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
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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`
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.
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^-.`
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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