The solubility product of PbCl_2 at 20 , ^oC | vedclass

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

Question

(B) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) ⇌ Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $S \, mol/L$.Then,$[Pb^{2+}] = S$ and $[Cl^-

Vedclass · Accepted Answer

$3.34 	imes 10^{-2}$

Vedclass · Answer

(B) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) ⇌ Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $S \, mol/L$.Then,$[Pb^{2+}] = S$ and $[Cl^-] = 2S$.The solubility product expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2$.Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 1.5 	imes 10^{-4}$,we have $4S^3 = 1.5 	imes 10^{-4}$.$S^3 = \frac{1.5 	imes 10^{-4}}{4} = 0.375 	imes 10^{-4} = 3.75 	imes 10^{-5}$.$S = \sqrt[3]{3.75 	imes 10^{-5}} = \sqrt[3]{37.5 	imes 10^{-6}} \approx 3.34 	imes 10^{-2} \, mol/L$.

The solubility product of $PbCl_2$ at $20 \, ^oC$ is $1.5 \times 10^{-4}$. Calculate the solubility.

Similar Questions

If the solubility of $AgCl$ (molar mass $= 143 \, g/mol$) in water at $25 \, ^\circ C$ is $1.43 \times 10^{-4} \, g/100 \, mL$ of solution,then the value of $K_{sp}$ is?

How many grams of $CaC_2O_4$ will dissolve in distilled water to make one litre of saturated solution (in $g$)? (Solubility product of $CaC_2O_4$ is $2.5 \times 10^{-9} \ mol^2 \ L^{-2}$ and its molecular weight is $128$)

Find out the solubility of $Ni(OH)_2$ in $0.1 \ M \ NaOH$.
Given that the solubility product constant $(K_{sp})$ of $Ni(OH)_2$ is $2 \times 10^{-15}$.

Precipitation of $CaF_2$ $(K_{sp} = 1.7 \times 10^{-10})$ occurs when equal volumes of which of the following are mixed?

If the solubility of lithium sodium hexafluoroaluminate $Li_3Na_3(AlF_6)_2$ is $S \ mol/L$,then its solubility product $K_{sp}$ is equal to?

Vedclass Test Series

Exam Paper Generator

Online Exam Module