If the solubility product (K_{sp}) of a spari | vedclass

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Question

(B) For a salt of the type $MX_2$,the dissociation equilibrium is: $MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^-(aq)$.Let the solubility of $MX_2$ be

Vedclass · Accepted Answer

$1.36 	imes 10^{-4}$

Vedclass · Answer

(B) For a salt of the type $MX_2$,the dissociation equilibrium is: $MX_2(s) ightleftharpoons M^{2+}(aq) + 2X^-(aq)$.Let the solubility of $MX_2$ be $s \, mol \, L^{-1}$.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 $K_{sp} = 1.0 	imes 10^{-11}$.So,$4s^3 = 1.0 	imes 10^{-11}$.$s^3 = 0.25 	imes 10^{-11} = 2.5 	imes 10^{-12}$.$s = (2.5 	imes 10^{-12})^{1/3} \approx 1.357 	imes 10^{-4} \approx 1.36 	imes 10^{-4} \, mol \, L^{-1}$.

Metal sulfide	Solubility product
$A$. $PbS$	$I$. $4.0 \times 10^{-53}$
$B$. $HgS$	$II$. $8.0 \times 10^{-28}$
$C$. $MnS$	$III$. $1.6 \times 10^{-24}$
$D$. $ZnS$	$IV$. $2.5 \times 10^{-13}$

If the solubility product $(K_{sp})$ of a sparingly soluble salt $MX_2$ at $25 \, ^\circ C$ is $1.0 \times 10^{-11}$,what is the solubility of the salt in $mol \, L^{-1}$ at this temperature?

Similar Questions

Match the following metal sulfides with their respective solubility product $(K_{sp})$ values:

Metal sulfide Solubility product

$A$. $PbS$ $I$. $4.0 \times 10^{-53}$

$B$. $HgS$ $II$. $8.0 \times 10^{-28}$

$C$. $MnS$ $III$. $1.6 \times 10^{-24}$

$D$. $ZnS$ $IV$. $2.5 \times 10^{-13}$

The $K_{sp}$ value for $Gd(OH)_3$ is $2.7 \times 10^{-23}$. At what $pH$ will $Gd(OH)_3$ begin to precipitate? (Given: $\log 3 = 0.48$)

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

If the solubility product of a base $M(OH)_3$ is $2.7 \times 10^{-11}$,find the concentration of $OH^-$.

Express the relationship between $K_{sp}$ and solubility $(S)$ for the salt $Ca_3(PO_4)_2$.

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