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BITSAT 2010

Exam: BITSAT
Year: 2010
Subject: Chemistry
Chapter: Ionic Equilibrium

BITSAT 2010 · Chemistry · Ionic Equilibrium

The K_ sp of CuS, Ag_ 2 S and HgS are 10^ -31 , 10^ -44 and 10^ -54...

The

\mathrm{K}_{\mathrm{sp}}

of

\mathrm{CuS}, \mathrm{Ag}_{2} \mathrm{ S}

and

\mathrm{HgS}

are

10^{-31}, 10^{-44}

and

10^{-54}

respectively. The solubility of these sulphides are in the order :

Options

A $\quad \mathrm{Ag}_{2} \mathrm{ S} > \mathrm{CuS} > \mathrm{HgS}$
B $\quad \mathrm{AgS} > \mathrm{HgS} > \mathrm{CuS}$
C $\mathrm{HgS} > \mathrm{Ag}_{2} \mathrm{ S} > \mathrm{CuS}$
D $\quad\mathrm{CuS} > \mathrm{Ag}_{2} \mathrm{ S} > \mathrm{HgS}$

Correct answer

A. $\quad \mathrm{Ag}_{2} \mathrm{ S} > \mathrm{CuS} > \mathrm{HgS}$

Step-by-step solution

For CuS, solubility

=\left(10^{-31}\right)^{1 / 2} ;

\begin{array}{l}\text{ For } \mathrm{Ag}_{2} \mathrm{ S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{\frac{1}{3}}=\left(\frac{10^{-44}}{4}\right)^{\frac{1}{3}} \text{ and }\\ \text{ for } \mathrm{HgS}=\left(10^{-54}\right)^{\frac{1}{2}} \end{array}

$$

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