A solution is 0.1M in Cl$$^-$$ and 0.001M in CrO$$_4^{2-}$$.
Solid AgNO$$_3$$ is gradually added to it. Assuming that the addition does not change in volume and
K$$_{sp}$$(AgCl) = 1.7 $$\times$$ 10$$^{-10}$$ M$$^2$$ and
K$$_{sp}$$(Ag$$_2$$CrO$$_4$$) = 1.9 $$\times$$ 10$$^{-12}$$ M$$^3$$.
Select correct statement from the following:

To determine which precipitate forms first, we need to find the minimum concentration of $$Ag^+$$ required to start precipitation of each ion.

For AgCl: $$K_{sp}(\text{AgCl}) = [\text{Ag}^+][\text{Cl}^-] = 1.7 \times 10^{-10}$$

With $$[\text{Cl}^-] = 0.1$$ M, precipitation begins when:

$$[\text{Ag}^+] = \frac{K_{sp}}{[\text{Cl}^-]} = \frac{1.7 \times 10^{-10}}{0.1} = 1.7 \times 10^{-9} \text{ M}$$

For $$Ag_2CrO_4$$: $$K_{sp}(\text{Ag}_2\text{CrO}_4) = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] = 1.9 \times 10^{-12}$$

With $$[\text{CrO}_4^{2-}] = 0.001$$ M, precipitation begins when:

$$[\text{Ag}^+]^2 = \frac{K_{sp}}{[\text{CrO}_4^{2-}]} = \frac{1.9 \times 10^{-12}}{10^{-3}} = 1.9 \times 10^{-9}$$

$$[\text{Ag}^+] = \sqrt{1.9 \times 10^{-9}} \approx 4.4 \times 10^{-5} \text{ M}$$

Since $$1.7 \times 10^{-9}$$ M $$\ll$$ $$4.4 \times 10^{-5}$$ M, a much smaller amount of $$Ag^+$$ is needed to precipitate $$Cl^-$$ as AgCl. Therefore, AgCl precipitates first because the amount of $$Ag^+$$ needed is lower.