JEE Advanced 2023 Paper 1 · Chemistry · Ionic Equilibrium · hard

A solution is 0.10 M in each of M^2+ and N^2+ and is saturated with H2S (concentration 0.10 M, kept constant by continuous bubbling). Given Ka1× Ka2…

Numerical answer

A solution is

0.10\ \mathrm{M}

in each of

\mathrm{M^{2+}}

and

\mathrm{N^{2+}}

and is saturated with

\mathrm{H_2S}

(concentration

0.10\ \mathrm{M}

, kept constant by continuous bubbling). Given

K_{a1}\times K_{a2}

for

\mathrm{H_2S}=1.0\times10^{-21}

K_{sp}(\mathrm{MS})=1.0\times10^{-15}

and

K_{sp}(\mathrm{NS})=1.0\times10^{-21}

(all in appropriate units). The pH is gradually raised by adding a base. Let

a

be the minimum pH at which the concentration of

\mathrm{N^{2+}}

in solution has fallen to

1.0\times10^{-5}\ \mathrm{M}

, and

b

the maximum pH up to which

\mathrm{M^{2+}}

(at

0.10\ \mathrm{M}

) remains in solution. The value of

a+b

Show answer and solution

Answer: 7

In a saturated

\mathrm{H_2S}

solution

[\mathrm{S^{2-}}]=\dfrac{K_{a1}K_{a2}[\mathrm{H_2S}]}{[\mathrm{H^+}]^2}=\dfrac{10^{-22}}{[\mathrm{H^+}]^2}

.
For

\mathrm{N^{2+}}

to fall to

10^{-5}\ \mathrm{M}

[\mathrm{S^{2-}}]=\dfrac{10^{-21}}{10^{-5}}=10^{-16}\ \mathrm{M}

, so

[\mathrm{H^+}]^2=\dfrac{10^{-22}}{10^{-16}}=10^{-6}

[\mathrm{H^+}]=10^{-3}

, giving

a=3

.
For

\mathrm{M^{2+}}

to just begin precipitating:

[\mathrm{S^{2-}}]=\dfrac{10^{-15}}{0.10}=10^{-14}\ \mathrm{M}

, so

[\mathrm{H^+}]^2=\dfrac{10^{-22}}{10^{-14}}=10^{-8}

[\mathrm{H^+}]=10^{-4}

, giving

b=4

.
Between pH

3

and

4

the two cations can be separated selectively. Hence

a+b=3+4=7

Original question written in the JEE Advanced 2023 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake