Question

Morpholine \(\left(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{NO}\right)\) is a weak organic base with \(\mathrm{p} K_{\mathrm{b}}=5.68\). Calculate the pH and the concentrations of all species present \(\left(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{NO}, \mathrm{HC}_{4} \mathrm{H}_{9} \mathrm{NO}^{+}\right.\), and \(\left.\mathrm{OH}^{-}\right)\) in a \(0.0100 \mathrm{M}\) morpholine solution.

Short answer

pH = 10.16, [C4H9NO] = 0.009855 M, [HC4H9NO+] = [OH−] = 1.45 × 10^-4 M.

Step-by-step solution

Understanding the Problem

Morpholine is a weak base, and we are given its base dissociation constant (kb) for calculating the equilibrium concentrations. We also know that the initial concentration of morpholine is 0.0100 M. Our task is to find the pH and the equilibrated concentrations of morpholine (\(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{NO}\)), its conjugate acid (\(\mathrm{HC}_{4} \mathrm{H}_{9} \mathrm{NO}^{+}\)), and hydroxide ions (\(\mathrm{OH}^{-}\)).

Determine the Base Dissociation Constant, \(K_b\)

Since we are given \(\mathrm{p}K_b = 5.68\), we can find the base dissociation constant \(K_b\) using the formula: \[ K_b = 10^{-\mathrm{p}K_b} \]Substituting the given \(\mathrm{p}K_b\) value:\[ K_b = 10^{-5.68} \approx 2.09 \times 10^{-6} \]

Set Up the Equilibrium Expression

The equilibrium expression for the reaction of morpholine with water can be written as:\[ \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{NO} + \mathrm{H}_2\mathrm{O} \leftrightarrow \mathrm{HC}_{4} \mathrm{H}_{9} \mathrm{NO}^{+} + \mathrm{OH}^{-} \]The expression for \(K_b\) is:\[ K_b = \frac{[\mathrm{HC}_{4} \mathrm{H}_{9} \mathrm{NO}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{NO}]} \]

Apply the Initial and Change in Concentration

Let \(x\) be the change in concentration of \([\mathrm{OH}^{-}]\) and \([\mathrm{HC}_{4} \mathrm{H}_{9} \mathrm{NO}^{+}]\) at equilibrium. Initially, we have:\[ [\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{NO}]_0 = 0.0100 \mathrm{M}, [\mathrm{HC}_{4} \mathrm{H}_{9} \mathrm{NO}^{+}]_0 = 0, [\mathrm{OH}^{-}]_0 = 0 \]Change:\[ [\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{NO}] = 0.0100 - x, [\mathrm{HC}_{4} \mathrm{H}_{9} \mathrm{NO}^{+}] = x, [\mathrm{OH}^{-}] = x \]

Solve for x Using the Quadratic Formula

By substituting these values into \(K_b\)'s expression, we have:\[ 2.09 \times 10^{-6} = \frac{x^2}{0.0100 - x} \]Assuming \(x\) is much smaller than 0.0100, we simplify:\[ 2.09 \times 10^{-6} = \frac{x^2}{0.0100} \]\[ x^2 = 2.09 \times 10^{-6} \times 0.0100 = 2.09 \times 10^{-8} \]\[ x \approx \sqrt{2.09 \times 10^{-8}} \approx 1.45 \times 10^{-4} \]

Calculate Equilibrium Concentrations and pH

Now that we have \(x\), we determine the concentrations:\[ [\mathrm{OH}^{-}] = x = 1.45 \times 10^{-4} \]\[ [\mathrm{HC}_{4} \mathrm{H}_{9} \mathrm{NO}^{+}] = x = 1.45 \times 10^{-4} \]\[ [\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{NO}] = 0.0100 - 1.45 \times 10^{-4} \approx 0.009855 \]Next, calculate the pOH:\[ \mathrm{pOH} = -\log[\mathrm{OH}^{-}] = -\log(1.45 \times 10^{-4}) \approx 3.84 \]Since \(\mathrm{pH} + \mathrm{pOH} = 14\), we calculate:\[ \mathrm{pH} = 14 - 3.84 = 10.16 \]

Key concepts

Base Dissociation Constant

The base dissociation constant (K_b) is a fundamental concept when dealing with weak bases. It tells us how well a base can attract and accept a proton in an aqueous solution. This is similar to the acid dissociation constant (K_a) for acids. The lower the K_b, the weaker the base.For weak organic bases like morpholine, we use the \(\mathrm{p}K_b\) given in problems to calculate K_b. The formula to find K_b from \(\mathrm{p}K_b\) is:\[ K_b = 10^{-\mathrm{p}K_b} \]Using the \(\mathrm{p}K_b\) value, we calculate K_b directly, as shown in the original solution:\[ K_b = 10^{-5.68} \approx 2.09 \times 10^{-6} \]This helps us in setting up equations to determine how much of the base ionizes in solutions, highlighting its weakness.

Equilibrium Concentration

Equilibrium concentration is about understanding how substance concentrations change during a reaction until they reach equilibrium. At equilibrium, concentrations of reactants and products no longer change, although the molecules continue to react back and forth.In the example problem, morpholine (\(\mathrm{C}_4\mathrm{H}_9\mathrm{NO}\)) reacts with water to form its conjugate acid (\(\mathrm{HC}_4\mathrm{H}_9\mathrm{NO}^+\)) and hydroxide ions (\(\mathrm{OH}^-\)). When setting up the equilibrium expression, the initial concentrations and changes must be considered:

• Start with initial concentrations: a known value for morpholine and zero for other species.
• Subtract and add 'x' to reflect changes as the equilibrium is reached (\(\mathrm{C}_4\mathrm{H}_9\mathrm{NO}\) decreases, \(\mathrm{HC}_4\mathrm{H}_9\mathrm{NO}^+\) and \(\mathrm{OH}^-\) increase).

Using the expression: \[ K_b = \frac{[\mathrm{HC}_4\mathrm{H}_9\mathrm{NO}^+][\mathrm{OH}^-]}{[\mathrm{C}_4\mathrm{H}_9\mathrm{NO}]} \]We plug in our equilibrium values to find "x", which represents the amount reacted, thus allowing us to determine the equilibrium concentrations.

pH Calculation

Calculating the pH of a solution indicates its acidity or basicity, with lower pH values reflecting more acidic conditions and higher values indicating basicity. Because morpholine is a weak base, we first calculate pOH to derive its pH.The steps involve:

• Finding [\(\mathrm{OH}^-\)] from the equilibrium reaction, using the value of x. Here [\(\mathrm{OH}^-\)] = 1.45×10^{-4} M.
• Calculating \(\mathrm{pOH}\) using: \[ \mathrm{pOH} = -\log([\mathrm{OH}^-]) \]
• Determining \(\mathrm{pH}\) knowing: \(\mathrm{pH} + \mathrm{pOH} = 14\)

This gives:\[ \mathrm{pOH} \approx 3.84 \]\[ \mathrm{pH} = 14 - 3.84 = 10.16 \]Thus, we know the solution is basic and understand how the hydroxide concentration plays a vital role in calculating the pH of a weak base solution.