A solution is 0.010 M in Mg^2+ and 0.10 M in NH3 (K_b = 1.8 x 10^-5). K_sp(Mg(OH)2) = 1.6 x 10^-11. Find the minimum mass of solid NH4Cl (molar mass 53.5 g mol^-1), in grams, to be dissolved in 500 mL of this solution (no volume change) to prevent precipitation of Mg(OH)2 (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3ie207

A solution is $0.010 M$ in $Mg X^{2 +}$ and $0.10 M$ in $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ). $K_{s p} (Mg (OH) X_{2}) = 1.6 \times 1 0^{- 11}$ . Find the minimum mass of solid $NH X_{4} Cl$ (molar mass $53.5 g mo l^{- 1}$ ), in grams, to be dissolved in $500 mL$ of this solution (no volume change) to prevent precipitation of $Mg (OH) X_{2}$ (two decimals).

Numerical answer type, answer in g, tolerance 0.02.

Precipitation is just avoided when

[OH X^{-}]^{2} \times 0.010 = K_{s p}

, giving

[OH X^{-}] = 4.0 \times 1 0^{- 5} M

. From

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

[NH X_{4} X^{+}] = 1.8 \times 1 0^{- 5} \times 0.10/4.0 \times 1 0^{- 5} = 0.0450 M

. Mass

= 0.0450 \times 0.5 \times 53.5 = 1.20 g