Question

Calculate the percentage of pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) that forms pyridinium ion, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), in a \(0.10 \mathrm{M}\) aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\)

Short answer

In a \(0.10 M\) aqueous solution of pyridine, approximately \(0.243\%\) of the pyridine forms the pyridinium ion.

Step-by-step solution

Establish base ionization equation

The ionization equation for pyridine in water looks like this: C5H5N + H2O ⇌ C5H5NH+ + OH- Here, pyridine reacts with water to form the pyridinium ion and hydroxide ions.

Write the Ka expression and substitute Kb

We can write the Ka expression for this reaction: Ka = [C5H5NH+][OH-] / [C5H5N] However, we are given Kb, the base dissociation constant. We can use the relationship Ka * Kb = Kw, where Kw is the ion-product constant for water, to find Ka: Kw = 1.0 x 10^{-14} Ka = Kw / Kb = 1.0 x 10^{-14} / 1.7 x 10^{-9} = 5.88 x 10^{-6} Now we can rewrite the Ka expression with this value: 5.88 x 10^{-6} = [C5H5NH+][OH-] / [C5H5N]

Plug in the initial concentrations and solve for equilibrium concentrations

The initial concentration of pyridine is 0.10 M. Let x be the concentration of C5H5NH+ and OH- at equilibrium. The equilibrium concentration of pyridine becomes (0.10 - x). Now we can plug these values into the Ka expression: 5.88 x 10^{-6} = (x)(x) / (0.10 - x)

Solve the quadratic equation for x

There are two possible approaches for solving this equation for x. One is to use the quadratic formula and the other is to make the approximation that x << 0.10, and therefore (0.10 - x) ≈ 0.10. Since the Kb value is very small, we can make this approximation to simplify the problem: 5.88 x 10^{-6} ≈ x^2 / 0.10 x^2 ≈ 5.88 x 10^{-7} x ≈ √(5.88 x 10^{-7}) ≈ 2.43 x 10^{-4} So the equilibrium concentration of C5H5NH+ is 2.43 x 10^{-4} M.

Calculate the percentage of pyridine that forms pyridinium ions

Now that we have the equilibrium concentration of C5H5NH+, we can find the percentage of pyridine that forms pyridinium ions: Percentage of pyridine ionized = (Concentration of C5H5NH+ / Initial concentration of C5H5N) * 100% Percentage of pyridine ionized ≈ (2.43 x 10^{-4} M / 0.10 M) * 100% ≈ 0.243% In a 0.10 M aqueous solution of pyridine, approximately 0.243% of the pyridine forms the pyridinium ion.

Key concepts

Pyridine Ionization

Pyridine, represented by the formula \( ext{C}_5 ext{H}_5 ext{N}\), acts as a weak base when dissolved in water. When it ionizes, it accepts a proton from water, leading to the formation of a pyridinium ion \( ext{C}_5 ext{H}_5 ext{NH}^+\) and hydroxide ions \( ext{OH}^-\). The ionization reaction can be represented as:

\[\text{C}_5\text{H}_5\text{N} + \text{H}_2\text{O} ⇌ \text{C}_5\text{H}_5\text{NH}^+ + \text{OH}^-\]

This reversible reaction demonstrates how pyridine can partially convert into its ionized form in solution. Understanding this equilibrium is crucial for determining how much pyridine ionizes under specific conditions.

Base Dissociation Constant

The base dissociation constant, denoted as \(K_b\), is a measure of a base's strength in solution. It provides an indication of the base's ability to dissociate in water and form ions. For pyridine, \(K_b = 1.7 \times 10^{-9}\), indicating that it is a weak base.

The significance of \(K_b\) lies in its relationship with the acid dissociation constant \(K_a\), which together maintain the water ion-product \(K_w\) through the expression:

\[K_a \times K_b = K_w\]

This relationship helps us find \(K_a\) when \(K_b\) is known, as shown in the solved exercise. Understanding \(K_b\) is vital for predicting how much a base is likely to dissociate in a given solution.

Pyridinium Ion

The pyridinium ion \(\text{C}_5\text{H}_5\text{NH}^+\) is formed when pyridine accepts a proton from water. This positively charged ion is a vital part of the acid-base equilibrium when pyridine is in solution.

Identifying the presence of pyridinium ions helps chemists understand how the balance between ionized and non-ionized forms of pyridine changes. The concentration of these ions at equilibrium can reveal the extent of ionization, and is used to compute the percentage of pyridine converted to its ionized form.

In the context of acid-base chemistry, recognizing such ions allows for a deeper insight into the dynamic nature of chemical equilibria in solutions.

Equilibrium Concentration

Equilibrium concentrations refer to the amounts of each species present when the reaction reaches a state where the rates of the forward and reverse reactions are equal. For pyridine ionization, you begin with an initial concentration of pyridine, which changes as equilibrium is established.

Using the expression \(5.88 \times 10^{-6} = \frac{[\text{C}_5\text{H}_5\text{NH}^+][\text{OH}^-]}{[\text{C}_5\text{H}_5\text{N}]}\), you can determine the concentration of pyridinium ions \(\text{C}_5\text{H}_5\text{NH}^+\) and hydroxide ions \(\text{OH}^-\). Solving this yields the equilibrium concentration, allowing for the calculation of how much of the original pyridine ionized.

This understanding helps predict the behavior of weak bases like pyridine in solution, providing clarity on their reactive potential.