Problem 6 - Analytical Chemistry

Question

The formation constant of the silver-ethylenediamine complex, \(\mathrm{Ag}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)^{+},\) is \(5.0 \times 10^{4}\). Calculate the concentration of \(\mathrm{Ag}^{+}\) in equilibrium with a \(0.10 \mathrm{M}\) solution of the complex. (Assume no higher order complexes.)

Step-by-Step Solution

Answer

The concentration of \( \mathrm{Ag}^{+} \) is approximately \( 4.47 \times 10^{-4} \text{ M} \).

Step 1: Write the Equilibrium Reaction

First, write the equilibrium expression for the formation of the silver-ethylenediamine complex:\[ \mathrm{Ag}^{+} + \mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2} \rightleftharpoons \mathrm{Ag} (\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2})^{+} \]In this reaction, one mole of \( \mathrm{Ag}^{+} \) reacts with one mole of ethylenediamine to form one mole of the complex.

Step 2: Write the Formation Constant Expression

The formation constant (\( K_f \)) for the complex is given by:\[ K_f = \frac{[\mathrm{Ag} (\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2})^{+}]}{[\mathrm{Ag}^{+}][\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}]} \]We are provided that \( K_f = 5.0 \times 10^{4} \).

Step 3: Assume Equilibrium Concentrations

Assume the initial concentration of the complex, \([\mathrm{Ag} (\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2})^{+} ] = 0.10 \, \text{M}\), and denote the change in concentration of \( \mathrm{Ag}^{+} \) as \( x \). Thus, at equilibrium:- \([\mathrm{Ag}^{+}] = x\)- \([\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}] = x\)- \([\mathrm{Ag} (\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2})^{+} ] = 0.10 - x\)

Step 4: Substitute into the Formation Constant Expression

Substitute the equilibrium concentrations into the formation constant expression:\[ 5.0 \times 10^{4} = \frac{0.10 - x}{x^2} \]This is a quadratic equation in terms of \( x \).

Step 5: Simplify and Solve the Quadratic Equation

Rearrange the equation to solve for \( x \):\[ 5.0 \times 10^{4} x^2 = 0.10 - x \]\[ 5.0 \times 10^{4} x^2 + x - 0.10 = 0 \]This quadratic can be solved using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 5.0 \times 10^{4}, b = 1, c = -0.10 \).

Step 6: Calculate the Values of x

Plug in the values into the quadratic formula:\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \times 5.0 \times 10^4 \times (-0.10)}}{2 \times 5.0 \times 10^{4}} \]\[ x = \frac{-1 \pm \sqrt{1 + 2.0 \times 10^{4}}}{10^5} \]\[ x = \frac{-1 \pm \sqrt{2.0 \times 10^{4} + 1}}{10^5} \]Calculate \( x \).

Step 7: Positive Root for x

The correct value for \( x \) must be positive, as it represents a concentration:- After solving the quadratic equation, find that \( x \approx 4.47 \times 10^{-4} \text{ M}\).This is the concentration of \( \mathrm{Ag}^{+} \) in the solution.

Key Concepts

Equilibrium Concentration

Equilibrium concentration refers to the steady-state concentration of reactants and products in a reaction that has reached equilibrium. In this context, a complex formation reaction of silver ions with ethylenediamine is being considered. At equilibrium, the rate of formation of the complex equals the rate at which it dissociates back to its components.

The equilibrium concentration can be influenced by the initial concentrations of the reactants. In our exercise, we start with a specific concentration of the silver-ethylenediamine complex and calculate the concentration of unreacted silver ions at equilibrium.

To do this, we establish a balance between the concentrations of all species involved in the equilibrium, using the formation constant. This constant, denoted as \( K_f \), is crucial as it quantifies the tendency of the complex to form versus dissociate, affecting the concentrations of the individual ions in the solution.

Complex Formation

Complex formation involves the reaction between metal ions and ligands to form a complex compound. Ligands, such as ethylenediamine, are molecules that can donate pairs of electrons to the metal ion, stabilizing the complex.

In our example, silver ions are complexed with ethylenediamine, changing their properties and reducing the concentration of free silver ions in solution. The formation constant \( K_f \) of \( 5.0 \times 10^4 \) indicates a strong tendency for the complex to form. The larger the \( K_f \), the more stable the complex, and the less likely it is to dissociate.

• Complex formation can affect equilibrium as more product is created, shifting balance toward formation.
• If competing species or different ligands are present, it can prevent or reduce the extent of the complex formation.

This is essential in many applications such as catalysis or metal ion sequestration in industrial and biochemical systems.

Quadratic Equation

When dealing with chemical equilibria, sometimes assumptions and simplifications lead us to solving quadratic equations. In our exercise, calculating the equilibrium concentrations required solving a quadratic equation derived from the formation constant expression:

\[ 5.0 \times 10^4 x^2 + x - 0.10 = 0 \]

Solving this equation helps find the equilibrium concentration \( x \) of silver ions \( \mathrm{Ag}^+ \). To solve quadratic equations, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a \), \( b \), and \( c \) are coefficients from the equation format \( ax^2 + bx + c = 0 \).

In chemical applications, the positive root is typically the one of interest, as concentrations cannot be negative. Understanding how to solve these equations is crucial in determining equilibrium compositions when simple stoichiometric calculations aren’t enough. This is a common task in chemistry studies, highlighting the intersection of algebra and chemical equilibrium.