Determine the concentration of `NH_(3)` solution whose one litre can dissolve 0.10 mole AgCl. `K_(sp)` of AgCl and` K_(f)` of `Ag(NH_(3))_(2)^(+)` are `1.0xx10^(-10)M^(2)` and `1.6xx10^(7)M^(-2)` respectively.

Answer

To determine the concentration of the `NH₃` solution that can dissolve `0.10` moles of `AgCl`, we will follow these steps:

### Step 1: Dissociation of AgCl
When `AgCl` dissolves in water, it dissociates into its ions:
\[ \text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \]

From the problem, `0.10` moles of `AgCl` will produce `0.10` moles of `Ag^+` and `0.10` moles of `Cl^-`.

### Step 2: Expression for Ksp of AgCl
The solubility product constant \( K_{sp} \) for `AgCl` is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Given \( K_{sp} = 1.0 \times 10^{-10} \, M^2 \).

Substituting the concentrations:
\[ K_{sp} = [\text{Ag}^+][0.10] = 1.0 \times 10^{-10} \]

### Step 3: Calculate [Ag+]
Rearranging the equation to find the concentration of `Ag^+`:
\[ [\text{Ag}^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.0 \times 10^{-10}}{0.10} = 1.0 \times 10^{-9} \, M \]

### Step 4: Formation of Ag(NH₃)₂⁺
The `Ag^+` ions will react with `NH₃` to form the complex ion `Ag(NH₃)₂⁺`:
\[ \text{Ag}^+ + 2 \text{NH}_3 \rightleftharpoons \text{Ag(NH}_3)_2^+ \]

### Step 5: Expression for Kf of Ag(NH₃)₂⁺
The formation constant \( K_f \) for the complex ion is given by:
\[ K_f = \frac{[\text{Ag(NH}_3)_2^+]}{[\text{Ag}^+][\text{NH}_3]^2} \]
Given \( K_f = 1.6 \times 10^7 \, M^{-2} \).

### Step 6: Substitute Known Values
Let \( [\text{Ag(NH}_3)_2^+] = 0.10 \, M \) (since it is formed from `0.10` moles of `Ag^+`):
\[ 1.6 \times 10^7 = \frac{0.10}{[Ag^+][NH_3]^2} \]

### Step 7: Substitute [Ag+]
Substituting \( [Ag^+] = 1.0 \times 10^{-9} \):
\[ 1.6 \times 10^7 = \frac{0.10}{(1.0 \times 10^{-9})[NH_3]^2} \]

### Step 8: Rearranging to Find [NH₃]
Rearranging to find \( [NH_3]^2 \):
\[ [NH_3]^2 = \frac{0.10}{1.6 \times 10^7 \times 1.0 \times 10^{-9}} \]

### Step 9: Calculate [NH₃]
Calculating \( [NH_3]^2 \):
\[ [NH_3]^2 = \frac{0.10}{1.6 \times 10^{-2}} = 6.25 \]

Taking the square root:
\[ [NH_3] = \sqrt{6.25} = 2.5 \, M \]

### Step 10: Total Concentration of NH₃
Since we need to account for the concentration of `NH₃` needed to dissolve `Ag^+`, we add:
\[ [NH_3] = 2.5 + 0.1 = 2.6 \, M \]

### Final Answer
The concentration of the `NH₃` solution is approximately **2.6 M**.

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