A solution is 0.010 M in Mg^2+ and 0.010 M in Fe^3+. It is made 0.10 M in NH3 and 1.00 M in NH4Cl. Take K_b(NH3) = 1.8x10^-5, K_sp[Mg(OH)2] = 1.2x10^-11, K_sp[Fe(OH)3] = 1.0x10^-38. What precipitates?

JEE Advanced levelSingle correct optionDifficulty 3 of 3qc089

A solution is $0.010 M$ in $Mg X^{2 +}$ and $0.010 M$ in $Fe X^{3 +}$ . It is made $0.10 M$ in $NH X_{3}$ and $1.00 M$ in $NH X_{4} Cl$ . Take $K_{b} (NH X_{3}) = 1.8 \times 1 0^{- 5}$ , $K_{s p} [Mg (OH) X_{2}] = 1.2 \times 1 0^{- 11}$ , $K_{s p} [Fe (OH) X_{3}] = 1.0 \times 1 0^{- 38}$ . What precipitates?

[OH X^{-}] = K_{b} [NH X_{3}] / [NH X_{4} X^{+}] = 1.8 \times 1 0^{- 5} (0.10) /1.00 = 1.8 \times 1 0^{- 6} M

. For Mg:

0.010 \times (1.8 \times 1 0^{- 6})^{2} = 3.2 \times 1 0^{- 14}

, below

1.2 \times 1 0^{- 11}

, so no precipitate. For Fe:

0.010 \times (1.8 \times 1 0^{- 6})^{3} = 5.8 \times 1 0^{- 20}

, far above

1 0^{- 38}

, so

Fe (OH) X_{3}

precipitates. This is the group III/VI separation.