Question

The equivalent conductivity of solutions of $BaC{l}_2,H_{2}S{O}_4\text{and}HCl$, are $x_1,x_2\text{and}{x}_3{\text{Scm}}^2{\text{eq}}^{-1}$ , respectively at infinite dilution. If the conductivity of saturated $BaS{O}_4$ solution is ${\text{x Scm}}^{-1}$ , then the $K_{sp}$ of $BaS{O}_4$ is:

• A. $\frac{500x}{(x_1+x_2-2x_3)}$
• B. $\frac{{10}^6{x}^2}{4{(x_1+x_2-2x_3)}^2}$
• C. $\frac{0.25x^2}{{(x_1+x_2-x_3)}^2}$
• D. $\frac{2.5\times {10}^5{x}^2}{{(x_1+x_2-x_3)}^2}$

Answer 1

The correct option is D $\frac{2.5\times {10}^5{x}^2}{{(x_1+x_2-x_3)}^2}$

From Kohlrausch law:
${\Lambda }_{e{q}_{BaS{O}_4}}^o={\Lambda }_{e{q}_{BaC{l}_2}}^o+{\Lambda }_{e{q}_{H_{2}S{O}_4}}^o-{\Lambda }_{e{q}_{HCl}}^o$
= $(x_1+x_2-x_3)Sc{m}^{2}e{q}^{-1}$
Since, we know that:
${\Lambda }_M^o={\Lambda }_{eq}^o\times n-factor$
$\therefore$
${\Lambda }_{M_{BaS{O}_4}}^o={\Lambda }_{e{q}_{BaS{O}_4}}^o\times 2$
= $2\times (x_1+x_2-x_3)$
For saturated solution of $BaS{O}_4$
$\kappa =xSc{m}^{-1}$ (given in question)
${\Lambda }_M^o=\frac{kappa\times 1000}{M}$
$\therefore$
$2\times (x_1+x_2-x_3)=\frac{\kappa \times 1000}{M}$
Solving for ‘M’ we get:
M = $\frac{x\times 1000}{2\times (x_1+x_2-x_3)}$
Now, as we know that at infinite dilution, mostly sparingly soluble salts are 100% dissociated therefore,
Solubility will be the same as its molarity
I.e. solubility(s) = $\frac{x\times 1000}{2\times (x_1+x_2-x_3)}$
The equation for solubility of $BaS{O}_4$ is:
$BaS{O}_4\rightleftharpoons B{a}^{2+}+S{O}_4^{2-}$
From the above equation we can say that the solubility of both $B{a}^{2+}$ and $S{O}_4^{2-}$ is the same and is equal to ‘s’.
Hence,
$K_{sp}=s^2$
= $(\frac{x\times 1000}{2\times (x_1+x_2-x_3)}{)}^2$
= $\frac{2.5\times {10}^5\times x^2}{(x_1+x_2-x_3{)}^2}$
Hence, the correct option is (d)