15.2 Lewis acids and bases  (Page 3/12)

Dissociation of a complex ion

Solution

1. Determine the direction of change. The complex ion $\text{Ag}{({\text{NH}}_3)}_2{}^{\mathrm{+}}$ is in equilibrium with its components, as represented by the equation:

   ${\text{Ag}}^{\text{+}}(aq)+2{\text{NH}}_3(aq)\rightleftharpoons \text{Ag}{({\text{NH}}_3)}_2{}^{\text{+}}(aq)$

   We write the equilibrium as a formation reaction because Appendix K lists formation constants for complex ions. Before equilibrium, the reaction quotient is larger than the equilibrium constant [ K _f = 1.7 $\times$ 10 ⁷ , and $Q=\frac{0.10}{0\times 0},$ it is infinitely large], so the reaction shifts to the left to reach equilibrium.

2. Determine x and equilibrium concentrations. We let the change in concentration of Ag ⁺ be x . Dissociation of 1 mol of $\text{Ag}{({\text{NH}}_3)}_2{}^{\mathrm{+}}$ gives 1 mol of Ag ⁺ and 2 mol of NH ₃ , so the change in [NH ₃ ] is 2 x and that of $\text{Ag}{({\text{NH}}_3)}_2{}^{\mathrm{+}}$ is – x . In summary:

   

3. Solve for x and the equilibrium concentrations. At equilibrium:

   $K_{\text{f}}=\frac{[\text{Ag}{({\text{NH}}_3)}_2{}^{\mathrm{+}}]}{[{\text{Ag}}^{\mathrm{+}}]{[{\text{NH}}_3]}^2}$
   $1.7\times {10}^7\text{=}\frac{0.10-x}{(x){(2x)}^2}$

   Both Q and K _f are much larger than 1, so let us assume that the changes in concentrations needed to reach equilibrium are small. Thus 0.10 – x is approximated as 0.10:

   $1.7\times {10}^7=\frac{0.10-x}{(x){(2x)}^2}$
   $x^3=\frac{0.10}{4(1.7\times {10}^7)}=1.5\times {10}^{-9}$
   $x=\sqrt[3]{1.5\times {10}^{-9}}=1.1\times {10}^{-3}$

   Because only 1.1% of the $\text{Ag}{({\text{NH}}_3)}_2{}^{\mathrm{+}}$ dissociates into Ag ⁺ and NH ₃ , the assumption that x is small is justified.

   Now we determine the equilibrium concentrations:

   $[{\text{Ag}}^{\mathrm{+}}]=0+x=1.1\times {10}^{-3}M$
   $[{\text{NH}}_3]=0+2x=2.2\times {10}^{-3}M$
   $[\text{Ag}{({\text{NH}}_3)}_2{}^{\mathrm{+}}]=0.10-x=0.10-0.0011=0.099$

   The concentration of free silver ion in the solution is 0.0011 M .

4. Check the work. The value of Q calculated using the equilibrium concentrations is equal to K _f within the error associated with the significant figures in the calculation.

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