The solubility product of a sparingly soluble | vedclass

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

Question

(C) The dissociation of the salt is given by: $AX_{2(s)} \rightleftharpoons A^{2+}_{(aq)} + 2X^{-}_{(aq)}$Let $S$ be the solubility of the salt.At equ

Vedclass · Accepted Answer

$2.0 	imes 10^{-3}$

Vedclass · Answer

(C) The dissociation of the salt is given by: $AX_{2(s)} ightleftharpoons A^{2+}_{(aq)} + 2X^{-}_{(aq)}$Let $S$ be the solubility of the salt.At equilibrium,$[A^{2+}] = S$ and $[X^{-}] = 2S$.The solubility product expression is: $K_{sp} = [A^{2+}][X^{-}]^2$$K_{sp} = (S)(2S)^2 = 4S^3$Given $K_{sp} = 3.2 	imes 10^{-8}$.$4S^3 = 3.2 	imes 10^{-8}$$S^3 = \frac{3.2}{4} 	imes 10^{-8} = 0.8 	imes 10^{-8} = 8 	imes 10^{-9}$$S = \sqrt[3]{8 	imes 10^{-9}} = 2 	imes 10^{-3} \ mol \ dm^{-3}$

The solubility product of a sparingly soluble salt $AX_2$ is $3.2 \times 10^{-8}$. What is its solubility (in $mol \ dm^{-3}$)?

Similar Questions

The low solubility of $BaSO_4$ in water can be attributed to

If the $pH$ of a $0.001 \, M \, Mg(NO_3)_2$ solution is $9$,will precipitation occur? Given the $K_{sp}$ of $Mg(OH)_2 = 8.9 \times 10^{-12}$.

Solubility of a salt $M_2X_3$ is $y \ mol \ dm^{-3}$. The solubility product of the salt will be

In $1 \ L$ saturated solution of $AgCl$ $[K_{sp}(AgCl)=1.6 \times 10^{-10}]$,$0.1 \ mol$ of $CuCl$ $[K_{sp}(CuCl)=1.0 \times 10^{-6}]$ is added. The resultant concentration of $Ag^{+}$ in the solution is $1.6 \times 10^{-x}$. The value of $x$ is

The solubility of $AgCl$ at $20\,^oC$ is $1.435 \times 10^{-3} \, g/L$. The solubility product of $AgCl$ is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module