Chapter 22, Problem 22.71QP

Commercial silver-plating operations frequently use a solution containing the complex Ag(CN)₂⁻ ion. Because the formation constant (K_f) is quite large, this procedure ensures that the free Ag⁺ concentration in solution is low for uniform electrodeposition. In one process, a chemist added 9.0 L of 5.0 M NaCN to 90.0 L of 0.20 M AgNO₃. Calculate the concentration of free Ag⁺ ions at equilibrium. See Table 17.5 for K_f value.

Interpretation Introduction

Interpretation: The concentration of free ${\text{Ag}}^{\text{+}}$ ions at equilibrium to be calculated in a given chemical equilibrium.

Concept Introduction:

Reaction of coordination compounds: Complex ion undergoes ligand exchange (or substitution) reactions in solution. The rate of reaction is depends on the nature of metal ions and the ligands.

Formation constant: ${\text{K}}_{\text{f}}\text{=}\frac{{\text{[product]}}_{\text{ eq}}}{{\text{[reactant] }}_{\text{eq}}}$ the formation constant of an equilibrium reaction is the ratio of equilibrium concentration of product by the equilibrium concentration of reactant.

No. of moles: $\text{number}\text{of}\text{moles}\text{=}\text{molarity}\text{×}\text{volume}$

To Identify: The concentration of free ${\text{Ag}}^{\text{+}}$ ions at equilibrium to be calculated in a given chemical equilibrium.

Answer to Problem 22.71QP

Answer

The calculated concentration of free ${\text{Ag}}^{\text{+}}$ ions at equilibrium is $\text{2}{\text{.2 × 10}}^{\text{-20}}\text{ M}\text{.}$

Explanation of Solution

Find the concentration of free ${\text{Ag}}^{\text{+}}$ ions at equilibrium.

${\text{Ag}}^{\text{+}}\text{(aq)+}{\text{2CN}}^{\text{-}}\text{(aq)}\rightleftharpoons {{\text{[Ag(CN)}}_{\text{2}}\text{]}}^{\text{-}}\text{(aq)}$

The formation constant for the above reaction: ${\text{K}}_{\text{f}}\text{=}\frac{{{\text{[Fe(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{NCS]}}^{\text{2+}}{}_{\text{ eq}}}{{{\text{[Fe(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}}{}_{\text{eq}}{\text{ [SCN}}^{\text{-}}{\text{] }}_{\text{eq}}}$

• Concentration of cyanide ions:

$\begin{array}{l}{\text{Concentration [CN}}^{\text{-}}\text{] = }\frac{\text{moles of cyanide ions}}{\text{volume of solution}}\text{.}\\ \text{= }\frac{\text{Molarity × volume (of cyanide ions)}}{\text{volume of solution}}\\ \text{= }\frac{\text{(5}\text{.0 mol/L)(9}\text{.0 L)}}{\text{(90 + 9}\text{.0) L}}\\ \text{= 0}\text{.455 M}\text{.}\end{array}$

The concentration of cyanide ions is calculated as shown above. The concentration of cyanide ions is used at the equilibrium formation constant.

• Concentration of silver ions:

$\begin{array}{l}{\text{Concentration [Ag}}^{+}\text{] = }\frac{\text{moles of silver ions}}{\text{volume of solution}}\text{.}\\ \text{= }\frac{\text{Molarity × volume (of silver ions)}}{\text{volume of solution}}\\ \text{= }\frac{\text{(0}\text{.20 mol/L)(90 L)}}{\text{(90 + 9}\text{.0) L}}\\ \text{= 0}\text{.182 M}\text{.}\end{array}$

The concentration of silver ions is calculated as shown above. The concentration of silver ion is used at the equilibrium formation constant.

• Determine the concentrations after complete reaction.

$\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{Ag}}^{\text{+}}\text{(aq) +}{\text{2CN}}^{\text{-}}\text{(aq)}\rightleftharpoons {{\text{[Ag(CN)}}_{\text{2}}}^{\text{]}}\text{(aq)}\\ \text{initial(M):}\text{0}\text{.182}\text{0}\text{.455}\text{0}\text{.00}\\ \text{change(M):}\text{-0}\text{.182}\text{-(2)(0}\text{.182)}\text{+0}\text{.182}\end{array}}\\ \text{Equilibrium(M):}\text{0}\text{0}\text{.0910}\text{0}\text{.182}\\ {\text{K}}_{\text{f}}\text{= }\frac{{{\text{[Ag(CN)}}_{\text{2}}\text{]}}^{\text{-}}{}_{\text{ eq}}}{{{\text{[Ag}}^{\text{+}}\text{]}}_{\text{eq}}{\text{ [CN}}^{\text{-}}{\text{]}}^{\text{2}}{}_{\text{eq}}}\\ \text{1}\text{.0}{\text{×10}}^{\text{21}}\text{=}\frac{\text{0}\text{.182}}{{\text{[Ag}}^{\text{+}}\text{](0}{\text{.0910 M)}}^{\text{2}}}\\ {\text{[Ag}}^{\text{+}}\text{]}\text{= 2}{\text{.2×10}}^{\text{-20}}\text{M}\end{array}$

In order to calculate the concentration of silver ions, all the known values are substituted in the equilibrium formation constant. Thus, the obtained concentration of silver ions is $\text{2}{\text{.2×10}}^{\text{-20}}\text{M}\text{.}$

Conclusion

The concentration of free ${\text{Ag}}^{\text{+}}$ ions at equilibrium is calculated in a given chemical equilibrium.