Question 16.14: What are the concentrations of Ag^+ , Ag(NH3)^+, and Ag(NH3)......
What are the concentrations of Ag^{+} , Ag(NH_{3})^{+}, and Ag(NH_{3})_{2} ^{+} in a solution prepared by adding 0.10 mol of AgNO_{3} to 1.0 L of 3.0 M NH_{3}? K_{f} = 1.7\times 10^{7}, K_{1} = 2.1\times 10^{3}, and K_{2} = 8.1\times 10^{3}.
STRATEGY
Because K_{1}, K_{2}, and K_{f} for Ag(NH_{3})_{2} ^{+} are all large numbers, we surmise that nearly all the Ag^{+} from AgNO_{3} will be converted to Ag(NH_{3})_{2} ^{+}:
Ag^{+}(aq)+2 NH_{3}(aq)\xrightleftharpoons{}Ag(NH_{3})_{2} ^{+}(aq) K_{f}=1.7\times 10^{7}To calculate the concentrations, it’s convenient to imagine that 100% of the Ag^{+} is converted to Ag(NH_{3})_{2} ^{+}, followed by a tiny amount of back-reaction [dissociation of Ag(NH_{3})_{2} ^{+}] to give a small equilibrium concentration of Ag^{+}.
| IDENTIFY | |
| Known | Unknown |
| Molarity and volume of NH_{3} (3.0 M, 1.0 L) | [Ag^{+}], [Ag(NH_{3})^{+}], [Ag(NH_{3})_{2} ^{+}] |
| Amount of AgNO_{3} (0.10 mol) | |
| Equilibrium constants and reactions | |
First determine concentrations assuming 100% conversion to Ag(NH_{3})_{2} ^{+}. The conversion of 0.10 mol/L of Ag^{+} to Ag(NH_{3})_{2} ^{+} consumes 0.20 mol/L of NH_{3} and the following concentrations are obtained:
[Ag^{+}]= 0 M[Ag(NH_{3})_{2} ^{+}]=0.10 M
[NH_{3}]=3.0-0.20=2.8 M
Next set up an equilibrium table for the dissociation of Ag(NH_{3})_{2} ^{+}:
Table 1
The dissociation of x mol/L of Ag(NH_{3})_{2} ^{+} in the back-reaction produces x mol/L of Ag^{+} and 2x mol/L of NH_{3}. Therefore, the equilibrium concentrations (in mol/L) are
[Ag(NH_{3})_{2} ^{+}]=0.10 – x[Ag^{+}]= x
[NH_{3}]=2.8 + 2x
Substituting the equilibrium concentrations into the expression for K_{f} and making the approximation that x is negligible compared to 0.10 (and to 2.8) gives
K_{1}=1.7\times 10^{7}=\frac{[Ag(NH_{3})_{2} ^{+}]}{[Ag^{+}][NH_{3}]^{2}}=\frac{0.10 – x}{(x)(2.8 + 2x)^{2}}=\frac{0.10}{(x)(2.8)^{2}}[Ag^{+}]=x=\frac{0.10}{(1.7\times 10^{7})(2.8)^{2}}=7.5\times 10^{-10} M
[Ag(NH_{3})_{2} ^{+}]=0.10 – x=0.10-(7.5\times 10^{-10})=0.10 M
The concentration of Ag(NH_{3})^{+} can be calculated from either of the stepwise equilibria. Let’s use the equilibrium equation for the formation of Ag(NH_{3})^{+} from Ag^{+}:
K_{1}=\frac{[Ag(NH_{3})^{+}]}{[Ag^{+}][NH_{3}]}=2.1\times 10^{3}[Ag(NH_{3})^{+}]=K_{1}[Ag^{+}][NH_{3}]=(2.1\times 10^{3})(7.5\times 10^{-10})(2.8)=4.4\times 10^{-6} M
Thus, nearly all the Ag^{+} is in the form of Ag(NH_{3})_{2} ^{+}.
Table 1
| Ag^{+}(aq) | + | 2 NH_{3}(aq) | \xrightleftharpoons{} | Ag(NH_{3})_{2} ^{+}(aq) | |
| Initial concentration (M) | 0 | 2.8 | 0.10 | ||
| Change (M) | + x | + 2x | – x | ||
| Equilibrium concentration (M) | x | 2.8 + 2x | 0.10 – x |