If the solution `0.5 M NH_(3)` and stability constant for `A^(+)(NH_(3))_(2)` is `K_(stb)=([Ag(NH_(3))_(2)]^(+))/([Ag^(+)(aq)][NH_(3)]^(2))=6.4xx10^(7),`
then find the solubility of AgCl in the above solution `K_(sp)of AgCl=2xx10^(-10)`

Explanation

To find the solubility of AgCl in a 0.5 M NH₃ solution given the stability constant for Ag(NH₃)₂⁺ and the solubility product of AgCl, we can follow these steps:

### Step 1: Write down the relevant equations
We know the relationship between the solubility product (Ksp) and the stability constant (Kstb):
\[ K_{equilibrium} = K_{sp} \times K_{stb} \]

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