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Question: Equivalent conductance of saturated \(BaS{O_4}\) is \(400oh{m^{ - 1}}c{m^2}equv{i^{ - 1}}\) and spec...

Equivalent conductance of saturated BaSO4BaS{O_4} is 400ohm−1cm2equvi−1400oh{m^{ - 1}}c{m^2}equv{i^{ - 1}} and specific conductance is 8×10−5ohm−1cm−18 \times {10^{ - 5}}oh{m^{ - 1}}c{m^{ - 1}} . Hence Ksp{K_{sp}} of BaSO4BaS{O_4} is.
A) 4×10−8M24 \times {10^{ - 8}}{M^2}
B) 1×10−8M21 \times {10^{ - 8}}{M^2}
C) 2×10−4M22 \times {10^{ - 4}}{M^2}
D) 1×10−4M21 \times {10^{ - 4}}{M^2}

Explanation

Solution

As all units are given in equivalence, we can use the equivalent conductivity formulaΛeq=κ×1000C{\Lambda _{eq}} = \dfrac{{\kappa \times 1000}}{C}. But we have to find Ksp{K_{sp}} value also so that we need to change the equivalent conductance and molar conductance. For calculation, Ksp{K_{sp}} we need concentration value. After getting C values we will proceed to find Ksp{K_{sp}} . Specific conductance is denoted by κ\kappa .

Complete answer:
We will solve this problem in a two-step.
Step 1: we will find the solubility value.
We know that reciprocal of resistivity is called specific conductance. It is denoted by κ\kappa . It is represented by κ=1ρ\kappa = \dfrac{1}{\rho } . So its unit is ohm−1cm−1oh{m^{ - 1}}c{m^{ - 1}}.
So, we have κ=1ρ\kappa = \dfrac{1}{\rho }
Here ρ\rho resistivity
Molar conductance- Λm=κC{\Lambda _m} = \dfrac{\kappa }{C}
Here C= concentration
If concentration values are given in equivalent/liter then it is called equivalent conductance and the above equation becomes
Λeq=κ×1000s{\Lambda _{eq}} = \dfrac{{\kappa \times 1000}}{s}
SI unit of equivalent conductivity ohm−1cm2mol−1oh{m^{ - 1}}c{m^2}mo{l^{ - 1}} .
Given information-
κ\kappa = 8×10−5ohm−1cm−18 \times {10^{ - 5}}oh{m^{ - 1}}c{m^{ - 1}}
Λeq{\Lambda _{eq}} = 400ohm−1cm2equvi−1400oh{m^{ - 1}}c{m^2}equv{i^{ - 1}}
We know that relation between equivalent conductance into molar conductance is-
Λm = Λeq×equivalent factor of the electrolyte{{{\Lambda }}_{\text{m}}}{\text{ = }}{{{\Lambda }}_{{\text{eq}}}}{{ \times \text{equivalent factor of the electrolyte}}}
Here the equivalent factor of electrolyte means the total charge on cations or anions. And we know in BaSO4BaS{O_4} there are 2 equivalent factors of the electrolyte.
Λm=2×400{\Lambda _m} = 2 \times 400
Λm=800{\Lambda _m} = 800
Put all given data in the above equation and we get
800=8×10−5×1000s800 = \dfrac{{8 \times {{10}^{ - 5}} \times 1000}}{s}
Rearrange equation
s=8×10−5×1000800s = \dfrac{{8 \times {{10}^{ - 5}} \times 1000}}{{800}}
s=10−4Ms = {10^{ - 4}}M

Step 2: Ionization of BaSO4BaS{O_4}
We know that BaSO4BaS{O_4} will break into ions like this-
BaSO4→Ba+2+SO4−2BaS{O_4} \to B{a^{ + 2}} + S{O_4}^{ - 2}
So, the solubility product of BaSO4BaS{O_4} is
Ksp=s2{K_{sp}} = {s^2}
Ksp=[10−4]2{K_{sp}} = {\left[ {{{10}^{ - 4}}} \right]^2}
Solubility product value is 10−8M2{10^{ - 8}}{M^2}
Hence the correct option is B.

Note:
Should be careful with all units, as in question all units are equivalent but to find out the solubility product we need concentration in Molarity. So do not forget to change equivalent conductance into molar conductance. Use this formula Λm = Λeq×equivalent factor of the electrolyte{{{\Lambda }}_{\text{m}}}{\text{ = }}{{{\Lambda }}_{{\text{eq}}}}{{ \times \text{equivalent factor of the electrolyte}}} .