JEE Advanced 2024 Paper 1 · Chemistry · Ionic Equilibrium · medium

20.0 mL of 0.20 M MgCl2 is mixed with 20.0 mL of a buffer containing NH3 and NH4Cl. Given Kb(NH3)=2.0×10^-5 and Ksp[Mg(OH)2]=1.0×10^-11, the minimum…

Numerical answer

20.0\ \mathrm{mL}

0.20\ \mathrm{M}

\mathrm{MgCl_2}

is mixed with

20.0\ \mathrm{mL}

of a buffer containing

\mathrm{NH_3}

and

\mathrm{NH_4Cl}

. Given

K_b(\mathrm{NH_3})=2.0\times10^{-5}

and

K_{sp}[\mathrm{Mg(OH)_2}]=1.0\times10^{-11}

, the minimum value of the ratio

[\mathrm{NH_4^+}]/[\mathrm{NH_3}]

in the buffer that just prevents precipitation of

\mathrm{Mg(OH)_2}

Show answer and solution

Answer: 2

After mixing,

[\mathrm{Mg^{2+}}]=0.20\times\tfrac{20}{40}=0.10

M. Precipitation begins when

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

, so the highest tolerable hydroxide concentration is

[\mathrm{OH^-}]=\sqrt{10^{-11}/0.10}=1.0\times10^{-5}

M. In the ammonia buffer,

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

. Equal-volume mixing dilutes

\mathrm{NH_3}

and

\mathrm{NH_4^+}

equally, so the ratio is unchanged. Requiring

[\mathrm{OH^-}]\le1.0\times10^{-5}

[\mathrm{NH_4^+}]/[\mathrm{NH_3}]\ge K_b/[\mathrm{OH^-}]=2.0\times10^{-5}/1.0\times10^{-5}=2

. The minimum ratio is 2.

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