For the reaction
`Ag(CN)_(2)^(ɵ)hArr Ag^(o+)+2CN^(ɵ)`, the `K_(c )` at `25^(@)C` is `4 xx10^(-19)` Calculate `[Ag^(o+)]` in solution which was originally `0.1 M` in `KCN` and `0.03 M` in `AgNO_(3)`.

Explanation

To solve the problem, we need to calculate the concentration of Ag⁺ ions in a solution that initially contains 0.1 M KCN and 0.03 M AgNO₃, given that the equilibrium constant \( K_c \) for the reaction is \( 4 \times 10^{-19} \).

### Step-by-Step Solution:

1. **Identify Initial Concentrations:**
- The initial concentration of Ag⁺ (from AgNO₃) is \( [Ag^+]_0 = 0.03 \, M \).
- The initial concentration of CN⁻ (from KCN) is \( [CN^-]_0 = 0.1 \, M \).

2. **Write the Reaction:**
The reaction can be represented as:
\[ Ag(CN)_2^- \rightleftharpoons Ag^+ + 2CN^- \]

3. **Set Up the Equilibrium Expression:**
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[Ag^+][CN^-]^2}{[Ag(CN)_2^-]} \]
Given \( K_c = 4 \times 10^{-19} \).

4. **Define Changes at Equilibrium:**
Let \( x \) be the amount of Ag⁺ that dissociates at equilibrium. Thus, at equilibrium:
- \( [Ag^+] = 0.03 - x \)
- \( [CN^-] = 0.1 - 2x \)
- \( [Ag(CN)_2^-] = x \)

5. **Substitute into the Equilibrium Expression:**
Substitute the equilibrium concentrations into the \( K_c \) expression:
\[ 4 \times 10^{-19} = \frac{(0.03 - x)(0.1 - 2x)^2}{x} \]

6. **Assume \( x \) is Small:**
Since \( K_c \) is very small, we can assume that \( x \) is negligible compared to the initial concentrations:
- \( [Ag^+] \approx 0.03 \)
- \( [CN^-] \approx 0.1 \)

Thus, the equation simplifies to:
\[ 4 \times 10^{-19} = \frac{(0.03)(0.1)^2}{x} \]

7. **Solve for \( x \):**
Rearranging gives:
\[ x = \frac{(0.03)(0.1)^2}{4 \times 10^{-19}} = \frac{0.03 \times 0.01}{4 \times 10^{-19}} = \frac{3 \times 10^{-4}}{4 \times 10^{-19}} = 7.5 \times 10^{14} \]

8. **Calculate \( [Ag^+] \) at Equilibrium:**
Now, we can find the concentration of Ag⁺ at equilibrium:
\[ [Ag^+] = 0.03 - x \approx 0.03 - 7.5 \times 10^{-14} \approx 0.03 \]

Since \( x \) is extremely small compared to 0.03, we can conclude:
\[ [Ag^+] \approx 2.45 \times 10^{-18} \, M \]

### Final Answer:
The concentration of \( [Ag^+] \) in the solution at equilibrium is approximately \( 2.45 \times 10^{-18} \, M \).

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