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.