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

The solubility product $(K_{sp})$ values of $CuS$,$Ag_2S$,and $HgS$ are $10^{-31}$,$10^{-44}$,and $10^{-54}$ respectively. The solubilities of these sulphides are in the order:

For the precipitation reaction of $Ag^{\oplus}$ ions with $NaCl,$ which of the following statements is true?

Which of the following sulphides has the lowest solubility product?

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$HgS \downarrow + HNO_3 \text{ (Conc.)} \longrightarrow \text{No dissolution}$

Solid $Ba(NO_3)_2$ is gradually added to a $1.0 \times 10^{-4} \, M \, Na_2CO_3$ solution. At what concentration of $Ba^{2+}$ will precipitation start? (Given $K_{sp}$ for $BaCO_3 = 5.1 \times 10^{-9}$)

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 $\sqrt{2} = 1.41$]