A solution is 0.10 M in Cl- and 1.0 x 10^-3 M in CrO4^2-. Solid AgNO3 is added slowly (no volume change). K_sp(AgCl) = 1.8 x 10^-10, K_sp(Ag2CrO4) = 1.1 x 10^-12. What percentage of the original Cl- has been precipitated when Ag2CrO4 just begins to precipitate?

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia067

A solution is $0.10 M$ in $Cl X^{-}$ and $1.0 \times 1 0^{- 3} M$ in $CrO X_{4} X^{2 -}$ . Solid $AgNO X_{3}$ is added slowly (no volume change). $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ , $K_{s p} (Ag X_{2} CrO X_{4}) = 1.1 \times 1 0^{- 12}$ . What percentage of the original $Cl X^{-}$ has been precipitated when $Ag X_{2} CrO X_{4}$ just begins to precipitate?

AgCl

starts first (

[Ag X^{+}] = 1.8 \times 1 0^{- 9} M

vs the larger value below for chromate).

Ag X_{2} CrO X_{4}

appears when

[Ag X^{+}] = 1.1 \times 1 0^{- 12} /1 0^{- 3} = 3.32 \times 1 0^{- 5} M

. Then

[Cl X^{-}] = 1.8 \times 1 0^{- 10} / (3.32 \times 1 0^{- 5}) = 5.43 \times 1 0^{- 6} M

, so the fraction precipitated is 99.9946 %.