The solubility of $\mathrm{CdSO}_{4}$ in water is $8.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$. Its solubility in $0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ solution is ______ $\times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$. (Round off to the nearest integer). (Assume that solubility is much less than 0.01 M)

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Equilibrium

The solubility of

CdSO_{4}

in water is

8.0 \times 1 0^{- 4} mol L^{- 1}

. Its solubility in

0.01 M H_{2} SO_{4}

solution is
______

\times 1 0^{- 6} mol L^{- 1}

.
(Round off to the nearest integer).
(Assume that solubility is much less than 0.01 M)

Answer: 64

In pure water the dissolved salt gives equal concentrations of the two ions, so

K_{s p} = [C d^{2 +}] [S O_{4}^{2 -}] = (8.0 \times 1 0^{- 4})^{2} = 6.4 \times 1 0^{- 7}

Sulphuric acid is fully ionised at this dilution, so in the

0.01

M acid the sulphate already present swamps the trace coming from the salt and

[S O_{4}^{2 -}] = 0.01

M. Writing the new solubility as

s

, the cadmium ion concentration is

s

and

s \times 0.01 = 6.4 \times 1 0^{- 7} ⟹ s = 6.4 \times 1 0^{- 5} mol L^{- 1}

The solubility in the acid is

64 \times 1 0^{- 6} mol L^{- 1}

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Integer answer