A solution is 0.050 M in both Cl- and Br-. AgNO3 is added until AgCl just begins to precipitate (no volume change). Given K_sp(AgCl) = 1.8 x 10^-10 and K_sp(AgBr) = 5.0 x 10^-13, find the remaining [Br-] in units of 10^-4 M (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3ia063

A solution is $0.050 M$ in both $Cl X^{-}$ and $Br X^{-}$ . $AgNO X_{3}$ is added until $AgCl$ just begins to precipitate (no volume change). Given $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ and $K_{s p} (AgBr) = 5.0 \times 1 0^{- 13}$ , find the remaining $[Br X^{-}]$ in units of $1 0^{- 4} M$ (two decimals).

Numerical answer type, answer in x 10^-4 M, tolerance 0.01.

AgCl

starts at

[Ag X^{+}] = 1.8 \times 1 0^{- 10} /0.050 = 3.6 \times 1 0^{- 9} M

. Then

[Br X^{-}] = 5.0 \times 1 0^{- 13} / (3.6 \times 1 0^{- 9}) = 1.39 \times 1 0^{- 4} M

, so the answer is 1.39 in units of

1 0^{- 4} M