The solubility product of a salt having gener | vedclass

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

Question

(B) For a salt of type $MX_2$,the dissociation is given by: $MX_2(s) ⇌ M^{2+}(aq) + 2X^-(aq)$.Let the solubility be $S \ M$. Then,$[M^{2+}] = S$ and $

Vedclass · Accepted Answer

$1.0 	imes 10^{-4} \ M$

Vedclass · Answer

(B) For a salt of type $MX_2$,the dissociation is given by: $MX_2(s) ⇌ M^{2+}(aq) + 2X^-(aq)$.Let the solubility be $S \ M$. Then,$[M^{2+}] = S$ and $[X^-] = 2S$.The solubility product expression is $K_{sp} = [M^{2+}][X^-]^2$.Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 4 	imes 10^{-12}$,we have $4S^3 = 4 	imes 10^{-12}$.$S^3 = 10^{-12}$,so $S = \sqrt[3]{10^{-12}} = 10^{-4} \ M$.Since $[M^{2+}] = S$,the concentration of $M^{2+}$ ions is $1.0 	imes 10^{-4} \ M$.

The solubility product of a salt having general formula $MX_2$ in water is $4 \times 10^{-12}$. The concentration of $M^{2+}$ ions in the aqueous solution of the salt is

Similar Questions

The solubility product of a sparingly soluble salt $AX$ is $4.9 \times 10^{-13}$. What is its solubility in $mol \ dm^{-3}$?

The best explanation for the solubility of $MnS$ in dil. $HCl$ is that

The solubility product of $Ag_2CO_3$ and $FeCO_3$ in water at $25 \, ^\circ C$ are $4 \times 10^{-12}$ and $2.5 \times 10^{-11}$ respectively. If $S_1$ and $S_2$ are their solubilities in water respectively when dissolved separately,then $S_1/S_2$ is

What is the molar solubility of $Al(OH)_3$ in $0.2 \ M \ NaOH$ solution? Given that,the solubility product of $Al(OH)_3 = 2.4 \times 10^{-24}$.

The solubility products of three sparingly soluble salts $AB$,$A_2B$ and $AB_3$ are respectively $4.0 \times 10^{-20}$,$3.2 \times 10^{-11}$ and $2.7 \times 10^{-31}$. The increasing order of their solubility is

Vedclass Test Series

Exam Paper Generator

Online Exam Module