Question

Question #61e40

Answer 1

`sf([TlCl_4^-]=0.125color(white)(x)"mol/l")`

`sf([Tl^(3+)]=0.025color(white)(x)"mol/l")`

Explanation:

`sf(Tl^(3+)+4Cl^(-)rightleftharpoonsTlCl_4^-)`

For which:

`sf(K_f=([TlCl_4^-])/([Tl^(3+)][Cl^-]^4)=10^(18)color(white)(x)"mol/l")`

Because the formation constant is so large we can say that the reaction has gone to completion.

An ICE table is not applicable in this case. It is just a reacting masses problem.

From the equation you can see that 1 mole of `sf(Tl^(3+))` requires 4 moles of `sf(Cl^-)` ions.

This means that 0.15 mol require 4 x 0.15 = 0.6 moles.

However, we are told we only have 1L of a 0.5M solution which = 0.5 moles.

0.5 moles of `sf(Cl^-)` ions will consume 0.5/4 = 0.125 moles of `sf(Tl^(3+))` ions to form 0.125 moles of `sf(TlCl_4^-)` ions.

This means that the number of moles of `sf(Tl^(3+))` left unreacted = 0.15-0.125 = 0.025.

So we can say:

`sf([TlCl_4^-]=0.125color(white)(x)"mol/l")`

`sf([Tl^(3+)]=0.025color(white)(x)"mol/l")`