Q.58. Consider the following data :
Electrolyte | (S cm mol)
BaCl | x
HSO | x
HCl | x
BaSO is spari

Q: What is the correct answer for JEE Main Chemistry Q58?

The correct answer is A. For a sparingly soluble salt like BaSO$_4$, the conductivity of its saturated solution is primarily due to the ions from the salt. Let the solubility of BaSO$_4$ be $S$ mol/L. BaSO$_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$. So, $[Ba^{2+}] = S$ and $[SO_4^{2-}] = S$. The molar conductivity at infinite dilution for BaSO$_4$ is $\Lambda_m^0(BaSO_4) = \lambda^0(Ba^{2+}) + \lambda^0(SO_4^{2

Q: Q.58. Consider the following data : Electrolyte | (S cm mol) BaCl | x HSO | x HCl | x BaSO is sparingly soluble in water. If the conductivity of the saturated BaSO solution is x S cm then the…

The correct answer is A. For a sparingly soluble salt like BaSO$_4$, the conductivity of its saturated solution is primarily due to the ions from the salt. Let the solubility of BaSO$_4$ be $S$ mol/L. BaSO$_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$. So, $[Ba^{2+}] = S$ and $[SO_4^{2-}] = S$. The molar conductivity at infinite dilution for BaSO$_4$ is $\Lambda_m^0(BaSO_4) = \lambda^0(Ba^{2+}) + \lambda^0(SO_4^{2

Q: What are the options for this JEE Main Chemistry question?

A. $\frac{10^6x^2}{(x_1+x_2-2x_3)^2}$ · B. $\frac{x^2}{(x_1+x_2-2x_3)^2}$ · C. $\frac{10^6x^2}{(x_1+x_2+2x_3)^2}$ · D. $\frac{x^2}{(x_1+x_2+2x_3)^2}$

Question

Q.58. Consider the following data :
Electrolyte |  (S cm mol)
BaCl | x
HSO | x
HCl | x
BaSO is sparingly soluble in water. If the conductivity of the saturated BaSO solution is x S cm then the solubility product of BaSO can be given as.
(Here )

FirstInTest · Accepted Answer

For a sparingly soluble salt like BaSO$_4$, the conductivity of its saturated solution is primarily due to the ions from the salt. Let the solubility of BaSO$_4$ be $S$ mol/L. BaSO$_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$. So, $[Ba^{2+}] = S$ and $[SO_4^{2-}] = S$. The molar conductivity at infinite dilution for BaSO$_4$ is $\Lambda_m^0(BaSO_4) = \lambda^0(Ba^{2+}) + \lambda^0(SO_4^{2-})$. Using Kohlrausch's Law: $\Lambda_m^0(BaCl_2) = \lambda^0(Ba^{2+}) + 2\lambda^0(Cl^-) = x_1$. $\Lambda_m^0(H_2SO_4) = 2\lambda^0(H^+) + \lambda^0(SO_4^{2-}) = x_2$. $\Lambda_m^0(HCl) = \lambda^0(H^+) + \lambda^0(Cl^-) = x_3$. We want $\lambda^0(Ba^{2+}) + \lambda^0(SO_4^{2-})$. We can combine the given values: $\Lambda_m^0(BaSO_4) = \Lambda_m^0(BaCl_2) + \Lambda_m^0(H_2SO_4) - 2\Lambda_m^0(HCl)$. This is because $(\lambda^0(Ba^{2+}) + 2\lambda^0(Cl^-)) + (2\lambda^0(H^+) + \lambda^0(SO_4^{2-})) - 2(\lambda^0(H^+) + \lambda^0(Cl^-)) = \lambda^0(Ba^{2+}) + \lambda^0(SO_4^{2-})$. So, $\Lambda_m^0(BaSO_4) = x_1 + x_2 - 2x_3$. The conductivity $\kappa = x$ S cm$^{-1}$. For a sparingly soluble salt, $\Lambda_m = \frac{1000 \kappa}{S}$. Since $\Lambda_m = \Lambda_m^0$ for sparingly soluble salts, $S = \frac{1000 \kappa}{\Lambda_m^0} = \frac{1000x}{x_1 + x_2 - 2x_3}$. The solubility product $K_{sp} = [Ba^{2+}][SO_4^{2-}] = S^2$. So, $K_{sp} = \left( \frac{1000x}{x_1 + x_2 - 2x_3} \right)^2 = \frac{1000^2 x^2}{(x_1 + x_2 - 2x_3)^2} = \frac{10^6 x^2}{(x_1 + x_2 - 2x_3)^2}$.

Answer

$\frac{10^6x^2}{(x_1+x_2-2x_3)^2}$

Answer

$\frac{x^2}{(x_1+x_2-2x_3)^2}$

Answer

$\frac{10^6x^2}{(x_1+x_2+2x_3)^2}$

Answer

$\frac{x^2}{(x_1+x_2+2x_3)^2}$

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