Three suggestion are made for ways to removes silver ions from solution:
a. Make the solution `0.01M`in `Nal`.
b. Buffer the solution at `pH = 13`.
c. Make the solution `0.01M Na_(2)S`. What will be the equilibrium silver ion concentartion in each case? which course of action is most effective in removing `Ag^(o+)` ions?
`K_(sp) (AgI) = 8.5 xx 10^(-17), K_(sp) (AgOH) = 2 xx 10^(-8)`,
`K_(sp) (Ag_(2)S) = 5.5 xx 10^(-51)`

Explanation

To solve the problem of determining the equilibrium silver ion concentration in each suggested method for removing silver ions from a solution, we will analyze each case step by step.

### Step 1: Case A - Making the solution 0.01 M in NaI

1. **Dissociation of NaI**: When NaI is added to the solution, it dissociates into Na⁺ and I⁻ ions. Thus, the concentration of I⁻ will be 0.01 M.

2. **Formation of AgI**: The I⁻ ions react with Ag⁺ ions to form AgI, which is a sparingly soluble salt. The solubility product (Ksp) expression for AgI is:
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