If the concentration of lead iodide in its sa | vedclass

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

Question

(D) The dissociation reaction for lead iodide is: $PbI_2(s) ⇌ Pb^{2+}(aq) + 2I^{-}(aq)$.For a salt of type $AB_2$,the solubility product $K_{sp}$ is g

Vedclass · Accepted Answer

$32 	imes 10^{-9}$

Vedclass · Answer

(D) The dissociation reaction for lead iodide is: $PbI_2(s) ⇌ Pb^{2+}(aq) + 2I^{-}(aq)$.For a salt of type $AB_2$,the solubility product $K_{sp}$ is given by $K_{sp} = [Pb^{2+}][I^{-}]^2$.If $S$ is the solubility,then $[Pb^{2+}] = S$ and $[I^{-}] = 2S$.Therefore,$K_{sp} = (S)(2S)^2 = 4S^3$.Given $S = 2 	imes 10^{-3} \ mol \ L^{-1}$.Substituting the value of $S$ into the expression: $K_{sp} = 4 	imes (2 	imes 10^{-3})^3 = 4 	imes (8 	imes 10^{-9}) = 32 	imes 10^{-9}$.

If the concentration of lead iodide in its saturated solution at $25 \, ^oC$ is $2 \times 10^{-3} \ mol \ L^{-1}$,then its solubility product is:

Similar Questions

When $H_2S$ gas is passed through a mixture containing $Mn^{2+}$,$Ni^{2+}$,$Cu^{2+}$,and $Hg^{2+}$ ions in an acidic aqueous solution,which of the following will precipitate?

On passing $H_2S$ gas through a solution of $Cu^{2+}$ and $Zn^{2+}$ ions,$CuS$ is precipitated first because

The $K_{sp}$ expression for $Ca_3(PO_4)_2$ is:

The solubility product expression for $Ca_3(PO_4)_2$ is represented as

The $K_{sp}$ for the following dissociation is $1.6 \times 10^{-5}$:
$PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}$
Which of the following choices is correct for a mixture of $300 \ mL$ of $0.134 \ M \ Pb(NO_3)_2$ and $100 \ mL$ of $0.4 \ M \ NaCl$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module