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Sure, let's go through the steps to calculate the solubility of AgCl in both pure water and a 0.0170 M AgNO_3 solution.
### Solubility in Pure Water
1. Identify the Solubility Product Constant (K_{sp}):
The solubility product constant for AgCl is given as [tex]\( K_{sp} = 1.77 \times 10^{-10} \)[/tex].
2. Set Up the Equilibrium Expression:
For the dissociation of AgCl in water:
[tex]\[ \text{AgCl}_{(s)} \leftrightarrow \text{Ag}^{+}_{(aq)} + \text{Cl}^{-}_{(aq)} \][/tex]
Let [tex]\( s \)[/tex] be the solubility of AgCl in [tex]\( \text{mol/L} \)[/tex].
At equilibrium, the concentrations of Ag⁺ and Cl⁻ ions will both be [tex]\( s \)[/tex].
3. Write the K_{sp} Expression:
[tex]\[ K_{sp} = [\text{Ag}^{+}] [\text{Cl}^{-}] = s \cdot s = s^2 \][/tex]
4. Solve for [tex]\( s \)[/tex]:
[tex]\[ s = \sqrt{K_{sp}} = \sqrt{1.77 \times 10^{-10}} = 1.33 \times 10^{-5} \, \text{mol/L} \][/tex]
5. Convert Solubility to g/L:
The molar mass of AgCl is 143.32 g/mol.
[tex]\[ \text{Solubility} = s \times \text{molar mass of AgCl} = 1.33 \times 10^{-5} \, \text{mol/L} \times 143.32 \, \text{g/mol} = 0.001905 \, \text{g/L} \][/tex]
6. Round to 2 Significant Digits:
[tex]\[ \text{Solubility in pure water} = 0.0 \, \text{g/L} \][/tex]
### Solubility in 0.0170 M AgNO_3 Solution
1. Set Up the Equilibrium Expression:
In the presence of AgNO_3, the concentration of [tex]\( \text{Ag}^+ \)[/tex] ions is already 0.0170 M.
2. Write the K_{sp} Expression:
[tex]\[ K_{sp} = [\text{Ag}^{+}] [\text{Cl}^{-}] \][/tex]
Given [Ag⁺] = 0.0170 M.
Let [tex]\( s' \)[/tex] be the solubility of AgCl in the AgNO_3 solution, representing the concentration of [tex]\( \text{Cl}^- \)[/tex].
3. Solve for [tex]\( [\text{Cl}^-] \)[/tex]:
[tex]\[ [\text{Cl}^-] = \frac{K_{sp}}{[\text{Ag}^{+}]} = \frac{1.77 \times 10^{-10}}{0.0170} = 1.04 \times 10^{-8} \, \text{mol/L} \][/tex]
4. Convert Solubility to g/L:
[tex]\[ \text{Solubility} = s' \times \text{molar mass of AgCl} = 1.04 \times 10^{-8} \, \text{mol/L} \times 143.32 \, \text{g/mol} = 0.000001492 \, \text{g/L} \][/tex]
5. Round to 2 Significant Digits:
[tex]\[ \text{Solubility in 0.0170 M AgNO_3 solution} = 0.0 \, \text{g/L} \][/tex]
To summarize:
- The solubility of AgCl in pure water is [tex]\( \boxed{0.0 \, \frac{g}{L}} \)[/tex].
- The solubility of AgCl in a 0.0170 M AgNO_3 solution is [tex]\( \boxed{0.0 \, \frac{g}{L}} \)[/tex].
### Solubility in Pure Water
1. Identify the Solubility Product Constant (K_{sp}):
The solubility product constant for AgCl is given as [tex]\( K_{sp} = 1.77 \times 10^{-10} \)[/tex].
2. Set Up the Equilibrium Expression:
For the dissociation of AgCl in water:
[tex]\[ \text{AgCl}_{(s)} \leftrightarrow \text{Ag}^{+}_{(aq)} + \text{Cl}^{-}_{(aq)} \][/tex]
Let [tex]\( s \)[/tex] be the solubility of AgCl in [tex]\( \text{mol/L} \)[/tex].
At equilibrium, the concentrations of Ag⁺ and Cl⁻ ions will both be [tex]\( s \)[/tex].
3. Write the K_{sp} Expression:
[tex]\[ K_{sp} = [\text{Ag}^{+}] [\text{Cl}^{-}] = s \cdot s = s^2 \][/tex]
4. Solve for [tex]\( s \)[/tex]:
[tex]\[ s = \sqrt{K_{sp}} = \sqrt{1.77 \times 10^{-10}} = 1.33 \times 10^{-5} \, \text{mol/L} \][/tex]
5. Convert Solubility to g/L:
The molar mass of AgCl is 143.32 g/mol.
[tex]\[ \text{Solubility} = s \times \text{molar mass of AgCl} = 1.33 \times 10^{-5} \, \text{mol/L} \times 143.32 \, \text{g/mol} = 0.001905 \, \text{g/L} \][/tex]
6. Round to 2 Significant Digits:
[tex]\[ \text{Solubility in pure water} = 0.0 \, \text{g/L} \][/tex]
### Solubility in 0.0170 M AgNO_3 Solution
1. Set Up the Equilibrium Expression:
In the presence of AgNO_3, the concentration of [tex]\( \text{Ag}^+ \)[/tex] ions is already 0.0170 M.
2. Write the K_{sp} Expression:
[tex]\[ K_{sp} = [\text{Ag}^{+}] [\text{Cl}^{-}] \][/tex]
Given [Ag⁺] = 0.0170 M.
Let [tex]\( s' \)[/tex] be the solubility of AgCl in the AgNO_3 solution, representing the concentration of [tex]\( \text{Cl}^- \)[/tex].
3. Solve for [tex]\( [\text{Cl}^-] \)[/tex]:
[tex]\[ [\text{Cl}^-] = \frac{K_{sp}}{[\text{Ag}^{+}]} = \frac{1.77 \times 10^{-10}}{0.0170} = 1.04 \times 10^{-8} \, \text{mol/L} \][/tex]
4. Convert Solubility to g/L:
[tex]\[ \text{Solubility} = s' \times \text{molar mass of AgCl} = 1.04 \times 10^{-8} \, \text{mol/L} \times 143.32 \, \text{g/mol} = 0.000001492 \, \text{g/L} \][/tex]
5. Round to 2 Significant Digits:
[tex]\[ \text{Solubility in 0.0170 M AgNO_3 solution} = 0.0 \, \text{g/L} \][/tex]
To summarize:
- The solubility of AgCl in pure water is [tex]\( \boxed{0.0 \, \frac{g}{L}} \)[/tex].
- The solubility of AgCl in a 0.0170 M AgNO_3 solution is [tex]\( \boxed{0.0 \, \frac{g}{L}} \)[/tex].
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