`0.01` mole of `AgNO_(3)` is added to 1 litre of a solution which is `0.1M` in `Na_(2)CrO_(4)` and `0.005M` in `NaIO_(3)`. Calculate the mole of precipitate formed at equilibrium and the concentrations of `Ag^(+), IO_(3)^(-)` and `CrO_(4)^(2-)`. `(K_(sP)` values of `Ag_(2)CrO_(4)` and `AgIO_(3)` are `10^(-8)` and `10^(-13)` respectively)

Explanation

To solve the problem step by step, we will follow the outlined method to determine the moles of precipitate formed and the concentrations of the ions in the solution.

### Step 1: Determine the initial concentrations
We have:
- 0.01 moles of AgNO₃ added to 1 L of the solution.
- The solution contains:
- 0.1 M Na₂CrO₄ (which provides 0.1 M CrO₄²⁻)
- 0.005 M NaIO₃ (which provides 0.005 M IO₃⁻)
...