
Solubility of $M{{X}_{2}}$ type electrolytes is $0.5\times {{10}^{-4}}$ mole/lit. Then find out solubility product of electrolytes.
A. \[5\times {{10}^{12}}\]
B. \[25\times {{10}^{12}}\]
C. \[1.25\times {{10}^{12}}\]
D. \[5\times {{10}^{-13}}\]
Answer
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Hint: Solubility product of a salt depends on the number of ions formed in a reaction or the concentration of the ions at equilibrium. A salt with chemical formula $M{{X}_{2}}$ forms one cation and two anions. Thus the solubility product depends on concentration of the cations and anions.
Complete Step by Step Answer:
Solubility product of a species is defined as the product of the concentration of the cation and anion formed from a species at equilibrium. At this stage the coefficients before the ions exist as a power to the concentration of the corresponding cation and anion.
$M{{X}_{2}}$ dissociate at equilibrium as follows:
$M{{X}_{2}}\rightleftharpoons {{M}^{+2}}+2{{X}^{-}}$
Thus solubility product is given as follows-
$Ksp={{(2S)}^{2}}\times S$; where $Ksp$ is defined as the solubility product and $S$ is defined as the solubility or concentration for the cation and anion.
The solubility of the salt $M{{X}_{2}}$is $0.5\times {{10}^{-4}}$. Putting the value of solubility in the above equation we get-
$K_{sp}=4\times {{(0.5\times {{10}^{-4}})}^{3}}$
$K_{sp}=5\times {{10}^{-13}}$
Thus the value of solubility of the salt $M{{X}_{2}}$ is $5\times {{10}^{-13}}$.
Thus the correct option is D.
Note: One $M{{X}_{2}}$type salt is $Pb{{\left( N{{O}_{3}} \right)}_{2}}$. The unit of solubility product depends on the unit of concentration of the ions. As the concentration is expressed in moles per litre or in M unit thus the solubility product also has the same unit.
Complete Step by Step Answer:
Solubility product of a species is defined as the product of the concentration of the cation and anion formed from a species at equilibrium. At this stage the coefficients before the ions exist as a power to the concentration of the corresponding cation and anion.
$M{{X}_{2}}$ dissociate at equilibrium as follows:
$M{{X}_{2}}\rightleftharpoons {{M}^{+2}}+2{{X}^{-}}$
Thus solubility product is given as follows-
$Ksp={{(2S)}^{2}}\times S$; where $Ksp$ is defined as the solubility product and $S$ is defined as the solubility or concentration for the cation and anion.
The solubility of the salt $M{{X}_{2}}$is $0.5\times {{10}^{-4}}$. Putting the value of solubility in the above equation we get-
$K_{sp}=4\times {{(0.5\times {{10}^{-4}})}^{3}}$
$K_{sp}=5\times {{10}^{-13}}$
Thus the value of solubility of the salt $M{{X}_{2}}$ is $5\times {{10}^{-13}}$.
Thus the correct option is D.
Note: One $M{{X}_{2}}$type salt is $Pb{{\left( N{{O}_{3}} \right)}_{2}}$. The unit of solubility product depends on the unit of concentration of the ions. As the concentration is expressed in moles per litre or in M unit thus the solubility product also has the same unit.
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