A solution contains 0.050 M Mg^2+, 0.10 M NH3 and 0.20 M NH4Cl. K_b(NH3) = 1.8 x 10^-5, K_sp(Mg(OH)2) = 1.2 x 10^-11, K_w = 1.0 x 10^-14. Select the correct statement(s).

Mg(OH)2 does not precipitate from this solution

If the NH4Cl were omitted, Mg(OH)2 would precipitate

Mg(OH)2 does not precipitate from this solution

If the NH4Cl were omitted, Mg(OH)2 would precipitate

Mg^2+ up to 0.22 M could be present before precipitation begins

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ia091

A solution contains $0.050 M$ $Mg X^{2 +}$ , $0.10 M$ $NH X_{3}$ and $0.20 M$ $NH X_{4} Cl$ . $K_{b} (NH X_{3}) = 1.8 \times 1 0^{- 5}$ , $K_{s p} (Mg (OH) X_{2}) = 1.2 \times 1 0^{- 11}$ , $K_{w} = 1.0 \times 1 0^{- 14}$ . Select the correct statement(s).

Buffer:

[OH X^{-}] = K_{b} [NH X_{3}] / [NH X_{4} X^{+}] = 9.0 \times 1 0^{- 6} M

, pH = 8.95.

Q = 0.050 \times (9.0 \times 1 0^{- 6})^{2} = 4.1 \times 1 0^{- 12} < 1.2 \times 1 0^{- 11}

, so no precipitate. Without

NH X_{4} Cl

[OH X^{-}] = K_{b} \times 0.10 = 1.3 \times 1 0^{- 3} M

Q = 9.0 \times 1 0^{- 8} > K_{s p}

, so it precipitates. The maximum

[Mg X^{2 +}]

K_{s p} / [OH X^{-}]^{2} = 0.148 M

, not 0.22 M.