The solubility proudct of AgCl at `25^(@)C` is `1xx10^(-10)` A solution of `Ag^(+)` at a concentration `4xx10^(-3)` M just fails to yield a prenciitate of AgCl with concentration of `1xx10^(-3) M Cl^(-)` when the concentration of `NH_(3)` in the solution is `2xx10^(-2)M.` Calculate the equlibrium constant for `[Ag(NH_(3))_(2))hArrAg^(+)+2NH_(3)`

Explanation

To solve the problem step by step, we will calculate the equilibrium constant for the reaction involving silver ions and ammonia.

### Step 1: Understanding the Given Information
We know:
- The solubility product (Ksp) of AgCl at 25°C is \(1 \times 10^{-10}\).
- The concentration of \(Ag^+\) is \(4 \times 10^{-3} \, M\).
- The concentration of \(Cl^-\) is \(1 \times 10^{-3} \, M\).
- The concentration of \(NH_3\) is \(2 \times 10^{-2} \, M\).
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