Question

`K_"sp"=7.9xx10^-9` for `PbI_2`. If `0.365*L` of `0.0054*mol*L^-1` `Pb(NO_3)_2(aq)` is added to `0.960*L` `6.34xx10^-3*mol*L^-1` `HI(aq)`, will a precipitate be observed?

Answer 1

Well let's see.....

Explanation:

We assess the solubility equilibrium...

`Pb^(2+) + 2I^(-) rightleftharpoonsPbI_2(s)darr`

Where `K_"sp"=[Pb^(2+)][I^(-)]^2=7.9xx10^-9`

Now we assess the ion product of the given reaction, and note that the final volume is `960*mL+365*mL`, and we must calculate `[Pb^(2+)]` and `[I^-]` appropriately.

`[Pb^(2+)]=(0.365*Lxx0.0054*mol*L^-1)/(0.960*L+0.365*L)=1.488xx10^-3*mol*L^-1`

`[I^(-)]=(0.595*Lxx6.34xx10^(-4)*mol*L^-1)/(0.960*L+0.365*L)=2.85xx10^-4*mol*L^-1`

And now we take the ion product `Q=[2.85xx10^-4][1.488xx10^-3]^2=6.31xx10^-10`

And since `Q` IS MANIFESTLY LESS THAN `K_(sp)`, precipitation of `PbI_2` SHOULD NOT OCCUR.....