The K at 25°C is 3.5 x 10-11 for the reaction below. Fill in the change portion of the ICE table below that would be required to calculate the molar solubility of CaF i water. CaF (s) + H20 (1) = Ca2+ (aq) + 2F (aq) CaF (s) H20 (1) = Ca2* (aq) 2F (aq) C 3.5 x 10-11 not applicable -2x +X +2x
The K at 25°C is 3.5 x 10-11 for the reaction below. Fill in the change portion of the ICE table below that would be required to calculate the molar solubility of CaF i water. CaF (s) + H20 (1) = Ca2+ (aq) + 2F (aq) CaF (s) H20 (1) = Ca2* (aq) 2F (aq) C 3.5 x 10-11 not applicable -2x +X +2x
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Chapter1: Chemical Foundations
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![### Calculating the Molar Solubility of Calcium Fluoride (CaF₂) in Water
**Objective:**
To determine the molar solubility of calcium fluoride (CaF₂) in water at 25°C using the solubility product constant (Ksp).
**Background:**
The solubility product constant, Ksp, for CaF₂ at 25°C is 3.5 x 10⁻¹¹. This constant helps us understand how much of the solid compound can dissolve in water to form a saturated solution.
**Reaction:**
\[ \text{CaF}_2 \text{(s)} + \text{H}_2\text{O (l)} \rightleftharpoons \text{Ca}^{2+} \text{(aq)} + 2\text{F}^- \text{(aq)} \]
**ICE Table:**
The ICE table (Initial, Change, Equilibrium) helps us organize the concentrations of reactants and products at the initial state, the change that occurs as the reaction progresses to equilibrium, and the concentrations at equilibrium.
| | \(\text{CaF}_2 \text{(s)}\) | \(\text{H}_2\text{O (l)}\) | \(\text{Ca}^{2+} \text{(aq)}\) | \(\text{2F}^- \text{(aq)}\) |
|---------|-----------------------------|----------------------------|-------------------------------|----------------------------|
| I | - | - | 0 | 0 |
| C | - | - | +x | +2x |
| E | - | - | x | 2x |
**Explanation of the ICE Table:**
- **Initial (I):**
- \( \text{CaF}_2 \text{(s)} \): An undissolved quantity of the solid is present.
- \( \text{H}_2\text{O (l)} \): Water is in excess and its concentration is practically constant (considered solvent).
- \( \text{Ca}^{2+} \text{(aq)} \): Initially, the concentration of calcium ions is 0.
- \( \text{2F}^- \text{(aq)} \): Initially,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc25f84cd-37fe-4eda-bb3a-0db4abed4038%2F5429feac-39e3-4acb-9c46-5d57069f9ff3%2Fhx73xyg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculating the Molar Solubility of Calcium Fluoride (CaF₂) in Water
**Objective:**
To determine the molar solubility of calcium fluoride (CaF₂) in water at 25°C using the solubility product constant (Ksp).
**Background:**
The solubility product constant, Ksp, for CaF₂ at 25°C is 3.5 x 10⁻¹¹. This constant helps us understand how much of the solid compound can dissolve in water to form a saturated solution.
**Reaction:**
\[ \text{CaF}_2 \text{(s)} + \text{H}_2\text{O (l)} \rightleftharpoons \text{Ca}^{2+} \text{(aq)} + 2\text{F}^- \text{(aq)} \]
**ICE Table:**
The ICE table (Initial, Change, Equilibrium) helps us organize the concentrations of reactants and products at the initial state, the change that occurs as the reaction progresses to equilibrium, and the concentrations at equilibrium.
| | \(\text{CaF}_2 \text{(s)}\) | \(\text{H}_2\text{O (l)}\) | \(\text{Ca}^{2+} \text{(aq)}\) | \(\text{2F}^- \text{(aq)}\) |
|---------|-----------------------------|----------------------------|-------------------------------|----------------------------|
| I | - | - | 0 | 0 |
| C | - | - | +x | +2x |
| E | - | - | x | 2x |
**Explanation of the ICE Table:**
- **Initial (I):**
- \( \text{CaF}_2 \text{(s)} \): An undissolved quantity of the solid is present.
- \( \text{H}_2\text{O (l)} \): Water is in excess and its concentration is practically constant (considered solvent).
- \( \text{Ca}^{2+} \text{(aq)} \): Initially, the concentration of calcium ions is 0.
- \( \text{2F}^- \text{(aq)} \): Initially,
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