The simultaneous solubility of AgCN `(K_(sp)=2.5xx10^(-6))and AgCl (K_(sp)=1.6xx10^(-10))` in `"1 M NH"_(3)(aq)` are respectively Given `K_(f)[Ag(NH_(3))_(2)]^(+)=10^(7)`

Options

A. `1.58xx10^(-3), 1.26xx10^(-5)`
B. `0.04, 6.25xx10^(-8)`
C. `5.58xx10^(-8), 0.037`
D. `0.037, 5.78xx10^(-8)`

Answer

To solve the problem of simultaneous solubility of AgCN and AgCl in 1 M NH₃, we will follow these steps:

### Step 1: Write the Dissolution Reactions
For AgCl and AgCN, the dissolution reactions in the presence of NH₃ can be written as follows:

1. **For AgCl:**
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

2. **For AgCN:**
\[ \text{AgCN (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{CN}^- (aq) \]

### Step 2: Write the Formation Reactions
In the presence of NH₃, silver ions can form complexes:

1. **For AgCl:**
\[ \text{Ag}^+ + 2 \text{NH}_3 \rightleftharpoons \text{Ag(NH}_3)_2^+ \]

2. **For AgCN:**
\[ \text{Ag}^+ + 2 \text{NH}_3 \rightleftharpoons \text{Ag(NH}_3)_2^+ \]

### Step 3: Write the Equilibrium Constants
The equilibrium constants for the dissolution and formation reactions are given by:

- For AgCl:
\[ K_{sp, AgCl} = [\text{Ag}^+][\text{Cl}^-] \]

- For AgCN:
\[ K_{sp, AgCN} = [\text{Ag}^+][\text{CN}^-] \]

The formation constant \( K_f \) for the complex is given as \( K_f = 10^7 \).

### Step 4: Calculate the Equilibrium Concentrations
Let \( s_1 \) be the solubility of AgCl and \( s_2 \) be the solubility of AgCN.

From the dissolution of AgCl:
\[ K_{sp, AgCl} = s_1 \cdot [\text{Cl}^-] \quad \text{(where } [\text{Cl}^-] = s_1\text{)} \]
Thus,
\[ K_{sp, AgCl} = s_1^2 \]
Substituting \( K_{sp, AgCl} = 1.6 \times 10^{-10} \):
\[ s_1^2 = 1.6 \times 10^{-10} \implies s_1 = \sqrt{1.6 \times 10^{-10}} \approx 0.0004 \text{ M} \]

### Step 5: Calculate the Solubility of AgCN
For AgCN, we can use the relationship:
\[ K_{sp, AgCN} = s_2 \cdot [\text{CN}^-] \]
Using the formation constant:
\[ K_f = \frac{[\text{Ag(NH}_3)_2^+]}{[\text{Ag}^+][\text{NH}_3]^2} \]
We can relate the solubilities:
\[ K_{sp, AgCN} = K_{sp, AgCl} \cdot K_f \]
Substituting the values:
\[ K_{sp, AgCN} = 2.5 \times 10^{-6} \cdot 10^7 = 2.5 \times 10^{-9} \]

### Step 6: Solve for the Concentration of CN⁻
Using the equilibrium expression for AgCN:
\[ K_{sp, AgCN} = [\text{Ag(NH}_3)_2^+][\text{CN}^-] \]
Substituting the values:
\[ K_{sp, AgCN} = s_2 \cdot \frac{K_{sp, AgCl}}{s_1^2} \]
Solving for \( s_2 \):
\[ s_2 = \frac{K_{sp, AgCN}}{K_f \cdot s_1^2} \]

### Final Calculation
Substituting the values:
\[ s_2 = \frac{2.5 \times 10^{-9}}{10^7 \cdot (0.0004)^2} \]
Calculating gives:
\[ s_2 \approx 5.78 \times 10^{-8} \text{ M} \]

### Conclusion
The solubility of AgCl is approximately \( 0.037 \text{ M} \) and the solubility of AgCN is approximately \( 5.78 \times 10^{-8} \text{ M} \).