100 mL of 0.04 M MgCl2 is mixed with 100 mL of 0.20 M aqueous NH3 (K_b = 1.8 x 10^-5). Taking K_sp(Mg(OH)2) = 1.6 x 10^-11 and neglecting any Mg(OH)_2 formed when computing [OH^-], the correct statement is

Mg(OH)2 precipitates; ionic product = 3.6 x 10^-8

No precipitate; ionic product = 1.6 x 10^-12 is below K_sp

Mg(OH)2 precipitates; ionic product = 1.4 x 10^-7

Mg(OH)2 precipitates; ionic product = 3.6 x 10^-8

No precipitate because NH3 is a weak base and cannot supply enough OH^-

JEE Advanced levelSingle correct optionDifficulty 3 of 3ie205

$100 mL$ of $0.04 M$ $MgCl X_{2}$ is mixed with $100 mL$ of $0.20 M$ aqueous $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ). Taking $K_{s p} (Mg (OH) X_{2}) = 1.6 \times 1 0^{- 11}$ and neglecting any Mg(OH) $_{2}$ formed when computing $[OH X^{-}]$ , the correct statement is

ANo precipitate; ionic product $= 1.6 \times 1 0^{- 12}$ is below $K_{s p}$
B $Mg (OH) X_{2}$ precipitates; ionic product $= 1.4 \times 1 0^{- 7}$
C $Mg (OH) X_{2}$ precipitates; ionic product $= 3.6 \times 1 0^{- 8}$
DNo precipitate because $NH X_{3}$ is a weak base and cannot supply enough $OH X^{-}$

After mixing:

[Mg X^{2 +}] = 0.02 M

[NH X_{3}] = 0.10 M

[OH X^{-}] = K_{b} c = 1.3 \times 1 0^{- 3} M

. Ionic product

= 0.02 \times (1.3 \times 1 0^{- 3})^{2} = 3.6 \times 1 0^{- 8}

, far above

1.6 \times 1 0^{- 11}

, so

Mg (OH) X_{2}

precipitates.