# Problem related to solubility product....1) a) calculate the minimum chloride ion concentration, [Cl-], required to initiate precipitation of lead (II) chloride from a solution that is initially...

Problem related to solubility product....

1) a) calculate the minimum chloride ion concentration, [Cl-], required to initiate precipitation of lead (II) chloride from a solution that is initially 0.02 M in Pb 2+?

b) What volume of 6M HCl must be added to 2 ml of 0.02 M Pb2+ to effect the [Cl-] you calculated above? Ksp=[Pb2+][Cl-]^2 = 2x10^-5

jeew-m | College Teacher | (Level 1) Educator Emeritus

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`PbCl_2 (s)rarr Pb^(2+)_(aq)+2Cl^-_(aq)`

The solubility product of `PbCl_2` is `K_(sp) = 2xx10^-5`

`K-(sp) = [Pb^(2+)][Cl^-]^2`

a)

It is given that in the solution `[Pb^(2+)] ` = 0.02 M

2x10^-5 = 0.02*`[Cl^-]^2`

`[Cl^-] ` =`sqrt[(2xx10^-5)/0.02]`

= 0.032 M

So the minimum `Cl^-` concentration to precipitate `PbCl_2` is 0.032 M.

b)

Amount of `Pb^(2+)` moles present    = 0.02/1000*2

= `4*10^(-5) mol`

In the solution `Pb^(2+):Cl^-` = 1:2

So to start precipitation we need at least `2*4*10^(-5)` mol of `Cl^-.`

`HCl_(aq)--gt H^+_(aq)+Cl^-_(aq)`

The volume of HCl where 6 moles present         = 1000

The volume needed to form `2*4*10^(-5)` mol  = `1000/6*8*10^(-5)`

= 0.013 ml

The volume of 6M HCl needed is 0.013 ml which is a very small volume. Even this cannot be measured. Even a drop of HCl will precipitate lead(ii) chloride.

But if you decrease the concentration of HCl to a lower value same as Pb^2+ you will get a measurable amount of HCl volume to initiate the precipitation.

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