Question

A saturated solution in $AgA{\text{ }}({K_{SP}} = 3 \times {10^{ - 14}})$ and $AgB{\text{ }}({K_{SP}} = 1 \times {10^{ - 14}})$ has a conductivity of $375 \times {10^{ - 10}}Scm{}^{ - 1}$ and the limiting molar conductivity of $A{g^ + }$ and ${A^ - }$ are $60{\text{ }}Sc{m^2}mo{l^{ - 3}}$ and $80{\text{ }}Sc{m^2}mo{l^{ - 1}}$ respectively then what will be the limiting molar conductivity of ${B^ - }$ ( in $Sc{m^2}mo{l^{ - 4}}$)
A.150
B.180
C.190
D.270

Answer 1

Hint:Limiting molar conductivity of an electrolyte also known as molar conductivity at finite dilution is equal to the conductivity of the electrolyte in the solution when its concentration reaches zero.

Complete step by step answer:
Here given that the conductivity =\[375 \times {10^{ - 10}}{\text{ Sc}}{{\text{m}}^{{\text{ - 1}}}}\]
 The limiting molar conductivity $A{g^ + }$ is given =$60{\text{ }}Sc{m^2}mo{l^{ - 3}}$
And the limiting molar conductivity ${A^ - }$ is given =$80{\text{ }}Sc{m^2}mo{l^{ - 1}}$
The solubility product constant for $AgA{\text{ }}$is given =$3 \times {10^{ - 14}}$
Similarly, the solubility product $AgB$ is given =$1 \times {10^{ - 14}}$
We have to find the limiting molar conductivity ${B^ - }$.
Since the solubility product constant is defined as the product of solubility of ions raised to the power of their stoichiometric coefficient.
As we know the dissociation $AgA$ is carried out as:
$AgA{\text{ }}\overset {} \leftrightarrows A{g^ + } + {A^ - }$
Similarly, the dissociation $AgB$ is carried out as:
$AgB\overset {} \leftrightarrows A{g^ + } + {B^ - }$
As we see one mole $AgA$ gives one mole of silver ion and one mole of ${A^ - }$ ion. Suppose the solubility of silver ion is ${S_1}$ and the solubility of ${A^ - }$ ion also same as that of silver ion i.e. ${S_1}$ .
Similarly, one mole $AgB$ gives one mole of silver ion and one mole of ${B^ - }$ ion. Again suppose the solubility of silver ion in this case is ${S_2}$ and the solubility of ${B^ - }$ ion is also ${S_2}$.
Since $A{g^ + }$ is a common ion hence the total solubility of silver ion is ${S_1} + {S_2}$
Since ${K_{S{P_1}}} = 3 \times {10^{ - 14}} = {[Ag]^ + } \times {[A]^ - }$ (1)
Similarly
${K_{S{P_2}}} = 1 \times {10^{ - 14}} = {[Ag]^ + } \times {[B]^ - }$ (2)

Now divide equation 1 by equation 2 we get:
$\dfrac{{{K_{S{P_1}}}}}{{{K_{S{P_2}}}}} = \dfrac{{{{[Ag]}^ + }{{[A]}^ - }}}{{{{[Ag]}^ + }{{[B]}^ - }}}$
Now put the value of solubility constants in the above equation we get:
$\dfrac{{3 \times {{10}^{ - 14}}}}{{1 \times {{10}^{ - 14}}}} = \dfrac{{{{[Ag]}^ + }{{[A]}^ - }}}{{{{[Ag]}^ + }{{[B]}^ - }}}$
On solving we get:
$\dfrac{{{{[A]}^ - }}}{{{{[B]}^ - }}} = 3$
As we know ${[A]^ - } = {S_1}$ and ${[B]^ - } = {S_2}$, the above equation will be:
$\dfrac{{{S_1}}}{{{S_2}}} = 3$
$ \Rightarrow {S_1} = 3{S_2}$ (3)
Now add ${K_{S{P_1}}}$ and ${K_{S{P_2}}}$ we get:
${K_{S{P_1}}} + {K_{S{P_2}}} = \{ {[Ag]^ + }{[A]^ - }\} + \{ {[Ag]^ + }{[B]^ - }\} $
Taking ${[Ag]^ + }$common we get:
${K_{S{P_1}}} + {K_{S{P_2}}} = {[Ag]^ + }\{ {[A]^ - } + {[B]^ - }\} $
On putting the value we get:
$(3 \times {10^{ - 14}}) + (1 \times {10^{ - 14}}) = ({S_1} + {S_2}) \times ({S_1} + {S_2})$
$ \Rightarrow (3 \times 1){10^{ - 14}} = {({S_1} + {S_2})^2}$
$ \Rightarrow 4 \times {10^{ - 14}} = {({S_1} + {S_2})^2}$
Taking the square root on both sides we get:
$ \Rightarrow \sqrt {4 \times {{10}^{ - 14}}} = ({S_1} + {S_2})$
Or
$ \Rightarrow 2 \times {10^{ - 7}} = ({S_1} + {S_2})$
Now put the value ${S_1}$from equation (3) we get:
\[2 \times {10^{ - 7}} = 4{S_2}\]
\[ \Rightarrow {S_2} = \dfrac{1}{2} \times {10^{ - 7}}\] and as we know ${S_1} = 3{S_2}$, So we get
${S_1} = \dfrac{3}{2} \times {10^{ - 7}}$
Now we will find the limiting molar conductivity $\lambda _m^o$ by using the given formula:
$\lambda _m^o = \dfrac{{1000 \times \kappa }}{c}$
Here $\kappa $= conductivity and c= molar concentration.
Now put the value of conductivity and molar concentration in the above equation we get:
$\lambda _m^o = \dfrac{{1000 \times 375 \times {{10}^{ - 10}}}}{{{S_1} + {S_2}}}$
Or
$\lambda _m^o = \dfrac{{1000 \times 375 \times {{10}^{ - 10}}}}{{2 \times {{10}^{ - 7}}}}$

$ \Rightarrow \lambda _m^o = 187.5$
According to Kohlaruch’s law, the limiting molar conductivity is the sum of conductivity of cations and anions, hence it can be calculated as shown below:
$\lambda _m^o = \lambda _{A{g^ + }}^o + \lambda _{{A^ - }}^o + \lambda _{{B^ - }}^o$
$ \Rightarrow 187.5 = 60 + 80 + \lambda _{{B^ - }}^o$
$ \Rightarrow 187.5 - 60 - 80 = \lambda _{{B^ - }}^o$
$ \Rightarrow \lambda _{{B^ - }}^o = 47.5Sc{m^2}/mol$

Note:
According to Kohlrausch’s law, for an electrolyte in the given solution, the conductivity at infinite dilution is equivalent to the sum of the conductance of its cations and anions. This law is also known as Kohlrausch law of independent migration of ions.