CaSO4(s) Ca²+ (aq) + SO3- (aq) K = [Ca ²+ ][SO²-] K= 3.2 x 10-5 R CaSO4(s) Ca2+(aq) + SO² (aq) I (M) SOLID 0 0 C (M) SOLID +X +X E(M) SOLID X X What is the concentration of the SO42- ion when the system reaches equilibrium? [SO²-] = [? ] × 10¹¹¹ M X Coefficient (green) Exponent (yellow) Enter

Chemistry: The Molecular Science
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Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter12: Chemical Equilibrium
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### Chemical Equilibrium and Solubility Product

**Given Reaction:**
\[ \text{CaSO}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{SO}_4^{2-}(aq) \]
**Equilibrium Constant (K):**
\[ K = \left[ \text{Ca}^{2+} \right] \left[ \text{SO}_4^{2-} \right] \]
\[ K = 3.2 \times 10^{-5} \]

### Initial, Change, and Equilibrium (ICE) Table:

| R      | CaSO<sub>4</sub>(s)   | Ca<sup>2+</sup>(aq)      | SO<sub>4</sub><sup>2-</sup>(aq)  |
|--------|----------------------|-------------------------|-------------------------------|
| I (M)  | SOLID                | 0                       | 0                             |
| C (M)  | SOLID                | +X                      | +X                            |
| E (M)  | SOLID                | X                       | X                             |

### Problem Statement:

What is the concentration of the SO<sub>4</sub><sup>2-</sup> ion when the system reaches equilibrium?

\[ \left[ \text{SO}_4^{2-} \right] = [ \boxed{\textcolor{green}{?}} ] \times 10^{\boxed{\textcolor{yellow}{?}}} \, \text{M} \]

### Instructions:
- Input the coefficient in the green box.
- Input the exponent in the yellow box.
- Click "Enter" to submit your answer.

### Explanation:
- **I (M)** represents the Initial molarity, where initially the concentrations of Ca<sup>2+</sup> and SO<sub>4</sub><sup>2-</sup> are zero.
- **C (M)** represents the Change in molarity, indicating how the concentrations of Ca<sup>2+</sup> and SO<sub>4</sub><sup>2-</sup> increase by X.
- **E (M)** represents the Equilibrium molarity, showing the concentrations of Ca<sup>2+</sup> and SO<sub>4</sub><sup>2-</sup> at equilibrium are equal to
Transcribed Image Text:### Chemical Equilibrium and Solubility Product **Given Reaction:** \[ \text{CaSO}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{SO}_4^{2-}(aq) \] **Equilibrium Constant (K):** \[ K = \left[ \text{Ca}^{2+} \right] \left[ \text{SO}_4^{2-} \right] \] \[ K = 3.2 \times 10^{-5} \] ### Initial, Change, and Equilibrium (ICE) Table: | R | CaSO<sub>4</sub>(s) | Ca<sup>2+</sup>(aq) | SO<sub>4</sub><sup>2-</sup>(aq) | |--------|----------------------|-------------------------|-------------------------------| | I (M) | SOLID | 0 | 0 | | C (M) | SOLID | +X | +X | | E (M) | SOLID | X | X | ### Problem Statement: What is the concentration of the SO<sub>4</sub><sup>2-</sup> ion when the system reaches equilibrium? \[ \left[ \text{SO}_4^{2-} \right] = [ \boxed{\textcolor{green}{?}} ] \times 10^{\boxed{\textcolor{yellow}{?}}} \, \text{M} \] ### Instructions: - Input the coefficient in the green box. - Input the exponent in the yellow box. - Click "Enter" to submit your answer. ### Explanation: - **I (M)** represents the Initial molarity, where initially the concentrations of Ca<sup>2+</sup> and SO<sub>4</sub><sup>2-</sup> are zero. - **C (M)** represents the Change in molarity, indicating how the concentrations of Ca<sup>2+</sup> and SO<sub>4</sub><sup>2-</sup> increase by X. - **E (M)** represents the Equilibrium molarity, showing the concentrations of Ca<sup>2+</sup> and SO<sub>4</sub><sup>2-</sup> at equilibrium are equal to
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