Question

Molar conductance of $BaC{l_2},{H_2}S{O_4}$and $HCL$are ${x_1},{x_2}$and ${x_3}\;Sc{m^2}mo{l^{ - 1}}$at infinite dilution. If specific conductance of saturated $BaS{O_4}$solution is of $y\;Sc{m^{ - 1}}$, then ${K_{sp}}$of $BaS{O_4}$ is:
(A) $\dfrac{{{{10}^3}y}}{{2({x_1} + {x_2} - 2{x_3})}}$
(B) $\dfrac{{{{10}^6}{y^2}}}{{{{({x_1} + {x_2} - 2{x_3})}^2}}}$
(C) $\dfrac{{{{10}^6}{y^2}}}{{4{{({x_1} + {x_2} - 2{x_3})}^2}}}$
(D) $\dfrac{{{x_1} + {x_2} - 2{x_3}}}{{{{10}^6}{y^2}}}$

Answer 1

Hint: As we have already learnt that Kohlrausch’s law states that at infinite dilution, the molar conductivity of an electrolyte is expressed as the sum of the contributions from its individual ions and we also know that molar conductance is the conductance offered by one mole of an electrolyte placed between two electrodes which are separated by a distance of $1cm$.

Complete step by step solution:
According to Kohlrausch’s law, we know that at infinite dilution, the molar conductivity of an electrolyte is expressed as the sum of the contributions from its individual ions and it can be represented as:
${\lambda _m} = {n_ + }{\lambda _ + } + {n_ - }{\lambda _ - }$where ${n_ + }$and ${n_ - }$ are the number of moles of cation and anion per mole of the electrolyte respectively and ${\lambda _ + }$and ${\lambda _ - }$ are the molar conductivities of cation and anion at infinite dilution respectively.
Similarly we can write the molar conductivity of $BaS{O_4}$which will be formed as shown below:
Molar conductance of ${\lambda _{BaC{l_2}}} = {x_1}$,\[{\lambda _{{H_2}S{O_4}}} = {x_2}\] and ${\lambda _{HCl}} = {x_3}$.
\[
  {\lambda _{BaS{O_4}}} = {\lambda _{BaC{l_2}}} + {\lambda _{{H_2}S{O_4}}} - 2{\lambda _{HCl}} \\
 \Rightarrow y = {x_1} + {x_2} - 2{x_3} \\
 \]
As we are given with Specific conductance ($K$)=$y$.
Now, we also know that solubility is obtained as the ratio of specific conductance to the molarity of the solution, hence:
 $
  S = \dfrac{{K \times 1000}}{M} \\
  \Rightarrow S = \dfrac{{y \times 1000}}{{{x_1} + {x_2} - 2{x_3}}} \\
 $
Where M is the molarity, and we know that Normality is half of Molarity, $N = \dfrac{M}{2}$ and we are given the solubility product constant ${K_{sp}}$ which is basically the square of the solubility, so using normality and putting the values in the formula we get:
$S = \dfrac{{y \times 1000 \times M}}{{2({x_1} + {x_2} - 2{x_3})}}$
Let us recall that $BaS{O_4}$ will dissociate into ions and give $B{a^{2 + }}$ and $SO_4^{2 - }$, so the solubility product${K_{sp}}$ of $BaS{O_4}$ will be calculated easily as possible:
$
  {K_{sp(BaS{O_4})}} = {([B{a^{2 + }}][SO_4^{2 - }] \times M)^2} \\
  {K_{sp(BaS{O_4})}} = \dfrac{{{{10}^6}{y^2}}}{{4{{({x_1} + {x_2} - 2{x_3})}^2}}} \\
 $
Therefore the correct answer is (C).

Note: The conductivity of all electrolytes increases with increase in temperature and varies slightly with the pressure due to change in the viscosity of the medium which decreases with increase in solution. Specific conductance of an electrolyte decreases upon dilution because the number of current carrying ions per unit volume of solution decreases. Equivalent or molar conductance of an electrolyte increases with increase in dilution and conductivity decreases with decrease in concentration.