At 298 , K,the solubility of PbCl_2 is 2 time | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^{-}(aq)$.If $S$ is the solubility of $PbCl_2$,then $[Pb^

Vedclass · Accepted Answer

$3.2 	imes 10^{-5}$

Vedclass · Answer

(D) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) ightleftharpoons Pb^{2+}(aq) + 2Cl^{-}(aq)$.If $S$ is the solubility of $PbCl_2$,then $[Pb^{2+}] = S$ and $[Cl^{-}] = 2S$.The solubility product constant is $K_{sp} = [Pb^{2+}][Cl^{-}]^2$.Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.Given $S = 2 	imes 10^{-2} \, mol/L$,we have:$K_{sp} = 4 	imes (2 	imes 10^{-2})^3 = 4 	imes (8 	imes 10^{-6}) = 3.2 	imes 10^{-5}$.

At $298 \, K$,the solubility of $PbCl_2$ is $2 \times 10^{-2} \, mol/L$,then $K_{sp} = ?$

Similar Questions

When equal volumes of the following solutions are mixed,precipitation of $AgCl$ $(K_{sp} = 1.8 \times 10^{-10})$ will occur only with:

An ionic compound is dissolved simultaneously in heavy water and simple water. Its solubility is

Calculate solubility $(mol \ dm^{-3})$ of a sparingly soluble electrolyte $AB$ at $298 \ K$ if its solubility product is $1.6 \times 10^{-5}$?

Which of the following formulas can be used to calculate the solubility of $Ag_2CO_3$?

The compound insoluble in water is

Vedclass Test Series

Exam Paper Generator

Online Exam Module