Question #61e40

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Answer 1 · 2016-08-10T17:36:26

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

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

#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")#