Question

Equivalent conductance of saturated $BaS{O_4}$ is $400oh{m^{ - 1}}c{m^2}equv{i^{ - 1}}$ and specific conductance is $8 \times {10^{ - 5}}oh{m^{ - 1}}c{m^{ - 1}}$ . Hence ${K_{sp}}$ of $BaS{O_4}$ is.
A) $4 \times {10^{ - 8}}{M^2}$
B) $1 \times {10^{ - 8}}{M^2}$
C) $2 \times {10^{ - 4}}{M^2}$
D) $1 \times {10^{ - 4}}{M^2}$

Answer 1

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

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 \[\kappa \]. It is represented by $\kappa = \dfrac{1}{\rho }$ . So its unit is $oh{m^{ - 1}}c{m^{ - 1}}$.
So, we have $\kappa = \dfrac{1}{\rho }$
Here $\rho $ resistivity
Molar conductance- ${\Lambda _m} = \dfrac{\kappa }{C}$
Here C= concentration
If concentration values are given in equivalent/liter then it is called equivalent conductance and the above equation becomes
${\Lambda _{eq}} = \dfrac{{\kappa \times 1000}}{s}$
SI unit of equivalent conductivity $oh{m^{ - 1}}c{m^2}mo{l^{ - 1}}$ .
Given information-
\[\kappa \] = $8 \times {10^{ - 5}}oh{m^{ - 1}}c{m^{ - 1}}$
${\Lambda _{eq}}$ = $400oh{m^{ - 1}}c{m^2}equv{i^{ - 1}}$
We know that relation between equivalent conductance into molar conductance is-
${{{\Lambda }}_{\text{m}}}{\text{ = }}{{{\Lambda }}_{{\text{eq}}}}{{ \times \text{equivalent factor of the electrolyte}}}$
Here the equivalent factor of electrolyte means the total charge on cations or anions. And we know in $BaS{O_4}$ there are 2 equivalent factors of the electrolyte.
${\Lambda _m} = 2 \times 400$
${\Lambda _m} = 800$
Put all given data in the above equation and we get
$800 = \dfrac{{8 \times {{10}^{ - 5}} \times 1000}}{s}$
Rearrange equation
$s = \dfrac{{8 \times {{10}^{ - 5}} \times 1000}}{{800}}$
$s = {10^{ - 4}}M$

Step 2: Ionization of $BaS{O_4}$
We know that $BaS{O_4}$ will break into ions like this-
$BaS{O_4} \to B{a^{ + 2}} + S{O_4}^{ - 2}$
So, the solubility product of $BaS{O_4}$ is
${K_{sp}} = {s^2}$
${K_{sp}} = {\left[ {{{10}^{ - 4}}} \right]^2}$
Solubility product value is ${10^{ - 8}}{M^2}$
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 ${{{\Lambda }}_{\text{m}}}{\text{ = }}{{{\Lambda }}_{{\text{eq}}}}{{ \times \text{equivalent factor of the electrolyte}}}$ .