JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Consider the following data.
| Electrolyte | \(\Lambda^{\circ}_m\) (S cm\(^2\) mol\(^{-1}\)) |
|---|---|
| \(\text{BaCl}_2\) | \(x_1\) |
| \(\text{H}_2\text{SO}_4\) | \(x_2\) |
| HCl | \(x_3\) |
\(\text{BaSO}_4\) is sparingly soluble in water. If the conductivity of the saturated \(\text{BaSO}_4\) solution is \(x\) S cm\(^{-1}\) then the solubility product of \(\text{BaSO}_4\) can be given as (Here \(\Lambda_m = \Lambda^{\circ}_m\))
- A \(\dfrac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}\)
- B \(\dfrac{x^2}{(x_1 + x_2 - 2x_3)^2}\)
- C \(\dfrac{\alpha^2(x_1 + x_2 - 2x_3)^2}{10^6 x^2}\)
- D \(\dfrac{x^2}{(x_1 + x_2 + 2x_3)^2}\)
Answer & Solution
Correct Answer
(A) \(\dfrac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}\)
Step-by-step Solution
Detailed explanation
Using Kohlrausch's law of independent migration of ions, the limiting molar conductivity of \(\text{BaSO}_4\) is given by: \(\Lambda^{\circ}_m(\text{BaSO}_4) = \Lambda^{\circ}_m(\text{BaCl}_2) + \Lambda^{\circ}_m(\text{H}_2\text{SO}_4) - 2\Lambda^{\circ}_m(\text{HCl})\)…
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