JEE Main 2023 · Chemistry · Ionic Equilibrium · hard

A solution contains 0.010 M Mg^2+ and 0.10 M NH3 (Kb=1.8×10^-5). Given Ksp(Mg(OH)2)=1.0×10^-12, the minimum concentration of NH4Cl to be added to…

Numerical answer

A solution contains

0.010\ \mathrm M

\mathrm{Mg^{2+}}

and

0.10\ \mathrm M

\mathrm{NH_3}

(

K_b=1.8\times10^{-5}

). Given

K_{sp}(\mathrm{Mg(OH)_2})=1.0\times10^{-12}

, the minimum concentration of

\mathrm{NH_4Cl}

to be added to prevent precipitation of

\mathrm{Mg(OH)_2}

, in

\mathrm{mmol\,L^{-1}}

(ignore volume change), is

Show answer and solution

Answer: 180

Precipitation begins when

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

[\mathrm{OH^-}]^2=10^{-12}/10^{-2}=10^{-10}

, so

[\mathrm{OH^-}]_{max}=10^{-5}\ \mathrm M

. For the buffer,

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

[\mathrm{NH_4^+}]=\frac{1.8\times10^{-5}\times0.10}{10^{-5}}=0.18\ \mathrm M=180\ \mathrm{mmol\,L^{-1}}

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