Question 17.SE.15: Calculate the concentration of Ag^+ present in solution at e...
Calculate the concentration of Ag^+ present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO_3 to give an equilibrium concentration of [NH_3] = 0.20 M. Neglect the small volume change that occurs when NH_3 is added.
Analyze Addition of NH_3(aq) to Ag^+(aq) forms Ag(NH_3)_2^+(aq), as shown in Equation 17.22. We are asked to determine what concentration of Ag^+(aq) remains uncombined when the NH_3 concentration is brought to 0.20 M in a solution originally 0.010 M in AgNO_3.
Ag ^{+}(a q)+2 NH _3(a q) \rightleftharpoons Ag \left( NH _3\right)_2^{+}(a q) (17.22)
Plan We assume that the AgNO_3 is completely dissociated, giving 0.010 M Ag^+. Because K_f for the formation of Ag(NH_3)_2^+ is quite large, we assume that essentially all the Ag^+ is converted to Ag(NH_3)_2^+ and approach the problem as though we are concerned with the dissociation of Ag(NH_3)_2^+ rather than its formation. To facilitate this approach, we need to reverse Equation 17.22 and make the corresponding change to the equilibrium constant:
\begin{gathered}Ag \left( NH _3\right)_2^{+}(a q) \rightleftharpoons Ag ^{+}(a q)+2 NH _3(a q) \\\frac{1}{K_f}=\frac{1}{1.7 \times 10^7}=5.9 \times 10^{-8}\end{gathered}
Solve If [Ag^+] is 0.010 M initially, [Ag(NH_3)_2^+] will be 0.010 M following addition of the NH_3. We construct a table to solve this equilibrium problem. Note that the NH_3 concentration given in the problem is an equilibrium concentration rather than an initial concentration.
\begin{array}{|l|c|r|c|}\hline&Ag \left( NH _3\right)_2^{+}(a q) \rightleftharpoons& Ag ^{+}(a q)+&2 NH _3(a q)\\\hline \text{Initial}(M) & 0.010 & 0 & – \\\hline \text{Change}(M) & -x & +x & – \\\hline \text{Equilibrium}(M) & (0.010-x) & x & 0.20 \\\hline\end{array}
Because [Ag^+] is very small, we can assume x is small compared to 0.010. Substituting these values into the equilibrium expression for the dissociation of Ag(NH_3)_2^+, we obtain
\begin{gathered}\frac{\left[ Ag ^{+}\right]\left[ NH _3\right]^2}{\left[ Ag \left( NH _3\right)_2^{+}\right]}=\frac{(x)(0.20)^2}{0.010}=5.9 \times 10^{-8}\\x=1.5 \times 10^{-8}\,M=\left[ Ag ^{+}\right]\end{gathered}
Formation of the Ag(NH_3)_2^+ complex drastically reduces the concentration of free Ag^+ ion in solution.