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Question: A saturated solution in \(AgA{\text{ }}({K_{SP}} = 3 \times {10^{ - 14}})\) and \(AgB{\text{ }}({K_{...

A saturated solution in AgA (KSP=3×10−14)AgA{\text{ }}({K_{SP}} = 3 \times {10^{ - 14}}) and AgB (KSP=1×10−14)AgB{\text{ }}({K_{SP}} = 1 \times {10^{ - 14}}) has a conductivity of 375×10−10Scm−1375 \times {10^{ - 10}}Scm{}^{ - 1} and the limiting molar conductivity of Ag+A{g^ + } and A−{A^ - } are 60 Scm2mol−360{\text{ }}Sc{m^2}mo{l^{ - 3}} and 80 Scm2mol−180{\text{ }}Sc{m^2}mo{l^{ - 1}} respectively then what will be the limiting molar conductivity of B−{B^ - } ( in Scm2mol−4Sc{m^2}mo{l^{ - 4}})
A.150
B.180
C.190
D.270

Explanation

Solution

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 =375×10−10 Scm - 1375 \times {10^{ - 10}}{\text{ Sc}}{{\text{m}}^{{\text{ - 1}}}}
The limiting molar conductivity Ag+A{g^ + } is given =60 Scm2mol−360{\text{ }}Sc{m^2}mo{l^{ - 3}}
And the limiting molar conductivity A−{A^ - } is given =80 Scm2mol−180{\text{ }}Sc{m^2}mo{l^{ - 1}}
The solubility product constant for AgA AgA{\text{ }}is given =3×10−143 \times {10^{ - 14}}
Similarly, the solubility product AgBAgB is given =1×10−141 \times {10^{ - 14}}
We have to find the limiting molar conductivity B−{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 AgAAgA is carried out as:
AgA ⇆Ag++A−AgA{\text{ }}\overset {} \leftrightarrows A{g^ + } + {A^ - }
Similarly, the dissociation AgBAgB is carried out as:
AgB⇆Ag++B−AgB\overset {} \leftrightarrows A{g^ + } + {B^ - }
As we see one mole AgAAgA gives one mole of silver ion and one mole of A−{A^ - } ion. Suppose the solubility of silver ion is S1{S_1} and the solubility of A−{A^ - } ion also same as that of silver ion i.e. S1{S_1} .
Similarly, one mole AgBAgB gives one mole of silver ion and one mole of B−{B^ - } ion. Again suppose the solubility of silver ion in this case is S2{S_2} and the solubility of B−{B^ - } ion is also S2{S_2}.
Since Ag+A{g^ + } is a common ion hence the total solubility of silver ion is S1+S2{S_1} + {S_2}
Since KSP1=3×10−14=[Ag]+×[A]−{K_{S{P_1}}} = 3 \times {10^{ - 14}} = {[Ag]^ + } \times {[A]^ - } (1)
Similarly
KSP2=1×10−14=[Ag]+×[B]−{K_{S{P_2}}} = 1 \times {10^{ - 14}} = {[Ag]^ + } \times {[B]^ - } (2)

Now divide equation 1 by equation 2 we get:
KSP1KSP2=[Ag]+[A]−[Ag]+[B]−\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:
3×10−141×10−14=[Ag]+[A]−[Ag]+[B]−\dfrac{{3 \times {{10}^{ - 14}}}}{{1 \times {{10}^{ - 14}}}} = \dfrac{{{{[Ag]}^ + }{{[A]}^ - }}}{{{{[Ag]}^ + }{{[B]}^ - }}}
On solving we get:
[A]−[B]−=3\dfrac{{{{[A]}^ - }}}{{{{[B]}^ - }}} = 3
As we know [A]−=S1{[A]^ - } = {S_1} and [B]−=S2{[B]^ - } = {S_2}, the above equation will be:
S1S2=3\dfrac{{{S_1}}}{{{S_2}}} = 3
⇒S1=3S2\Rightarrow {S_1} = 3{S_2} (3)
Now add KSP1{K_{S{P_1}}} and KSP2{K_{S{P_2}}} we get:
KSP1+KSP2=[Ag]+[A]−+[Ag]+[B]−{K_{S{P_1}}} + {K_{S{P_2}}} = \\{ {[Ag]^ + }{[A]^ - }\\} + \\{ {[Ag]^ + }{[B]^ - }\\}
Taking [Ag]+{[Ag]^ + }common we get:
KSP1+KSP2=[Ag]+[A]−+[B]−{K_{S{P_1}}} + {K_{S{P_2}}} = {[Ag]^ + }\\{ {[A]^ - } + {[B]^ - }\\}
On putting the value we get:
(3×10−14)+(1×10−14)=(S1+S2)×(S1+S2)(3 \times {10^{ - 14}}) + (1 \times {10^{ - 14}}) = ({S_1} + {S_2}) \times ({S_1} + {S_2})
⇒(3×1)10−14=(S1+S2)2\Rightarrow (3 \times 1){10^{ - 14}} = {({S_1} + {S_2})^2}
⇒4×10−14=(S1+S2)2\Rightarrow 4 \times {10^{ - 14}} = {({S_1} + {S_2})^2}
Taking the square root on both sides we get:
⇒4×10−14=(S1+S2)\Rightarrow \sqrt {4 \times {{10}^{ - 14}}} = ({S_1} + {S_2})
Or
⇒2×10−7=(S1+S2)\Rightarrow 2 \times {10^{ - 7}} = ({S_1} + {S_2})
Now put the value S1{S_1}from equation (3) we get:
2×10−7=4S22 \times {10^{ - 7}} = 4{S_2}
⇒S2=12×10−7\Rightarrow {S_2} = \dfrac{1}{2} \times {10^{ - 7}} and as we know S1=3S2{S_1} = 3{S_2}, So we get
S1=32×10−7{S_1} = \dfrac{3}{2} \times {10^{ - 7}}
Now we will find the limiting molar conductivity λmo\lambda _m^o by using the given formula:
λmo=1000×κc\lambda _m^o = \dfrac{{1000 \times \kappa }}{c}
Here κ\kappa = conductivity and c= molar concentration.
Now put the value of conductivity and molar concentration in the above equation we get:
λmo=1000×375×10−10S1+S2\lambda _m^o = \dfrac{{1000 \times 375 \times {{10}^{ - 10}}}}{{{S_1} + {S_2}}}
Or
λmo=1000×375×10−102×10−7\lambda _m^o = \dfrac{{1000 \times 375 \times {{10}^{ - 10}}}}{{2 \times {{10}^{ - 7}}}}

⇒λmo=187.5 \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:
λmo=λAg+o+λA−o+λB−o\lambda _m^o = \lambda _{A{g^ + }}^o + \lambda _{{A^ - }}^o + \lambda _{{B^ - }}^o
⇒187.5=60+80+λB−o\Rightarrow 187.5 = 60 + 80 + \lambda _{{B^ - }}^o
⇒187.5−60−80=λB−o\Rightarrow 187.5 - 60 - 80 = \lambda _{{B^ - }}^o
⇒λB−o=47.5Scm2/mol\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.