Question

Calculate the \(\mathrm{pH}\) of a \(0.20-M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.5.6 \times 10^{-4}\right)\)

Short answer

The pH of the 0.20-M ethylamine solution is approximately 12.02. To obtain this value, we calculated the concentration of OH- ions using the given Kb value and the ICE table method, found the pOH, and then used the relationship between pH and pOH.

Step-by-step solution

Write the ionization reaction for ethylamine.

Ethylamine (C2H5NH2) is a weak base and it will react with water to form its conjugate acid (C2H5NH3+) and hydroxide ion (OH-). The reaction can be written as: C2H5NH2 + H2O ⟶ C2H5NH3+ + OH-

Write the Kb expression.

Using the ionization reaction, we can write the Kb expression for ethylamine as: Kb = \(\frac{[C_{2}H_{5}NH_{3}^{+}][OH^{-}]}{[C_{2}H_{5}NH_{2}]}\)

Set up an ICE table.

To find the concentrations of the reacting species, we can create an ICE (Initial, Change, Equilibrium) table as follows: | | C2H5NH2 | C2H5NH3+ | OH- | |--------------|---------|----------|------| | Initial | 0.20 M | 0 M | 0 M | | Change | -x | +x | +x | | Equilibrium | 0.20-x | x | x |

Substitute the equilibrium concentrations in the Kb expression.

Substituting the equilibrium concentrations from the ICE table into the Kb expression: \(5.6 \times 10^{-4}\) = \(\frac{x \cdot x}{0.20-x}\)

Make an assumption to simplify the equation.

Since ethylamine is a weak base, we can assume that the change in its concentration (x) is small compared to its initial concentration (0.20 M). So, we have: 0.20 - x ≈ 0.20 Now, our equation becomes: \(5.6 \times 10^{-4}\) = \(\frac{x \cdot x}{0.20}\)

Solve for x to find the [OH-] concentration.

Solving the equation for x: x^2 = \(5.6 \times 10^{-4} \cdot 0.20\) x^2 = \(1.12 \times 10^{-4}\) x ≈ \(\sqrt{1.12 \times 10^{-4}}\) x ≈ \(1.06 \times 10^{-2}\) So, [OH-] = \(1.06 \times 10^{-2} M\)

Calculate pOH.

Now, we can calculate the pOH using the formula: pOH = -log([OH-]) pOH = -log(\(1.06 \times 10^{-2}\)) pOH ≈ 1.98

Calculate pH of the solution.

Finally, we can use the relationship between pH and pOH to find the pH of the solution: pH = 14 - pOH pH = 14 - 1.98 pH ≈ 12.02 The pH of the 0.20-M ethylamine solution is approximately 12.02.

Key concepts

Ethylamine

Ethylamine is an organic compound with the formula \( ext{C}_2 ext{H}_5 ext{NH}_2\). It is a colorless gas with a pungent odor and is highly flammable. In chemistry, ethylamine acts as a weak base. This means it doesn’t completely dissociate in water to produce hydroxide ions \( ext{OH}^-\).
Instead, ethylamine undergoes partial ionization when dissolved in water to form its conjugate acid ethylammonium ion \( ext{C}_2 ext{H}_5 ext{NH}_3^+\) and hydroxide ions. This is expressed in the ionization reaction:
\[\text{C}_2\text{H}_5\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{C}_2\text{H}_5\text{NH}_3^+ + \text{OH}^-\]
Understanding this reaction is crucial when calculating the pH of ethylamine solutions.

Weak Base

A weak base only partially ionizes in solution. This is contrasted with strong bases which fully dissociate. Ethylamine is a classic example of a weak base. When a base is weak, only a small fraction of its molecules react with water to form hydroxide ions.
This incomplete ionization is why weak bases have a much lower concentration of \([ ext{OH}^-]\) ions compared to strong bases. As a result, weak bases generally have higher pH values than weak acids but are still not as basic as a solution with a strong base.
Understanding the concept of weak bases is key to solving pH calculation problems, as it influences how we set up equations and assumptions.

Kb Expression

The base dissociation constant, \(K_b\), quantifies the strength of a base in solution. For ethylamine, the \(K_b\) represents the equilibrium constant for the reaction:
\[\text{C}_2\text{H}_5\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{C}_2\text{H}_5\text{NH}_3^+ + \text{OH}^-\]
The expression for \(K_b\) is formulated like so:
\[K_b = \frac{[\text{C}_2\text{H}_5\text{NH}_3^+][\text{OH}^-]}{[\text{C}_2\text{H}_5\text{NH}_2]}\]
The \(K_b\) value helps measure how much of the base dissociates to form \([ ext{OH}^-]\). A smaller \(K_b\) value indicates a weaker base. In the exercise, we used \(K_b = 5.6 \times 10^{-4}\) for ethylamine to calculate the pH, illustrating its weak base nature.

ICE Table

The ICE table is a useful tool in chemistry to keep track of changes in concentrations during a reaction. ICE stands for Initial, Change, and Equilibrium stages of a reaction.
To calculate the pH of an ethylamine solution, we carefully set up an ICE table to track the concentration of reactants and products. Initially, we start with a set concentration of ethylamine and none of its ionized products.

• Initial: Before reaction occurs.
• Change: As the reaction progresses, ethylamine concentration decreases while its products form.
• Equilibrium: When the reaction achieves balance, meaning no further changes in concentrations.

By applying the ICE table values to the \(K_b\) expression, one can solve for \([ ext{OH}^-]\) concentration, crucial for the subsequent pH calculation. This method simplifies complex equilibrium reactions and assists in accurate pH determination of weak bases like ethylamine.