Question

the solubility product of a sparingly soluble salt ax2 is 3.2 — 10“11. its solubility (in mol/l) is : 1. 5.6 — 10“6 2. 3.1 — 10“4 3. 2 — 10“4 4. 4 — 10“4

Answer 1

Answer: The correct answer is Option 3.

Explanation: The equation for the reaction will be as follows:

$AX_2\leftrightharpoons A^{2+}+2X^-$

By Stoichiometry,

1 mole of $AX_2$ gives 2 moles of $X^-}$ and 1 mole of $A^{2+}$.

Thus if solubility of $AX_2$ is s moles/liter, solubility of $A^{2+}$ is s moles\liter and solubility of $X^{-}$ is 2s moles/liter

Therefore,  

$K_sp=[A^{2+}][X^{-}]^2$

$3.2\times 10^{-11}=[s][2s]^2$

$4s^3=3.2\times 10^{-11}$

$s=2.0\times 10^{-4}moles/liter$

Hence, the solubility of $AX_2$ is $2.0\times 10^{-4}moles/liter$

Answer 2

Answer: 3.  $2\times 10^{-4}moles/L$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of the $AX_2$ is given as:

$AX_2\leftrightharpoons A^{2+}+2X^{-}$

We are given:

Solubility product of $AX_2$ = $3.2\times 10^{-11}$

By stoichiometry of the reaction:

1 mole of $AgX_2$ gives 1 mole of $A^{2+}$ and 2 moles of $X^{-}$.

When the solubility of $AgX_2$ is S moles/liter, then the solubility of $A^{2+}$ will be S moles\liter and solubility of $X^{-}$ will be 2S moles/liter.

$K_{sp}=[A^{2+}][X^{-}]^2$

$K_{sp}=[s][(2s)^2]=4s^3$

$3.2\times 10^{-11}=4s^3$

$s=2\times 10^{-4}M$

Hence, the solubility is $2\times 10^{-4}moles/L$