Answer :
Let's break down the calculation to find the solubility of AgCl in both pure water and in a 0.0140 M Mg(NO₃)₂ solution step-by-step:
### Solubility in Pure Water:
1. Determine the Solubility Product Constant (Ksp):
- The solubility product constant for AgCl at 25°C is [tex]\( Ksp_{AgCl} = 1.77 \times 10^{-10} \)[/tex].
2. Understand the Dissolution and Solubility Calculation:
- AgCl dissociates in water as follows:
[tex]\[ AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) \][/tex]
- Let [tex]\( s \)[/tex] be the solubility of AgCl in mol/L.
- Thus, [tex]\([Ag^+] = s\)[/tex] and [tex]\([Cl^-] = s\)[/tex].
- The Ksp expression can be written as:
[tex]\[ Ksp_{AgCl} = [Ag^+][Cl^-] = s \cdot s = s^2 \][/tex]
- Solving for [tex]\( s \)[/tex]:
[tex]\[ s = \sqrt{Ksp_{AgCl}} = \sqrt{1.77 \times 10^{-10}} \][/tex]
3. Calculate the Solubility in mol/L:
- Evaluating the square root:
[tex]\[ s = \sqrt{1.77 \times 10^{-10}} \approx 1.33 \times 10^{-5} \text{ mol/L} \][/tex]
4. Convert to g/L:
- Molar mass of AgCl is approximately 143.32 g/mol.
[tex]\[ \text{Solubility in g/L} = s \times \text{molar mass of AgCl} \approx 1.33 \times 10^{-5} \times 143.32 \][/tex]
- Simplifying:
[tex]\[ \text{Solubility in pure water} \approx 0.0 \frac{g}{L} \][/tex]
### Solubility in 0.0140 M Mg(NO₃)₂ solution:
1. Determine the Contribution of Mg(NO₃)₂:
- Mg(NO₃)₂ dissociates completely in water:
[tex]\[ Mg(NO_3)_2 \rightarrow Mg^{2+}(aq) + 2 NO_3^-(aq) \][/tex]
- Therefore, a 0.0140 M [tex]\( Mg(NO_3)_2 \)[/tex] solution provides:
[tex]\[ \left[ Mg^{2+} \right] = 0.0140 \text{ M} \][/tex]
2. Account for the Common Ion Effect:
- In the presence of [tex]\( Mg^{2+} \)[/tex], the solubility [tex]\( s \)[/tex] of [tex]\( AgCl \)[/tex] is affected.
- The Ksp expression in this context:
[tex]\[ Ksp_{AgCl} = [Ag^+][Cl^-] = (0.0140) \cdot s \][/tex]
- Solving for [tex]\( s \)[/tex]:
[tex]\[ s = \frac{Ksp_{AgCl}}{\left[ Mg^{2+} \right]} = \frac{1.77 \times 10^{-10}}{0.0140} \][/tex]
3. Calculate the Solubility in mol/L:
- Simplifying:
[tex]\[ s \approx 1.26 \times 10^{-8} \text{ mol/L} \][/tex]
4. Convert to g/L:
- Molar mass of AgCl is approximately 143.32 g/mol.
[tex]\[ \text{Solubility in g/L} = s \times \text{molar mass of AgCl} \approx 1.26 \times 10^{-8} \times 143.32 \][/tex]
- Simplifying:
[tex]\[ \text{Solubility in} \ 0.0140 \text{ M } Mg(NO_3)_2 \approx 0.0 \frac{g}{L} \][/tex]
### Summary:
- Solubility in pure water: [tex]\( 0.0 \frac{g}{L} \)[/tex]
- Solubility in 0.0140 M Mg(NO₃)₂ solution: [tex]\( 0.0 \frac{g}{L} \)[/tex]
Both the solubility numbers are rounded to two significant digits as instructed.
### Solubility in Pure Water:
1. Determine the Solubility Product Constant (Ksp):
- The solubility product constant for AgCl at 25°C is [tex]\( Ksp_{AgCl} = 1.77 \times 10^{-10} \)[/tex].
2. Understand the Dissolution and Solubility Calculation:
- AgCl dissociates in water as follows:
[tex]\[ AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) \][/tex]
- Let [tex]\( s \)[/tex] be the solubility of AgCl in mol/L.
- Thus, [tex]\([Ag^+] = s\)[/tex] and [tex]\([Cl^-] = s\)[/tex].
- The Ksp expression can be written as:
[tex]\[ Ksp_{AgCl} = [Ag^+][Cl^-] = s \cdot s = s^2 \][/tex]
- Solving for [tex]\( s \)[/tex]:
[tex]\[ s = \sqrt{Ksp_{AgCl}} = \sqrt{1.77 \times 10^{-10}} \][/tex]
3. Calculate the Solubility in mol/L:
- Evaluating the square root:
[tex]\[ s = \sqrt{1.77 \times 10^{-10}} \approx 1.33 \times 10^{-5} \text{ mol/L} \][/tex]
4. Convert to g/L:
- Molar mass of AgCl is approximately 143.32 g/mol.
[tex]\[ \text{Solubility in g/L} = s \times \text{molar mass of AgCl} \approx 1.33 \times 10^{-5} \times 143.32 \][/tex]
- Simplifying:
[tex]\[ \text{Solubility in pure water} \approx 0.0 \frac{g}{L} \][/tex]
### Solubility in 0.0140 M Mg(NO₃)₂ solution:
1. Determine the Contribution of Mg(NO₃)₂:
- Mg(NO₃)₂ dissociates completely in water:
[tex]\[ Mg(NO_3)_2 \rightarrow Mg^{2+}(aq) + 2 NO_3^-(aq) \][/tex]
- Therefore, a 0.0140 M [tex]\( Mg(NO_3)_2 \)[/tex] solution provides:
[tex]\[ \left[ Mg^{2+} \right] = 0.0140 \text{ M} \][/tex]
2. Account for the Common Ion Effect:
- In the presence of [tex]\( Mg^{2+} \)[/tex], the solubility [tex]\( s \)[/tex] of [tex]\( AgCl \)[/tex] is affected.
- The Ksp expression in this context:
[tex]\[ Ksp_{AgCl} = [Ag^+][Cl^-] = (0.0140) \cdot s \][/tex]
- Solving for [tex]\( s \)[/tex]:
[tex]\[ s = \frac{Ksp_{AgCl}}{\left[ Mg^{2+} \right]} = \frac{1.77 \times 10^{-10}}{0.0140} \][/tex]
3. Calculate the Solubility in mol/L:
- Simplifying:
[tex]\[ s \approx 1.26 \times 10^{-8} \text{ mol/L} \][/tex]
4. Convert to g/L:
- Molar mass of AgCl is approximately 143.32 g/mol.
[tex]\[ \text{Solubility in g/L} = s \times \text{molar mass of AgCl} \approx 1.26 \times 10^{-8} \times 143.32 \][/tex]
- Simplifying:
[tex]\[ \text{Solubility in} \ 0.0140 \text{ M } Mg(NO_3)_2 \approx 0.0 \frac{g}{L} \][/tex]
### Summary:
- Solubility in pure water: [tex]\( 0.0 \frac{g}{L} \)[/tex]
- Solubility in 0.0140 M Mg(NO₃)₂ solution: [tex]\( 0.0 \frac{g}{L} \)[/tex]
Both the solubility numbers are rounded to two significant digits as instructed.
