Question

The formation constant of the TICI₄^- complex ion is $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{18}}$. Suppose 0.15 mol of TI(NO₃)₃ is dissolved in 1.00 L of a 0.50 M solution of NaCl. Calculate the concentration at equilibrium of TICI₄^- and of TI³⁺.

Short answer

As K_f is large, the equilibrium concentrations of TI³⁺(aq) andTICI₄^- (aq) do not change significantly from the values, the concentration of TI³⁺(aq) would be 0.025mol.L^-1 , and the concentration of TICI₄^- (aq) would be 0.125mol.L^-1.

Step-by-step solution

The chemical equation involved

In key chemical equilibrium reaction involved is as shown below:

$$\begin{gathered} T{I}^{3+}\left(aq\right)+4C{I}^{-}\left(aq\right)\square TiC{I_4}^{-}\left(aq\right) \\ \frac{\left[TiC{I_4}^{-}\right]}{\left[T{I}^{3+}\right]{\left[C{I}^{-}\right]}^4}=K_f=1\times {10}^{18} \end{gathered}$$

Explanation

The large formation constant K_f means that the complex ion is heavily favored at equilibrium.

Suppose that all of the TI³⁺(aq) ion from the dissolution TI(NO₃)₃ reacts with the 0.50 mol of CI^-(aq) in solution to form the complex ion.

TheTI³⁺(aq) is in excess, so the concentration of CI^- would fall to zero, the concentration of TI³⁺(aq) would be 0.025mol.L^-1 ,and the concentration of TiCI₄^- (aq) would be 0.125mol.L^-1.

Then imagine that equilibrium is attained by the partial break up of the complex.

Because K_f is large, the equilibrium concentrations of TI³⁺(aq) and TiCI₄^-(aq) do not change significantly from the above values.

No calculations are needed except to verify that the equilibrium concentration of is indeed small.(it equals $5\times {10}^{-5}mol.L^{-1}$).