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Question: We have taken a saturated solution of AgBr. \({K_{sp}}\) of AgBr is \(12 \times {10^{ - 14}}\) . If ...

We have taken a saturated solution of AgBr. Ksp{K_{sp}} of AgBr is 12×10−1412 \times {10^{ - 14}} . If 10−7{10^{ - 7}} mol of AgNO3AgN{O_3}​ are added to 1 litre of this solution then the conductivity of this solution in terms of 10−7Sm−1{10^{ - 7}}S{m^{ - 1}} units will be: [Given: λAg+0=4×10−3Sm2mol−1,λBr−0=6×10−3Sm2mol−1,λNO3−0=5×10−3Sm2mol−1\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

Explanation

Solution

This question we need to know about the Molar conductivity (λm)({\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 λm=κc{\lambda _m} = \dfrac{\kappa }{c}
Where κ\kappa is the specific conductivity and c is the concentration.

Complete answer:
In this question we are given that AgNO3AgN{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 Ag+A{g^ + } . Hence the concentration of Ag+A{g^ + } will increase in the resultant solution.
We are given the value of Ksp{K_{sp}} of AgBr as 12×10−1412 \times {10^{ - 14}} . The concentration of AgNO3AgN{O_3} is given as 10−7{10^{ - 7}} . Let us consider the solubility of AgBr to be ‘s’. The dissociation of both the salts can be given as:
AgBr ⇌ Ag+ + Br−AgBr{\text{ }} \rightleftharpoons {\text{ }}A{g^ + }{\text{ }} + {\text{ }}B{r^ - }

T=0a--
T=equilibriuma−sa - ss+10−7s + {10^{ - 7}}ss

AgNO3 ⇌ Ag+ + NO3−AgN{O_3}{\text{ }} \rightleftharpoons {\text{ }}A{g^ + }{\text{ }} + {\text{ }}NO_3^ -

T=010−7{10^{ - 7}}--
T=equilibrium10−7{10^{ - 7}}-xs+10−7{10^{ - 7}}10−7{10^{ - 7}}

Therefore, the total concentration of [Ag+]=s+10−7M[A{g^ + }] = s + {10^{ - 7}}M
The Ksp{K_{sp}} for AgBr is 12×10−1412 \times {10^{ - 14}} . Therefore, Ksp(AgBr)=[Ag+][Br−]{K_{sp}}(AgBr) = [A{g^ + }][B{r^ - }]
Ksp(AgBr)=[s+10−7][s]=12×10−14{K_{sp}}(AgBr) = [s + {10^{ - 7}}][s] = 12 \times {10^{ - 14}}
Solving the brackets and finding the quadratic equation, s2+10−7s=12×10−14{s^2} + {10^{ - 7}}s = 12 \times {10^{ - 14}}
s2+10−7s−12×10−14=0{s^2} + {10^{ - 7}}s - 12 \times {10^{ - 14}} = 0
On solving the quadratic equation we get, s=3×10−7Ms = 3 \times {10^{ - 7}}M
Therefore, the concentration of [Br−]=s=3×10−7M[B{r^ - }] = s = 3 \times {10^{ - 7}}M
On converting it to m3{m^3} we get the answer as [Br−]=s=3×10−7M=3×10−7×103=3×10−4m3[B{r^ - }] = s = 3 \times {10^{ - 7}}M = 3 \times {10^{ - 7}} \times {10^3} = 3 \times {10^{ - 4}}{m^3}
The concentration of [Ag+]=s+10−7M[A{g^ + }] = s + {10^{ - 7}}M . Substituting the values for s and converting into m3{m^3} we get,
[Ag+]=3×10+10−7=4×10−7M=4×−7×103m3=4×10−4m3[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 NO3−NO_3^ - has already be found to be [NO3−]=10−7M=10−7×103=10−4m3[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 λm=κc{\lambda _m} = \dfrac{\kappa }{c} . The specific conductance will be equal to κ=λm×c\kappa = {\lambda _m} \times c
κAg+=6×10−3×4×−4Sm−1=24×10−7Sm−1{\kappa _{A{g^ + }}} = 6 \times {10^{ - 3}} \times 4 \times {{ - 4}}S{m^{ - 1}} = 24 \times {10^{ - 7}}S{m^{ - 1}}
κBr−=8×10−3×3×10−4Sm−1=24×10−7Sm−1{\kappa _{B{r^ - }}} = 8 \times {10^{ - 3}} \times 3 \times {10^{ - 4}}S{m^{ - 1}} = 24 \times {10^{ - 7}}S{m^{ - 1}}
κNO3−=7×10−3×1×10−4Sm−1=7×10−7Sm−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. κsolution{\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:
κsolution=κBr−+κAg++κNO3−{\kappa _{solution}} = {\kappa _{B{r^ - }}} + {\kappa _{A{g^ + }}} + {\kappa _{NO_3^ - }}
κsolution=24×10−7+24×10−7+1×10−7Sm−1{\kappa _{solution}} = 24 \times {10^{ - 7}} + 24 \times {10^{ - 7}} + 1 \times {10^{ - 7}}S{m^{ - 1}}
κsolution=55×10−7Sm−1{\kappa _{solution}} = 55 \times {10^{ - 7}}S{m^{ - 1}}
Therefore, in terms of 10−7Sm−1{10^{ - 7}}S{m^{ - 1}} the answer obtained will be =55Sm−1 = 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 κ\kappa (Kappa) . It is an additive value, i.e. the values of κ\kappa can be added to find the total specific conductance. On the other hand the molar conductance (λm)({\lambda _m}) is multiplicative.