If the solubility product of PbS is 8 times 1 | vedclass

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

Question

(B) The dissociation of $PbS$ is given by: $PbS(s) \rightleftharpoons Pb^{2+}(aq) + S^{2-}(aq)$.The solubility product expression is $K_{sp} = [Pb^{2+

Vedclass · Accepted Answer

$282$

Vedclass · Answer

(B) The dissociation of $PbS$ is given by: $PbS(s) ightleftharpoons Pb^{2+}(aq) + S^{2-}(aq)$.The solubility product expression is $K_{sp} = [Pb^{2+}][S^{2-}] = S 	imes S = S^{2}$.Given $K_{sp} = 8 	imes 10^{-28}$.Therefore,$S = \sqrt{8 	imes 10^{-28}} = \sqrt{4 	imes 2 	imes 10^{-28}} = 2 \sqrt{2} 	imes 10^{-14} \ mol \ L^{-1}$.Using the given value $\sqrt{2} = 1.41$,we get $S = 2 	imes 1.41 	imes 10^{-14} = 2.82 	imes 10^{-14} \ mol \ L^{-1}$.To express this in the form $x 	imes 10^{-16} \ mol \ L^{-1}$,we write $2.82 	imes 10^{-14} = 282 	imes 10^{-16}$.Thus,the value of $x$ is $282$.

If the solubility product of $PbS$ is $8 \times 10^{-28}$,then the solubility of $PbS$ in pure water at $298 \ K$ is $x \times 10^{-16} \ mol \ L^{-1}$. The value of $x$ is $\dots$.
[Given $\sqrt{2} = 1.41$]

Similar Questions

Explain the equilibrium involving the dissolution of a solid in a liquid.

If $[S^{2-}] = 0.6 \times 10^{-2} \ mol \ m^{-3}$ and $[Hg^{2+}] = [Mn^{2+}] = [Fe^{2+}] = [Zn^{2+}] = 1 \times 10^{-16} \ mol \ m^{-3}$,which of the following will precipitate first in an aqueous solution?

The ${K_{sp}}$ of $BaSO_4$ is $1.1 \times 10^{-10}$. Will a precipitate form when equal volumes of $2 \times 10^{-4} \ M \ BaCl_2$ and $5.0 \times 10^{-3} \ M \ H_2SO_4$ solutions are mixed? Explain by calculation.

The solubility product of $AgCl$ is $1.5625 \times 10^{-10}$ at $25\ ^oC$. Its solubility in grams per litre will be

What is the minimum concentration of $SO_4^{2-}$ required to precipitate $BaSO_4$ in a solution containing $1.0 \times 10^{-4} \ M \ Ba^{2+}$ ($K_{sp}$ for $BaSO_4$ is $4 \times 10^{-10}$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If the solubility product of $PbS$ is $8 \times 10^{-28}$,then the solubility of $PbS$ in pure water at $298 \ K$ is $x \times 10^{-16} \ mol \ L^{-1}$. The value of $x$ is $\dots$.[Given $\sqrt{2} = 1.41$]