Question

We have taken a saturated solution of AgBr. ${K_{sp}}$ of AgBr is $12 \times {10^{ - 14}}$ . If ${10^{ - 7}}$ mol of $AgN{O_3}$​ are added to 1 litre of this solution then the conductivity of this solution in terms of ${10^{ - 7}}S{m^{ - 1}}$ units will be: [Given: $\lambda _{A{g^ + }}^0 = 4 \times {10^{ - 3}}S{m^2}mo{l^{ - 1}},\lambda _{B{r^ - }}^0 = 6 \times {10^{ - 3}}S{m^2}mo{l^{ - 1}},\lambda _{NO_3^ - }^0 = 5 \times {10^{ - 3}}S{m^2}mo{l^{ - 1}}$ ]
A) 39
B) 55
C) 15
D) 41

Answer 1

Hint: This question we need to know about the Molar conductivity $({\lambda _m})$ . It is defined as the conductivity of a solution containing one mole of electrolyte. In other words, it can also be said as the conducting power of a solution, when all the ions in the electrolyte are dissociated. Molar conductivity is not a constant value. The molar conductivity can be given as ${\lambda _m} = \dfrac{\kappa }{c}$
Where $\kappa $ is the specific conductivity and c is the concentration.

Complete answer:
In this question we are given that $AgN{O_3}$ is added to the saturated solution of AgBr. Here the common ion effect will come into play. The common ion in both the solutions is $A{g^ + }$ . Hence the concentration of $A{g^ + }$ will increase in the resultant solution.
We are given the value of ${K_{sp}}$ of AgBr as $12 \times {10^{ - 14}}$ . The concentration of $AgN{O_3}$ is given as ${10^{ - 7}}$ . Let us consider the solubility of AgBr to be ‘s’. The dissociation of both the salts can be given as:
$AgBr{\text{ }} \rightleftharpoons {\text{ }}A{g^ + }{\text{ }} + {\text{ }}B{r^ - }$

T=0	a	-	-
T=equilibrium	$a - s$	$s + {10^{ - 7}}$	$s$

  $AgN{O_3}{\text{ }} \rightleftharpoons {\text{ }}A{g^ + }{\text{ }} + {\text{ }}NO_3^ - $

T=0	${10^{ - 7}}$	-	-
T=equilibrium	${10^{ - 7}}$-x	s+${10^{ - 7}}$	${10^{ - 7}}$

Therefore, the total concentration of $[A{g^ + }] = s + {10^{ - 7}}M$
The ${K_{sp}}$ for AgBr is $12 \times {10^{ - 14}}$ . Therefore, ${K_{sp}}(AgBr) = [A{g^ + }][B{r^ - }]$
${K_{sp}}(AgBr) = [s + {10^{ - 7}}][s] = 12 \times {10^{ - 14}}$
Solving the brackets and finding the quadratic equation, ${s^2} + {10^{ - 7}}s = 12 \times {10^{ - 14}}$
${s^2} + {10^{ - 7}}s - 12 \times {10^{ - 14}} = 0$
On solving the quadratic equation we get, $s = 3 \times {10^{ - 7}}M$
Therefore, the concentration of $[B{r^ - }] = s = 3 \times {10^{ - 7}}M$
On converting it to ${m^3}$ we get the answer as $[B{r^ - }] = s = 3 \times {10^{ - 7}}M = 3 \times {10^{ - 7}} \times {10^3} = 3 \times {10^{ - 4}}{m^3}$
The concentration of $[A{g^ + }] = s + {10^{ - 7}}M$ . Substituting the values for s and converting into ${m^3}$ we get,
$[A{g^ + }] = 3 \times {10} + {10^{ - 7}} = 4 \times {10^{ - 7}}M = 4 \times {{ - 7}} \times {10^3}{m^3} = 4 \times {10^{ - 4}}{m^3}$
The concentration of $NO_3^ - $ has already be found to be $[NO_3^ - ] = {10^{ - 7}}M = {10^{ - 7}} \times {10^3} = {10^{ - 4}}{m^3}$
Now, the specific conductance of each ion can be found from the formula ${\lambda _m} = \dfrac{\kappa }{c}$ . The specific conductance will be equal to $\kappa = {\lambda _m} \times c$
${\kappa _{A{g^ + }}} = 6 \times {10^{ - 3}} \times 4 \times {{ - 4}}S{m^{ - 1}} = 24 \times {10^{ - 7}}S{m^{ - 1}}$
${\kappa _{B{r^ - }}} = 8 \times {10^{ - 3}} \times 3 \times {10^{ - 4}}S{m^{ - 1}} = 24 \times {10^{ - 7}}S{m^{ - 1}}$
${\kappa _{NO_3^ - }} = 7 \times {10^{ - 3}} \times 1 \times {10^{ - 4}}S{m^{ - 1}} = 7 \times {10^{ - 7}}S{m^{ - 1}}$
Now, the specific conductance of the solution i.e. ${\kappa _{solution}}$ can be found as the sum of the specific conductance of all the ions present in the solution. Therefore, the specific conductance can be given as:
${\kappa _{solution}} = {\kappa _{B{r^ - }}} + {\kappa _{A{g^ + }}} + {\kappa _{NO_3^ - }}$
${\kappa _{solution}} = 24 \times {10^{ - 7}} + 24 \times {10^{ - 7}} + 1 \times {10^{ - 7}}S{m^{ - 1}}$
${\kappa _{solution}} = 55 \times {10^{ - 7}}S{m^{ - 1}}$
Therefore, in terms of ${10^{ - 7}}S{m^{ - 1}}$ the answer obtained will be $ = 55S{m^{ - 1}}$.

The correct option is Option (B)

Note:
Specific conductance is the ability of the material to conduct electricity. This is denoted by the symbol $\kappa $ (Kappa) . It is an additive value, i.e. the values of $\kappa $ can be added to find the total specific conductance. On the other hand the molar conductance $({\lambda _m})$ is multiplicative.