Question

At 291 K, the conductivity of saturated solution of \[Ca{{F}_{2}}\]is \[3.86\times {{10}^{-5}}\]$Sc{{m}^{-1}}$ and that water used for solution is\[0.15\times {{10}^{-5}}\]$Sc{{m}^{-1}}$ . The ionic conductance of \[C{{a}^{2+}}\]and\[{{F}^{-}}\]at infinite dilution are 51.0 and 47.0 \[Sc{{m}^{2}}mo{{l}^{-1}}\] respectively. The solubility of \[Ca{{F}_{2}}\] in solution is \[X\times {{10}^{-2}}\]g/Litre. Value of X is:

Answer 1

Hint: Try to recall that Kohlrausch law states that “the molar conductance at infinite dilution for a given salt can be expressed as the sum of individual contribution from the ions of the salt”. Now by using the formula below you can easily answer the given question.
Formula: \[\wedge =\dfrac{1000k}{C}\]where, k is conductivity of solution, C is molarity of solution and\[\wedge \] is molar conductivity of solution

Complete step by step answer:
In the question given, conductivity of saturated solution of\[Ca{{F}_{2}}\], \[{{k}_{sol}}\]=\[3.86\times {{10}^{-5}}\]$Sc{{m}^{-1}}$
Also, given conductivity of water used for solution, \[{{k}_{water}}\]=\[0.15\times {{10}^{-5}}\]$Sc{{m}^{-1}}$

So, now you can easily calculate the conductivity of sparingly soluble salt \[Ca{{F}_{2}}\], \[{{k}_{Ca{{F}_{2}}}}\]by deducting the conductivity of water from that of saturated solution of \[Ca{{F}_{2}}\] i.e.
\[\begin{align}
 & {{k}_{sol}}={{k}_{Ca{{F}_{2}}}}+{{k}_{water}} \\
 & \Rightarrow {{k}_{Ca{{F}_{2}}}}={{k}_{sol}}-{{k}_{water}} \\
 & \Rightarrow {{k}_{Ca{{F}_{2}}}}=3.86\times {{10}^{-5}}-0.15\times {{10}^{-5}} \\
 & or,{{k}_{Ca{{F}_{2}}}}=3.71\times {{10}^{-5}}Sc{{m}^{-1}} \\
\end{align}\]

Now, since Kohlrausch law states that “at infinite dilution, molar conductivity of an electrolyte is equal to the sum of conductivity of its cations and anions”.
So, by Kohlrausch law:
\[{{\wedge }^{\infty }}(Ca{{F}_{2}})=n{{\lambda }^{\infty }}(C{{a}^{2+}})+m{{\lambda }^{\infty }}({{F}^{-}})\]
Where,\[{{\wedge }^{\infty }}(Ca{{F}_{2}})\] is molar conductivity of \[Ca{{F}_{2}}\] electrolyte at infinite dilution,
\[{{\lambda }^{\infty }}(C{{a}^{2+}})\]= ionic conductivity of \[C{{a}^{2+}}\] at infinite dilution
n = number of \[C{{a}^{2+}}\]in \[Ca{{F}_{2}}\].
\[{{\lambda }^{\infty }}({{F}^{-}})\]=ionic conductivity of \[{{F}^{-}}\] at infinite dilution
m = number of\[{{F}^{-}}\]in\[Ca{{F}_{2}}\].

Calculation:
Given, \[{{\lambda }^{\infty }}(C{{a}^{2+}})\]=51.0 \[Sc{{m}^{2}}mo{{l}^{-1}}\]
\[{{\lambda }^{\infty }}({{F}^{-}})\]= 47.0 \[Sc{{m}^{2}}mo{{l}^{-1}}\]
n = 1, m = 2

From Kohlrausch law,
\[\begin{align}
 & {{\wedge }^{\infty }}(Ca{{F}_{2}})=n{{\lambda }^{\infty }}(C{{a}^{2+}})+m{{\lambda }^{\infty }}({{F}^{-}}) \\
& {{\wedge }^{\infty }}(Ca{{F}_{2}})=(1)(51)+(2)(47) \\
& \Rightarrow {{\wedge }^{\infty }}(Ca{{F}_{2}})=(51)+(94) \\
& or,{{\wedge }^{\infty }}(Ca{{F}_{2}})=145\text{ }Sc{{m}^{2}}mo{{l}^{-1}} \\
\end{align}\]
For sparingly soluble salt, the molar conductance of the saturated solution is taken to be equal to as the saturated solution of a sparingly soluble salt when it is extremely dilute.

Using above formula,
\[{{\wedge }^{\infty }}=\dfrac{1000{{k}_{Ca{{F}_{2}}}}}{C}\]
Where, \[{{k}_{Ca{{F}_{2}}}}=3.71\times {{10}^{-5}}Sc{{m}^{-1}}\],
\[{{\wedge }^{\infty }}(Ca{{F}_{2}})=145Sc{{m}^{2}}mo{{l}^{-1}}\] and C is the solubility of \[Ca{{F}_{2}}\] in solution.
 So, \[C=\dfrac{1000{{k}_{Ca{{F}_{2}}}}}{{{\wedge }^{\infty }}}\]
\[\begin{align}
 & \Rightarrow C=\frac{(1000)(3.71\times {{10}^{-5}}Sc{{m}^{-1}})}{(145Sc{{m}^{2}}mo{{l}^{-1}})} \\
 & or,C=2.56\times {{10}^{-4}}mol/L \\
\end{align}\]

Now, we know that the molar mass of $Ca{{F}_{2}}$ is 78 g/mol.
Therefore, solubility in g/L will be $78\times 2.56\times {{10}^{-4}}g/litre$ i.e. 0.02 g/L.
But, according to the question, C is \[X\times {{10}^{-2}}g/L\] .
So, comparing the values from calculation and the above value we get that X = 2.

Note: It should be remembered that for a sparingly soluble salt, molar conductivity of any concentration is equal to the molar conductivity of salt at infinite dilution.
Also, you should remember that Kohlrausch’s law is used for the determination of molar conductivity of a weak electrolyte at infinite dilution.