Mg(OH)2 dissolves in ammonium chloride by Mg(OH)2 + 2NH4+ Mg^2+ + 2NH3 + 2H2O. Take K_sp(Mg(OH)2) = 1.2x 10^-11 and K_b(NH3) = 1.8x 10^-5. Treat the equilibrium above as the only reaction, so [NH3] = 2[Mg^2+]. Which statements are correct?

The equilibrium constant of the dissolution reaction is 0.037.

0.010 mol of Mg(OH)2 dissolves completely in 1.0 L of 0.50 M NH4Cl.

The equilibrium constant of the dissolution reaction is 0.037.

0.010 mol of Mg(OH)2 dissolves completely in 1.0 L of 0.50 M NH4Cl.

The solubility of Mg(OH)2 in 0.10 M NH4Cl exceeds 0.05 mol/L.

At saturation in 0.50 M NH4Cl the pH is above 10.

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ia165

$Mg (OH) X_{2}$ dissolves in ammonium chloride by $Mg (OH) X_{2} + 2 NH X_{4} X^{+} Mg X^{2 +} + 2 NH X_{3} + 2 H X_{2} O$ . Take $K_{s p} (Mg (OH) X_{2}) = 1.2 \times 1 0^{- 11}$ and $K_{b} (NH X_{3}) = 1.8 \times 1 0^{- 5}$ . Treat the equilibrium above as the only reaction, so $[NH X_{3}] = 2 [Mg X^{2 +}]$ . Which statements are correct?

AThe equilibrium constant of the dissolution reaction is 0.037.
B $0.010 mol$ of $Mg (OH) X_{2}$ dissolves completely in $1.0 L$ of $0.50 M$ $NH X_{4} Cl$ .
CThe solubility of $Mg (OH) X_{2}$ in $0.10 M$ $NH X_{4} Cl$ exceeds $0.05 mol/L$ .
DAt saturation in $0.50 M$ $NH X_{4} Cl$ the pH is above 10.

K = K_{s p} / K_{b}^{2} = 0.037

. With

[Mg X^{2 +}] = s

[NH X_{3}] = 2 s

[NH X_{4} X^{+}] = C - 2 s

4 s^{3} / (C - 2 s)^{2} = K

. For

s = 0.010

0.50 M

Q = 1.74 e - 05 < K

, so it all dissolves. Saturation in

0.10 M

s = 0.0270 M

(below 0.05). In

0.50 M

s = 0.0958 M

[OH X^{-}] = K_{b} (2 s) / (0.50 - 2 s) = 1.12 e - 05 M

pH = 9.05

, below 10.