# Question #61e40

Aug 10, 2016

$\textsf{\left[T l C {l}_{4}^{-}\right] = 0.125 \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{\left[T {l}^{3 +}\right] = 0.025 \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

$\textsf{T {l}^{3 +} + 4 C {l}^{-} r i g h t \le f t h a r p \infty n s T l C {l}_{4}^{-}}$

For which:

$\textsf{{K}_{f} = \frac{\left[T l C {l}_{4}^{-}\right]}{\left[T {l}^{3 +}\right] {\left[C {l}^{-}\right]}^{4}} = {10}^{18} \textcolor{w h i t e}{x} \text{mol/l}}$

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 $\textsf{T {l}^{3 +}}$ requires 4 moles of $\textsf{C {l}^{-}}$ 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 $\textsf{C {l}^{-}}$ ions will consume 0.5/4 = 0.125 moles of $\textsf{T {l}^{3 +}}$ ions to form 0.125 moles of $\textsf{T l C {l}_{4}^{-}}$ ions.

This means that the number of moles of $\textsf{T {l}^{3 +}}$ left unreacted = 0.15-0.125 = 0.025.

So we can say:

$\textsf{\left[T l C {l}_{4}^{-}\right] = 0.125 \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{\left[T {l}^{3 +}\right] = 0.025 \textcolor{w h i t e}{x} \text{mol/l}}$