JEE Main 2020 · Chemistry · Ionic Equilibrium · hard

A solution contains 0.05 M Mg^2+ and 0.10 M NH3 (Kb=1.8×10^-5). Ksp of Mg(OH)2=1.2×10^-11. The minimum concentration of NH4Cl needed to prevent…

Numerical answer

A solution contains

0.05\ \mathrm{M\ Mg^{2+}}

and

0.10\ \mathrm{M\ NH_3}

(

K_b=1.8\times10^{-5}

K_{sp}

\mathrm{Mg(OH)_2}=1.2\times10^{-11}

. The minimum concentration of

\mathrm{NH_4Cl}

needed to prevent precipitation of

\mathrm{Mg(OH)_2}

x\times10^{-2}\ \mathrm{M}

. The value of

x

is (nearest tenth)

Show answer and solution

Answer: 11.6

Precipitation starts when

[\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2=K_{sp}

[\mathrm{OH^-}]_{max}=\sqrt{1.2\times10^{-11}/0.05}=\sqrt{2.4\times10^{-10}}=1.549\times10^{-5}

M. In the buffer,

[\mathrm{OH^-}]=K_b\dfrac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]}

, so

[\mathrm{NH_4^+}]=\dfrac{1.8\times10^{-5}\times0.10}{1.549\times10^{-5}}=0.116

=11.6\times10^{-2}

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