A solution contains 0.010 M each of Br- and CrO4^2-. Solid AgNO3 is added slowly (no volume change). K_sp(AgBr) = 5.0 x 10^-13, K_sp(Ag2CrO4) = 1.1 x 10^-12. Select the correct statement(s).

AgBr precipitates before Ag2CrO4

Ag2CrO4 starts to precipitate at [Ag+] = 1.05 x 10^-5 M

At that stage more than 99.99 % of the bromide has been precipitated

AgBr precipitates before Ag2CrO4

Ag2CrO4 starts to precipitate at [Ag+] = 1.05 x 10^-5 M

When Ag2CrO4 just starts, [Br-] left is 1.4 x 10^-6 M

At that stage more than 99.99 % of the bromide has been precipitated

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ia072

A solution contains $0.010 M$ each of $Br X^{-}$ and $CrO X_{4} X^{2 -}$ . Solid $AgNO X_{3}$ is added slowly (no volume change). $K_{s p} (AgBr) = 5.0 \times 1 0^{- 13}$ , $K_{s p} (Ag X_{2} CrO X_{4}) = 1.1 \times 1 0^{- 12}$ . Select the correct statement(s).

A $AgBr$ precipitates before $Ag X_{2} CrO X_{4}$
B $Ag X_{2} CrO X_{4}$ starts to precipitate at $[Ag X^{+}] = 1.05 \times 1 0^{- 5} M$
CWhen $Ag X_{2} CrO X_{4}$ just starts, $[Br X^{-}]$ left is $1.4 \times 1 0^{- 6} M$
DAt that stage more than $99.99 %$ of the bromide has been precipitated

AgBr

needs

[Ag X^{+}] = 5.0 \times 1 0^{- 13} /0.010 = 5.0 \times 1 0^{- 11} M

, far below the

1.1 \times 1 0^{- 12} /0.010 = 1.0 \times 1 0^{- 5} M

needed for

Ag X_{2} CrO X_{4}

. At the latter value

[Br X^{-}] = 4.8 \times 1 0^{- 8} M

(99.9995 % precipitated), so the third statement (wrong value) is false.