Solid AgNO3 is added to a solution that is 0.10 M in Cl- and 1.0x10^-3 M in I-. K_sp(AgCl) = 1.6x10^-10, K_sp(AgI) = 8.5x10^-17. At the moment AgCl just starts to precipitate, find [I-] in units of 10^-8 M (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3ia211

Solid $AgNO X_{3}$ is added to a solution that is $0.10 M$ in $Cl X^{-}$ and $1.0 \times 1 0^{- 3} M$ in $I X^{-}$ . $K_{s p} (AgCl) = 1.6 \times 1 0^{- 10}$ , $K_{s p} (AgI) = 8.5 \times 1 0^{- 17}$ . At the moment $AgCl$ just starts to precipitate, find $[I X^{-}]$ in units of $1 0^{- 8} M$ (two decimals).

Numerical answer type, answer in x 10^-8 M, tolerance 0.01.

AgCl

starts at

[Ag X^{+}] = 1.6 \times 1 0^{- 10} /0.10 = 1.6 \times 1 0^{- 9} M

. Then

[I X^{-}] = 8.5 \times 1 0^{- 17} /1.6 \times 1 0^{- 9} = 5.31 \times 1 0^{- 8} M

, i.e. only about 0.0053% of the iodide remains.