknown. **ICE Table:** | | Ag₂CrO₄(s) | ⇌ | 2 Ag⁺(aq) | + | CrO₄²⁻(aq) | |-------------------|--------------------|-------|----------------------|------|----------------------| | Initial (M) | | | | | | | Change (M) | | | | | | | Equilibrium (M) | | | | | | **Explanation:** - **Initial (M)**: Represents the initial concentrations of the reactants and products. Typically, the initial concentration of the solid, Ag₂CrO₄(s), is not included in calculations as it does not affect the equilibrium. - **Change (M)**: Indicates the change in concentration as the system reaches equilibrium. - **Equilibrium (M)**: Shows the concentrations of reactants and products at equilibrium. **Interactive Options:** - Interactive buttons are present below the table for inputting values. These buttons show various mathematical expressions involving x, such as \(-2x\), \(+2x\), \(1.2 \times 10^{-12}\), and expressions like \(2.4 \times 10^{-6} - 2x\). - A "Reset" button is available to clear inputs and start again. **Objective:** Use the ICE table to solve for the molar solubility by employing the solubility product constant (Ksp). Adjust the concentrations based on the stoichiometry of the dissolution reaction of Ag₂CrO₄.
Transcribed Image Text:**Title: Calculating Molar Solubility Using an ICE Table** **Problem:** Determine the molar solubility for Ag₂CrO₄ (Ksp = 1.2 × 10⁻¹²). **Instructions:** Based on the given values, set up an ICE table in order to determine the unknown. **ICE Table:** | | Ag₂CrO₄(s) | ⇌ | 2 Ag⁺(aq) | + | CrO₄²⁻(aq) | |-------------------|--------------------|-------|----------------------|------|----------------------| | Initial (M) | | | | | | | Change (M) | | | | | | | Equilibrium (M) | | | | | | **Explanation:** - **Initial (M)**: Represents the initial concentrations of the reactants and products. Typically, the initial concentration of the solid, Ag₂CrO₄(s), is not included in calculations as it does not affect the equilibrium. - **Change (M)**: Indicates the change in concentration as the system reaches equilibrium. - **Equilibrium (M)**: Shows the concentrations of reactants and products at equilibrium. **Interactive Options:** - Interactive buttons are present below the table for inputting values. These buttons show various mathematical expressions involving x, such as \(-2x\), \(+2x\), \(1.2 \times 10^{-12}\), and expressions like \(2.4 \times 10^{-6} - 2x\). - A "Reset" button is available to clear inputs and start again. **Objective:** Use the ICE table to solve for the molar solubility by employing the solubility product constant (Ksp). Adjust the concentrations based on the stoichiometry of the dissolution reaction of Ag₂CrO₄.
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