Question

[Ag⁺]_min for the start of precipitation of

$AgBr = \frac{K_{sp}(AgBr)}{\left\lbrack Br^{-} \right\rbrack} = \frac{5 \times 10^{- 13}}{0.1} = 5 \times 10^{- 12}\text{ M}$

For Cl^–: [Ag⁺]_min = ${\left\lbrack Ag^{+} \right\rbrack\frac{1.8 \times 10^{- 10}}{0.1}^{- 9}\text{ M}}_{\min}$

For $CO_{3}^{2 -}$: [Ag⁺]_min = $\sqrt{\frac{8.1 \times 10^{- 12}}{0.1}} = 9 \times 10^{- 6}\text{ M}$

For $AsO_{4}^{3 -}$: [Ag⁺]_min = $\left( \frac{1 \times 10^{- 22}}{0.1} \right) = 1 \times 10^{- 7}\text{ M}$

As [Ag⁺] needed is minimum for AgBr, it will precipitate first.