Question

Oxycodone \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{4}\right)\), a narcotic analgesic, is a weak base with \(\mathrm{pK}_{\mathrm{b}}=5.47\). Calculate the \(\mathrm{pH}\) and the concentrations of all species present \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{4}, \mathrm{HC}_{18} \mathrm{H}_{21} \mathrm{NO}_{4}^{+}, \mathrm{H}_{3} \mathrm{O}^{+}\right.\), and \(\left.\mathrm{OH}^{-}\right)\) in a \(0.00250 \mathrm{M}\) oxycodone solution.

Short answer

The pH is 9.46; concentrations are [C_18H_21NO_4] = 0.00247 M, [HC_18H_21NO_4^+] = 2.91 × 10^{-5} M, [OH^-] = 2.91 × 10^{-5} M, [H_3O^+] = 3.47 × 10^{-10} M.

Step-by-step solution

Convert pK_b to K_b

Use the equation \( K_b = 10^{-pK_b} \). Substitute \( pK_b = 5.47 \) to find \( K_b \).\[K_b = 10^{-5.47} = 3.39 \times 10^{-6}\]

Write the Base Ionization Equation

Oxycodone (B) ionizes in water as:\[\text{C}_{18}\text{H}_{21}\text{NO}_4 + \text{H}_2\text{O} \rightleftharpoons \text{HC}_{18}\text{H}_{21}\text{NO}_4^{+} + \text{OH}^{-}\]

Set Up ICE Table

Initial, Change, Equilibrium (ICE) table:- Initial: \([\text{C}_{18}\text{H}_{21}\text{NO}_4] = 0.00250\) M, \([\text{HC}_{18}\text{H}_{21}\text{NO}_4^{+}] = 0\), \([\text{OH}^-] = 0\)- Change: \(-x, +x, +x\)- Equilibrium: \(0.00250 - x, x, x\)

Apply Equilibrium Expression for K_b

Using \( K_b = \frac{[\text{HC}_{18}\text{H}_{21}\text{NO}_4^{+}][\text{OH}^-]}{[\text{C}_{18}\text{H}_{21}\text{NO}_4]} \), substitute the expressions:\[3.39 \times 10^{-6} = \frac{x^2}{0.00250 - x}\]Assume \( 0.00250 - x \approx 0.00250 \) for simplicity:\[3.39 \times 10^{-6} = \frac{x^2}{0.00250}\]

Solve for x

Rearrange and solve for \( x \):\[x^2 = 3.39 \times 10^{-6} \times 0.00250\]\[x^2 = 8.475 \times 10^{-9}\]\[x = \sqrt{8.475 \times 10^{-9}} = 2.91 \times 10^{-5}\]

Calculate [OH^-] and [HC_18H_21NO_4^+]

From the solved value of \( x \) as \([\text{OH}^-]\) and \([\text{HC}_{18}\text{H}_{21}\text{NO}_4^{+}]\):\[\text{[OH]^-} = 2.91 \times 10^{-5} \]\[\text{[HC}_{18}\text{H}_{21}\text{NO}_4^{+}] = 2.91 \times 10^{-5} \]

Calculate pOH and pH

Find \( \text{pOH} \) using \( \text{pOH} = -\log[\text{OH}^-] \):\[\text{pOH} = -\log(2.91 \times 10^{-5}) = 4.54\]Then, find \( \text{pH} \) using \( \text{pH} = 14 - \text{pOH} \):\[\text{pH} = 14 - 4.54 = 9.46\]

Calculate H_3O^+ and Final Concentration of C_18H_21NO_4

Find \([\text{H}_3\text{O}^+]\) using \( \text{pH} = -\log[\text{H}_3\text{O}^+] \):\[[\text{H}_3\text{O}^+] = 10^{-9.46} = 3.47 \times 10^{-10} \text{ M}\]Final concentration of \( \text{C}_{18}\text{H}_{21}\text{NO}_4 \) is:\([\text{C}_{18}\text{H}_{21}\text{NO}_4] = 0.00250 - 2.91 \times 10^{-5} \approx 0.00247 \text{ M}\)

Key concepts

Weak Bases

Unlike strong bases which completely dissociate in water, weak bases only partially ionize. This means that only a small fraction of the weak base molecules dissociate to form ions in solution. For example, when oxycodone, a weak base, is dissolved in water, it does not fully ionize. This partial ionization is important to consider when calculating the

• pH of the solution,
• concentrations of various species,
• and the equilibrium state of the reaction.

The degree of ionization of a weak base can often be quantified using the base ionization constant, known as the equilibrium constant for the base reaction.

Equilibrium Constant

The equilibrium constant, specifically the base constant (\( K_b \), indicates how favorably a base ionizes in solution. It quantifies the strength of a weak base. For oxycodone, a calculated \( K_b \) value \( 3.39 \times 10^{-6} \), suggests that its tendency to accept protons is weak compared to strong bases.

• This means, less oxycodone molecules will convert to its ionic form in water.
• We use this constant to establish the equilibrium state of the chemical reaction.

When at equilibrium, the rate of the forward and reverse reactions is equal, and the concentrations of reactants and products remain constant over time.

Ionization Equation

For oxycodone, the equilibrium ionization equation in water is:\[\text{C}_{18}\text{H}_{21}\text{NO}_4 + \text{H}_2\text{O} \rightleftharpoons \text{HC}_{18}\text{H}_{21}\text{NO}_4^{+} + \text{OH}^{-}\] This equation describes how oxycodone (\( \text{C}_{18}\text{H}_{21}\text{NO}_4 \)) reacts with water to form its conjugate acid (\( \text{HC}_{18}\text{H}_{21}\text{NO}_4^{+} \)) and hydroxide ions (\( \text{OH}^- \)).

pH Calculation

The pH of a solution is a measure of the concentration of hydrogen ions (\( \text{H}_3\text{O}^+ \)) in solution and indicates how acidic or basic the solution is. For weak bases like oxycodone, calculating the pH involves determining the degree of ionization to find the hydroxide ion concentration, \( \text{OH}^- \).

• The relationship between pH and pOH is given by the equation: \( \text{pH} + \text{pOH} = 14 \).
• First, find \( \text{pOH} \) using \( -\log[\text{OH}^-] \).
• Then calculate pH: \( \text{pH} = 14 - \text{pOH} \)

For the given oxycodone solution, \( \text{pOH} \) is found to be \( 4.54 \), thus the pH is \( 9.46 \), indicating a basic solution due to the presence of hydroxide ions formed by the ionization of oxycodone.

Chemical Equilibrium

Chemical equilibrium in the context of weak bases involves balancing the reactants and products' concentrations in a reversible reaction.

ICE Table Setup

To understand equilibrium, we use an ICE (Initial, Change, Equilibrium) table to calculate the concentrations at equilibrium. For instance, with oxycodone:

• Initial concentrations: \( [\text{C}_{18}\text{H}_{21}\text{NO}_4] = 0.00250 \) M, \( [\text{HC}_{18}\text{H}_{21}\text{NO}_4^{+}] = 0 \), and \( [\text{OH}^-] = 0 \).
• Change in concentration defines the shift in amounts as the reaction approaches equilibrium.
• Equilibrium concentrations reflect the balance and demonstrate that reactant is only partially converted to products.

In this example, oxycodone, a weak base, reacts to reach a state where the formation of its conjugate acid and hydroxide ions ceases to progress noticeably. Understanding the equilibrium concept allows us to accurately determine the concentrations of all species involved after the reaction reaches stability.