Answer :
To determine the concentration of [tex]\( NH_3(aq) \)[/tex] required to dissolve 725 mg of [tex]\( AgCl(s) \)[/tex] in 100.0 mL of solution, we need to go through the following steps:
### Step-by-Step Solution
1. Determine the molar mass of [tex]\( AgCl \)[/tex]:
[tex]\[ M_{\text{AgCl}} = M_{\text{Ag}} + M_{\text{Cl}} = 107.87 \, \text{g/mol} + 35.45 \, \text{g/mol} = 143.32 \, \text{g/mol} \][/tex]
2. Convert the mass of [tex]\( AgCl \)[/tex] to moles:
[tex]\[ \text{mass of } AgCl = 725 \, \text{mg} = 725 \times 10^{-3} \, \text{g} \][/tex]
[tex]\[ \text{moles of } AgCl = \frac{725 \times 10^{-3} \, \text{g}}{143.32 \, \text{g/mol}} \approx 5.06 \times 10^{-3} \, \text{mol} \][/tex]
3. Calculate the molarity of [tex]\( AgCl \)[/tex] in the 100.0 mL solution:
[tex]\[ \text{volume of solution} = 100.0 \, \text{mL} = 0.1000 \, \text{L} \][/tex]
[tex]\[ [AgCl] = \frac{5.06 \times 10^{-3} \, \text{mol}}{0.1000 \, \text{L}} = 0.0506 \, \text{M} \][/tex]
4. Write the relevant expressions using the constants provided:
For the dissolution of [tex]\( AgCl \)[/tex]:
[tex]\[ AgCl(s) \leftrightarrow Ag^+(aq) + Cl^-(aq) \][/tex]
The solubility product ([tex]\( K_{sp} \)[/tex]) expression:
[tex]\[ K_{sp} = [Ag^+][Cl^-] = 1.77 \times 10^{-10} \][/tex]
For the complexation with [tex]\( NH_3 \)[/tex]:
[tex]\[ Ag^+(aq) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) \][/tex]
The formation constant ([tex]\( K_f \)[/tex]) expression:
[tex]\[ K_f = \frac{[[Ag(NH_3)_2]^+]}{[Ag^+][NH_3]^2} = 2.00 \times 10^7 \][/tex]
5. Set up the equilibrium expressions:
Assume all [tex]\( Ag^+ \)[/tex] forms the complex ion [tex]\( [Ag(NH_3)_2]^+ \)[/tex]:
[tex]\[ [Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{K_{sp}}{0.0506} \][/tex]
Substitute [tex]\( [Ag^+] \)[/tex] into the [tex]\( K_f \)[/tex] expression:
[tex]\[ K_f = \frac{0.0506}{\left(\frac{K_{sp}}{0.0506}\right)\times[NH_3]^2} \][/tex]
[tex]\[ K_f = \frac{0.0506 \times [NH_3]^2}{K_{sp}} \][/tex]
6. Solve for [tex]\( [NH_3] \)[/tex]:
[tex]\[ 2.00 \times 10^7 = \frac{0.0506 \times [NH_3]^2}{1.77 \times 10^{-10}} \][/tex]
[tex]\[ [NH_3]^2 = \frac{2.00 \times 10^7 \times 1.77 \times 10^{-10}}{0.0506} \][/tex]
[tex]\[ [NH_3]^2 = \frac{3.54 \times 10^{-3}}{0.0506} \][/tex]
[tex]\[ [NH_3]^2 \approx 6.992 \times 10^{-2} \][/tex]
[tex]\[ [NH_3] \approx \sqrt{6.992 \times 10^{-2}} \][/tex]
[tex]\[ [NH_3] \approx 0.2645 \, \text{M} \][/tex]
Thus, the concentration of [tex]\( NH_3(aq) \)[/tex] required to dissolve 725 mg of [tex]\( AgCl(s) \)[/tex] in 100.0 mL of solution is approximately:
[tex]\[ \boxed{0.2645} \][/tex]
### Step-by-Step Solution
1. Determine the molar mass of [tex]\( AgCl \)[/tex]:
[tex]\[ M_{\text{AgCl}} = M_{\text{Ag}} + M_{\text{Cl}} = 107.87 \, \text{g/mol} + 35.45 \, \text{g/mol} = 143.32 \, \text{g/mol} \][/tex]
2. Convert the mass of [tex]\( AgCl \)[/tex] to moles:
[tex]\[ \text{mass of } AgCl = 725 \, \text{mg} = 725 \times 10^{-3} \, \text{g} \][/tex]
[tex]\[ \text{moles of } AgCl = \frac{725 \times 10^{-3} \, \text{g}}{143.32 \, \text{g/mol}} \approx 5.06 \times 10^{-3} \, \text{mol} \][/tex]
3. Calculate the molarity of [tex]\( AgCl \)[/tex] in the 100.0 mL solution:
[tex]\[ \text{volume of solution} = 100.0 \, \text{mL} = 0.1000 \, \text{L} \][/tex]
[tex]\[ [AgCl] = \frac{5.06 \times 10^{-3} \, \text{mol}}{0.1000 \, \text{L}} = 0.0506 \, \text{M} \][/tex]
4. Write the relevant expressions using the constants provided:
For the dissolution of [tex]\( AgCl \)[/tex]:
[tex]\[ AgCl(s) \leftrightarrow Ag^+(aq) + Cl^-(aq) \][/tex]
The solubility product ([tex]\( K_{sp} \)[/tex]) expression:
[tex]\[ K_{sp} = [Ag^+][Cl^-] = 1.77 \times 10^{-10} \][/tex]
For the complexation with [tex]\( NH_3 \)[/tex]:
[tex]\[ Ag^+(aq) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) \][/tex]
The formation constant ([tex]\( K_f \)[/tex]) expression:
[tex]\[ K_f = \frac{[[Ag(NH_3)_2]^+]}{[Ag^+][NH_3]^2} = 2.00 \times 10^7 \][/tex]
5. Set up the equilibrium expressions:
Assume all [tex]\( Ag^+ \)[/tex] forms the complex ion [tex]\( [Ag(NH_3)_2]^+ \)[/tex]:
[tex]\[ [Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{K_{sp}}{0.0506} \][/tex]
Substitute [tex]\( [Ag^+] \)[/tex] into the [tex]\( K_f \)[/tex] expression:
[tex]\[ K_f = \frac{0.0506}{\left(\frac{K_{sp}}{0.0506}\right)\times[NH_3]^2} \][/tex]
[tex]\[ K_f = \frac{0.0506 \times [NH_3]^2}{K_{sp}} \][/tex]
6. Solve for [tex]\( [NH_3] \)[/tex]:
[tex]\[ 2.00 \times 10^7 = \frac{0.0506 \times [NH_3]^2}{1.77 \times 10^{-10}} \][/tex]
[tex]\[ [NH_3]^2 = \frac{2.00 \times 10^7 \times 1.77 \times 10^{-10}}{0.0506} \][/tex]
[tex]\[ [NH_3]^2 = \frac{3.54 \times 10^{-3}}{0.0506} \][/tex]
[tex]\[ [NH_3]^2 \approx 6.992 \times 10^{-2} \][/tex]
[tex]\[ [NH_3] \approx \sqrt{6.992 \times 10^{-2}} \][/tex]
[tex]\[ [NH_3] \approx 0.2645 \, \text{M} \][/tex]
Thus, the concentration of [tex]\( NH_3(aq) \)[/tex] required to dissolve 725 mg of [tex]\( AgCl(s) \)[/tex] in 100.0 mL of solution is approximately:
[tex]\[ \boxed{0.2645} \][/tex]
