The solubility of a sparingly soluble salt AX | vedclass

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

Question

(B) For a sparingly soluble salt of the type $AX_2$,the dissociation equilibrium is given by: $AX_2(s) \rightleftharpoons A^{2+}(aq) + 2X^-(aq)$ If $s

Vedclass · Accepted Answer

$4 	imes 10^{-12}$

Vedclass · Answer

(B) For a sparingly soluble salt of the type $AX_2$,the dissociation equilibrium is given by: $AX_2(s) ightleftharpoons A^{2+}(aq) + 2X^-(aq)$ If $s$ is the solubility of the salt in $mol \ dm^{-3}$,then the concentration of $A^{2+}$ is $s$ and the concentration of $X^-$ is $2s$. The solubility product expression is: $K_{sp} = [A^{2+}][X^-]^2 = (s)(2s)^2 = 4s^3$ Given $s = 1 	imes 10^{-4} \ mol \ dm^{-3}$. Substituting the value of $s$: $K_{sp} = 4 	imes (1 	imes 10^{-4})^3 = 4 	imes 10^{-12}$ Therefore,the correct option is $B$.

The solubility of a sparingly soluble salt $AX_2$ is $1 \times 10^{-4} \ mol \ dm^{-3}$ at $298 \ K$. Calculate its solubility product $(K_{sp})$?

Similar Questions

The solubility of $BaSO_4$ in water is $2.33 \times 10^{-3} \ g/L$. Its solubility product will be (molecular weight of $BaSO_4 = 233$).

Solubility (in molarity) of sparingly soluble salts $MX, MX_2$ and $MX_3$ in water are same. The order of $K_{sp}$ of $MX, MX_2$ and $MX_3$ is

Calculate the molar solubility of $Ni(OH)_2$ in $0.10 \ M \ NaOH$. The solubility product of $Ni(OH)_2$ is $2 \times 10^{-15}$.

According to the reaction $PbCl_2 \rightleftharpoons Pb^{2+} + 2Cl^-$,the solubility product constant $(K_{sp})$ of $PbCl_2$ is

$MY$ and $NY_3$,two nearly insoluble salts,have the same $K_{sp}$ values of $6.2 \times 10^{-13}$ at room temperature. Which statement would be true in regard to $MY$ and $NY_3$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module