Question 19.13: Calculating the Concentration of a Complex Ion Problem An in......

Plan We write the complex-ion formation equation and the K_f expression and use a reaction table to calculate the equilibrium concentrations. To set up the table, we must first find [Zn(H_2O)_4^{2+}]_{\text{init}}. We are given the individual volumes and molar concentrations, so we find the number of moles and divide by the total volume because the solutions are mixed. With the large excess of NH_3 and the high K_f, we assume that almost all the Zn(H_2O)_4^{2+} is converted to Zn(NH_3)_4^{2+}. Because [Zn(H_2O)_4^{2+}] at equilibrium is very small, we use x to represent it.

Solution Writing the equation and the K_f expression:
           Zn(H_2O)_4^{2+}(aq)  +  4NH_3(aq)  \xrightleftharpoons[]{}  Zn(NH_3)_4^{2+}(aq)  +  4H_2O(l)
           K_f  =  \frac{[Zn(NH_3)_4^{2+}]}{[Zn(H_2O)_4^{2+}][NH_3]^4}
Finding the initial reactant concentrations:

           [Zn(H_2O)_4^{2+}]_{\text{init}}  =  \frac{50.0  L  ×  0.0020  M}{50.0  L  +  25.0  L}  =  1.3×10^{−3}  M
           [NH_3]_{\text{init}}  =  \frac{25.0  L  ×  0.15  M}{50.0  L  +  25.0  L}  =  5.0×10^{−2}  M

Setting up a reaction table (Table 1): We assume that nearly all the Zn(H_2O)_4^{2+} is converted to Zn(NH_3)_4^{2+}, so we set up the table with x  =  [Zn(H_2O)_4^{2+}] at equilibrium. Because 4 mol of NH_3 is needed per mole of Zn(H_2O)_4^{2+}, the change in [NH_3] is

Solving for x, the [Zn(H_2O)_4^{2+}] remaining at equilibrium:
           K_f  =  \frac{[Zn(NH_3)_4^{2+}]}{[Zn(H_2O)_4^{2+}][NH_3]^4}  =  7.8×10^8  ≈  \frac{1.3×10^{−3}}{x(4.5×10^{−2})^4}
           x  =  [Zn(H_2O)_4^{2+}]  ≈  4.1×10^{−7}  M
Check The K_f is large, so we expect the [Zn(H_2O)_4^{2+}] remaining to be very low.