Let the concentration of $Cl^-$, $Br^-$, $I^-$, and $CrO_4^{2-}$ be $c$. The concentration of $Ag^+$ required to precipitate each salt is calculated as follows: For $AgCl$: $[Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8 \times 10^{-10}}{c}$ For $AgBr$: $[Ag^+] = \frac{K_{sp}}{[Br^-]} = \frac{5.0 \times 10^{-13}}{c}$ For $AgI$: $[Ag^+] = \frac{K_{sp}}{[I^-]} = \frac{8.3 \times 10^{-17}}{c}$ For $Ag_2CrO_4$: $[Ag^+] = \sqrt{\frac{K_{sp}}{[CrO_4^{2-}]}} = \sqrt{\frac{1.1 \times 10^{-12}}{c}}$ Assuming $c = 0.1\text{ M}$ for comparison: $[Ag^+]$ for $AgI = 8.3 \times 10^{-16}\text{ M}$ $[Ag^+]$ for $AgBr = 5.0 \times 10^{-12}\text{ M}$ $[Ag^+]$ for $AgCl = 1.8 \times 10^{-9}\text{ M}$ $[Ag^+]$ for $Ag_2CrO_4 = \sqrt{1.1 \times 10^{-11}} \approx 3.32 \times 10^{-6}\text{ M}$ The salt that requires the highest concentration of $Ag^+$ to exceed its ionic product over $K_{sp}$ will precipitate last. Here, $Ag_2CrO_4$ requires the maximum $[Ag^+]$, so it will precipitate last.