Question

Equivalent conductance of $BaC{l_2}$, ${H_2}S{O_4}$ and $HCl$ are ${x_1}$, ${x_2}$ and ${x_3}$ $Sc{m^2}equi{v^{ - 1}}$ at infinite dilution. If specific conduction of saturated $BaS{O_4}$ solution is of $ySc{m^{ - 1}}$ the ${K_{sp}}$ of $BaS{O_4}$ is
A.$\dfrac{{{{10}^3}y}}{{2\left( {{x_1} + {x_2} - 2{x_3}} \right)}}$
B.$\dfrac{{{{10}^6}{y^2}}}{{4{{\left( {{x_1} + {x_2} - 2{x_3}} \right)}^2}}}$
C.$\dfrac{{{{10}^6}{y^2}}}{{2{{\left( {{x_1} + {x_2} - 2{x_3}} \right)}^2}}}$
D.$\dfrac{{{x_1} + {x_2} - 2{x_3}}}{{{{10}^3}y}}$

Answer 1

We can define specific conductance as the ability of the material to conduct electricity. The specific conductance is expressed as k. The reciprocal of resistance is conductance.

Complete step by step answer:
We can discuss the following points about solubility product ${{\text{K}}_{{\text{sp}}}}$:
The equilibrium constant for a chemical reaction in which a solid ionic compound dissolves to yield its ions in solution is known as “solubility product”. It is denoted by ‘${{\text{K}}_{{\text{sp}}}}$’.
Solubility products are also known as “ion products”.
The value of solubility products usually increases with an increase in temperature due increased solubility.
Some of the factors that affect the value of solubility products are the common-ion effect, the diverse-ion effect and presence of ion-pairs.
If the ionic concentrations have less value than the solubility product, the solution isn’t saturated. No precipitate would be formed.
If the ionic concentration has more value than the solubility product, sufficient precipitate would be formed to decrease concentrations to give an answer similar to the solubility product.
Let us calculate the ${ \wedge _{eq}}\left( {BaS{O_4}} \right)$ by summing up the equivalent conductance of barium chloride, sulfuric acid and twice the difference of hydrochloric acid.
The calculation of ${ \wedge _{eq}}\left( {BaS{O_4}} \right)$ is as follows,
${ \wedge _{eq}}\left( {BaS{O_4}} \right) = { \wedge _{eq}}\left( {BaC{l_2}} \right) + { \wedge _{eq}}\left( {{H_2}S{O_4}} \right) - 2{ \wedge _{eq}}\left( {HCl} \right)$
${ \wedge _{eq}}\left( {BaS{O_4}} \right) = \left( {{x_1} + {x_2} - 2{x_3}} \right)Sc{m^2}e{q^{ - 1}}$
Let us substitute the values of specific conductance and equivalent conductance in the equation to find the solubility,
${ \wedge _{{\text{eq}}}}\left( {{\text{BaS}}{{\text{O}}_{\text{4}}}} \right){\text{ = }}\dfrac{{{\text{1000 X specific}}\,{\text{conductance}}}}{{{\text{Solubility}}}}$
$\left( {{x_1} + {x_2} - 2{x_3}} \right){\text{ = }}\dfrac{{{\text{1000 X y}}}}{{{\text{Solubility}}}}$

${\text{Solubility}} = \dfrac{{{\text{1000 X y}}}}{{\left( {{x_1} + {x_2} - 2{x_3}} \right)}}N$
In dibasic acid, normality is twice the molarity.
${\text{Solubility}} = \dfrac{{{\text{1000 X y}}}}{{2\left( {{x_1} + {x_2} - 2{x_3}} \right)}}M$
The dissociation equation of $BaS{O_4}$ is,
$BaS{O_4} \rightleftharpoons B{a^{2 + }} + S{O_4}^{2 - }$
The solubility product is given by,
${K_{sp}}\left( {BaS{O_4}} \right) = \left[ {B{a^{2 + }}} \right]\left[ {S{O_4}^{2 - }} \right]{M^2}$
${\text{Solubility}} = \dfrac{{{\text{1000 X y}}}}{{2\left( {{x_1} + {x_2} - 2{x_3}} \right)}}M$
${K_{sp}}\left( {BaS{O_4}} \right) = \dfrac{{{{10}^6}{y^2}}}{{4{{\left( {{x_1} + {x_2} - 2{x_3}} \right)}^2}}}$
Option (B) is correct.
On taking the square root of the final solubility product equation of barium sulfate, we get,
${K_{sp}}\left( {BaS{O_4}} \right) = \dfrac{{{{10}^3}y}}{{2\left( {{x_1} + {x_2} - 2{x_3}} \right)}}$
Option (A) is also correct.

Note:
In simple words, the reciprocal of resistivity is called specific conductivity. We could define specific conductivity as the conductance between the opposite faces of the centimeter cube of a conductor. We can represent it as $k$ (kappa). We can write the expression of specific conductivity as,
$k = \dfrac{1}{\rho }$
Here, $\rho$ is the resistivity of the material.