Question

An amount of \(0.01\) moles of solid \(\mathrm{AgCN}\) is rendered soluble in 11 by adding just sufficient excess cyanide ion to form \(\mathrm{Ag}(\mathrm{CN})_{2}^{-}\) and the concentration of free cyanide ion is \(2.5 \times 10^{-7} \mathrm{M}\). Determine \(\left[\mathrm{Ag}^{+}\right]\) in the solution neglecting hydrolysis of cyanide ion. \(K_{\text {diss }}\) for \(\mathrm{Ag}(\mathrm{CN})_{2}^{-}\) \(=1.0 \times 10^{-20}\) (a) \(6.25 \times 10^{-9} \mathrm{M}\) (b) \(1.6 \times 10^{-9} \mathrm{M}\) (c) \(1.6 \times 10^{-7} \mathrm{M}\) (d) \(6.25 \times 10^{-7} \mathrm{M}\)

Short answer

\( [Ag^{+}] \approx 1.6 \times 10^{-9} M \)

Step-by-step solution

Identify the Reaction and Relevant Equations

The dissolution of AgCN in the presence of excess cyanide ion is given by the equation: \(AgCN(s) + CN^{-} \rightarrow Ag(CN)_{2}^{-}\).The equilibrium expression associated with this reaction is: \(K_{diss} = [Ag(CN)_{2}^{-}] / [Ag^{+}][CN^{-}]^2\).

Calculate the Concentration of Ag(CN)2-

Since 0.01 moles of AgCN is completely converted into Ag(CN)2-, and assuming the volume of the solution is 1 liter for simplicity, the concentration of Ag(CN)2- is equal to the number of moles of AgCN, which is 0.01 M.

Express [Ag+] in Terms of Known Quantities

The concentration of free cyanide ion is given as \(2.5 \times 10^{-7} M\). Using the equilibrium expression and the given dissociation constant, we can write \(1.0 \times 10^{-20} = [Ag(CN)_{2}^{-}] / [Ag^{+}][CN^{-}]^2\). Solving for [Ag+], we get \([Ag^{+}] \approx [Ag(CN)_{2}^{-}] / (K_{diss} \times [CN^{-}]^2)\).

Calculate [Ag+]

Substitute the known values into the equation: \([Ag^{+}] \approx 0.01 / (1.0 \times 10^{-20} \times (2.5 \times 10^{-7})^2) \approx \frac{0.01}{1.0 \times 10^{-20} \times 6.25 \times 10^{-14}} \approx 1.6 \times 10^{-9} M\).

Key concepts

Dissociation Constant

Understanding the dissociation constant (\(K_{diss}\)) is essential for grasping the nature of chemical equilibrium in solutions. It is a particular form of the equilibrium constant specific to the dissociation of a complex ion into its components. For instance, when silver cyanide (\(AgCN\)) dissociates in the presence of excess cyanide ions (\(CN^{-}\)), it forms the complex ion \(Ag(CN)_{2}^{-}\).

The dissociation constant for this process quantifies the extent to which the complex ion is formed and denotes the ratio of the concentration of the product (\(Ag(CN)_{2}^{-}\)) to the concentrations of the reactants (\(Ag^{+}\) and \([CN^{-}]^2\)). The higher the value of \(K_{diss}\), the more the complex ion tends to form. In the given exercise, a very low value of \(K_{diss}\) suggests that the complex ion is very stable and does not readily dissociate back into its constituent ions.

Equilibrium Constant Calculations

The process of equilibrium constant calculations involves setting up and solving an expression that describes the state of equilibrium for a chemical reaction. This expression is derived from the stoichiometry of the balanced chemical equation and the concentrations or partial pressures of reactants and products.

In our example, the equilibrium constant expression is set up as \(K_{diss} = [Ag(CN)_{2}^{-}] / ([Ag^{+}][CN^{-}]^2)\). To find the concentration of \(Ag^{+}\), we rearrange this equation to isolate \(Ag^{+}\) on one side and substituting the known values into the equation. By carefully following this method, the calculation becomes straightforward, which makes understanding and solving equilibrium problems much easier.

Importance of Precision

It's critical to pay close attention to the concentrations and their units, as well as the arithmetic, which often deals with very small numbers expressed in scientific notation. Any mistake here can lead to a large error in the final result, leading to the wrong answer choice.

Solubility and Complex Ion Formation

Solubility is a measure of how much solute can be dissolved in a solvent at a given temperature to form a homogeneous mixture, or solution. The formation of complex ions can significantly affect solubility. Complex ions are species formed from a metal ion and one or more ligands, which can be anions like \(CN^{-}\) in our exercise.

Complex ion formation often increases the solubility of certain salts in solution. In the case of the exercise, the excess \(CN^{-}\) increases the solubility of \(AgCN\) by forming the complex ion \(Ag(CN)_{2}^{-}\). This is because the formation of the complex ion removes \(Ag^{+}\) ions from the solution, shifting the equilibrium toward the dissolved state of \(AgCN\).

Role of Ligands

Ligands like \(CN^{-}\) are critical since they are responsible for the formation of complex ions. By understanding the role of ligands and the fundamentals of solubility, students can better grasp how complex ions form in solution and predict the effects on the solubility of various compounds.