The solubility product of a sparingly soluble | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a sparingly soluble salt $AX_2$,the dissociation is: $AX_2(s) \rightleftharpoons A^{2+}(aq) + 2X^{-}(aq)$.Let the solubility be $s \ mol/L$.Th

Vedclass · Accepted Answer

$2 	imes 10^{-4}$

Vedclass · Answer

(A) For a sparingly soluble salt $AX_2$,the dissociation is: $AX_2(s) ightleftharpoons A^{2+}(aq) + 2X^{-}(aq)$.Let the solubility be $s \ mol/L$.Then,$[A^{2+}] = s$ and $[X^{-}] = 2s$.The solubility product expression is: $K_{sp} = [A^{2+}][X^{-}]^2 = (s)(2s)^2 = 4s^3$.Given $K_{sp} = 3.2 	imes 10^{-11}$.$4s^3 = 3.2 	imes 10^{-11}$.$s^3 = \frac{3.2 	imes 10^{-11}}{4} = 0.8 	imes 10^{-11} = 8 	imes 10^{-12}$.$s = \sqrt[3]{8 	imes 10^{-12}} = 2 	imes 10^{-4} \ mol/L$.

The solubility product of a sparingly soluble salt $AX_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles/litres) is

Similar Questions

The solubility product of salt $B_2A$ is $3.2 \times 10^{-11}$ at $298 \ K$. What is the solubility of the salt at the same temperature?

The $K_{sp}$ of $Mg(OH)_2$ is $1.2 \times 10^{-11}$. Calculate its solubility in pure water.

When equal volumes of the following solutions are mixed,precipitation of $AgCl$ $(K_{sp} = 1.8 \times 10^{-10})$ will occur only with:

If the solubility product of $AgCl$ is $10^{-10}$,what is the solubility of $AgCl$ in $0.001 \ M \ NaCl$?

If the ${K_{sp}}$ for $HgSO_4$ is $6.4 \times 10^{-5}$,then the solubility of the salt is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module