$50 mL$ of $0.20 M$ $MgCl X_{2}$ is mixed with $50 mL$ of $0.40 M$ $NH X_{3}$ . What minimum mass of solid $NH X_{4} Cl$ (molar mass $53.5 g mo l^{- 1}$ ) must be dissolved in this mixture (no volume change) to just prevent precipitation of $Mg (OH) X_{2}$ ? $K_{s p} (Mg (OH) X_{2}) = 1.2 \times 1 0^{- 11}$ , $K_{b} (NH X_{3}) = 1.8 \times 1 0^{- 5}$ . Give the answer in grams (two decimals).

Numerical answer type, answer in g, tolerance 0.01.

Answer: 1.76 g

After mixing,

[Mg X^{2 +}] = 0.10 M

and

[NH X_{3}] = 0.20 M

. The largest tolerable

[OH X^{-}] = 1.2 \times 1 0^{- 11} /0.10 = 1.10 \times 1 0^{- 5} M

. From

K_{b}

[NH X_{4} X^{+}] = K_{b} [NH X_{3}] / [OH X^{-}] = 0.3286 M

. Moles in

0.100 L

= 0.03286, mass = 0.03286 x 53.5 = 1.76 g.