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:

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

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