Given: `Ag(NH_(3))_(2)^(+)hArrAg^(+)2NH_(3), K_(C)=6.2xx10^(-8)` and `K_(SP)` of `AgCI=1.8xx10^(-10)` at 298 K. Calculate the concentration of the complex in `1.0M` aqueous ammonia.

Explanation

To solve the problem, we need to calculate the concentration of the complex ion \( \text{Ag(NH}_3)_2^+ \) in a 1.0 M aqueous ammonia solution, given the equilibrium constant \( K_c \) for the dissociation of the complex and the solubility product \( K_{sp} \) of \( \text{AgCl} \).

### Step-by-Step Solution:

1. **Write the Dissociation Reaction**:
The dissociation of the complex ion can be represented as:
\[ \text{Ag(NH}_3)_2^+ \rightleftharpoons \text{Ag}^+ + 2 \text{NH}_3 \]

2. **Write the Expression for \( K_c \)**:
The equilibrium constant \( K_c \) for the above reaction is given by:
\[ K_c = \frac{[\text{Ag}^+][\text{NH}_3]^2}{[\text{Ag(NH}_3)_2^+]} \]
Given \( K_c = 6.2 \times 10^{-8} \).

3. **Write the Expression for \( K_{sp} \)**:
The solubility product \( K_{sp} \) for \( \text{AgCl} \) is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Given \( K_{sp} = 1.8 \times 10^{-10} \).

4. **Set Up the Equilibrium**:
Assume that the concentration of \( \text{Ag(NH}_3)_2^+ \) at equilibrium is \( x \). The concentration of \( \text{Ag}^+ \) will then be \( \alpha \) (which we will find) and the concentration of \( \text{NH}_3 \) will be \( 1 - 2\alpha \) (since we start with 1.0 M of \( \text{NH}_3 \)).

5. **Substituting into the \( K_c \) Expression**:
Substitute the equilibrium concentrations into the \( K_c \) expression:
\[ 6.2 \times 10^{-8} = \frac{\alpha (1 - 2\alpha)^2}{x} \]

6. **Relate \( \alpha \) to \( K_{sp} \)**:
From the \( K_{sp} \) expression, we can express \( \alpha \) as:
\[ K_{sp} = \alpha [\text{Cl}^-] \]
Assuming \( [\text{Cl}^-] = \alpha \) (since we assume the formation of \( \text{AgCl} \) consumes \( \alpha \)), we can substitute this into the \( K_c \) equation.

7. **Assume \( 1 - 2\alpha \approx 1 \)**:
Since \( \alpha \) is expected to be small, we can approximate \( 1 - 2\alpha \approx 1 \):
\[ 6.2 \times 10^{-8} = \frac{\alpha}{x} \]

8. **Combine the Equations**:
Now we can combine the equations:
\[ 6.2 \times 10^{-8} = \frac{1.8 \times 10^{-10}}{x^2} \]

9. **Solve for \( x \)**:
Rearranging gives:
\[ x^2 = \frac{1.8 \times 10^{-10}}{6.2 \times 10^{-8}} \]
\[ x^2 = 2.9032 \times 10^{-3} \]
\[ x = \sqrt{2.9032 \times 10^{-3}} \approx 0.0538 \]

10. **Conclusion**:
The concentration of the complex ion \( \text{Ag(NH}_3)_2^+ \) in the solution is approximately:
\[ [\text{Ag(NH}_3)_2^+] \approx 0.0538 \, \text{M} \]