Question

The value of Ksp for PbCl2 is 1.6 ×10-5. What is the lowest concentration of Cl-(aq) that would be needed to begin precipitation of PbCl2(s) in 0.010 M Pb(NO3)2 ?

Answer 1

Answer:

0.04 M is the lowest concentration of chloride ions that would be needed to begin precipitation.

Explanation:

Concentration of lead nitarte = $[Pb(NO_3)_2]=0.010 M$

$Pb(NO_3)_2(aq)\rightleftharpoons Pb^{2+}(aq)+2NO_{3}^{-}(aq0$

1 Mole of lead nirate gives 1 mole of lead ion.

Concentration of lead ion in the solution = $1\times 0.010 M= 0.010 M$

$Pb(Cl)_2(aq)\rightleftharpoons  Pb^{2+}(aq)+2Cl^{-}(aq0$

Concentration of chloride ions = $[Cl^-]$

The value of $K_{sp} for [tex]PbCl_2= 1.6\times 10^{-5}$

$K_{sp}=[Pb^{2+}][Cl^{-}]^2$

$1.6\times 10^{-5}=0.010 M\times [Cl^{-}]^2$

$[Cl^-]=0.04 M$

0.04 M is the lowest concentration of chloride ions that would be needed to begin precipitation.

Answer 2

The minimum concentration of Chloride ions required for the precipitation of lead chloride has been 0.04 M.

Ksp has been the solubility product constant. It has been required for the solubility of the cation and anion in a chemical solution.

The Ksp for Lead chloride has been given by:

$\rm PbCl_2\;\rightarrow\;Pb^2^+\;+\;Cl^-$

Ksp = [$\rm Pb^2^+$] [$\rm Cl^-$]

The concentration of lead nitrate = 0.01 M

Dissociation of lead nitrate:

$\rm Pb(NO_3)_2\;\rightarrow\;Pb^2^+\;+\;2\;NO_3^-$

1 mole of lead nitrate = 1 mole of lead ions

0.01 M lead nitrate = 0.01 M Lead ions

Given, Ksp = $\rm 1.6\;\times\;10^-^5$

$\rm 1.6\;\times\;10^-^5$ = 0.01 $\times$  [$\rm Cl^-$]

[$\rm Cl^-$] = 0.04 M.

The minimum concentration of Chloride ions required for the precipitation of lead chloride has been 0.04 M.

For more information about solubility product constant, refer to the link:

https://brainly.com/question/13949246