Question

Calculate the \(\mathrm{pH}\) of each of the following buffered solutions. a. \(0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} / 0.25 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) b. \(0.25 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NH}_{2} / 0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) c. \(0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} / 0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\)

Short answer

Using the Henderson-Hasselbalch equation and the calculated \(\mathrm{p}K_\mathrm{a}\) value, we find the following \(\mathrm{pH}\) values for each buffered solution: a. \(0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2} / 0.25\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl}\): \(\mathrm{pH} \approx 10.17\) b. \(0.25\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{NH}_{2} / 0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl}\): \(\mathrm{pH} \approx 9.30\) c. \(0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2} / 0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl}\): \(\mathrm{pH} \approx 9.74\)

Step-by-step solution

Identify the weak acid and weak base

For each case, identify the weak acid and weak base present in the solution. In all three cases here, ethylamine (\( \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2} \)) is the weak base and the ethylamine hydrochloride (\( \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl} \)) is the weak acid.

Calculate the \(\mathrm{p}K_\mathrm{a}\) values

The \(\mathrm{p}K_\mathrm{a}\) of the weak acid can be found using the \(\mathrm{K}_{\mathrm{b}}\) of its conjugate base, which is ethylamine. Given that the \(\mathrm{K}_{\mathrm{b}}\) for ethylamine is \(5.6\times10^{-4}\), we can use the relationship between \(\mathrm{K}_{\mathrm{a}}\) and \(\mathrm{K}_{\mathrm{b}}\): \[\mathrm{K}_{\mathrm{a}}\times\mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}}\] Where \(\mathrm{K}_{\mathrm{w}}=1.0\times10^{-14}\) at 25°C. Calculate the \(\mathrm{p}K_\mathrm{a}\) value of the weak acid: \[\mathrm{p}K_\mathrm{a}=-\log(\mathrm{K}_\mathrm{a})\]

Apply the Henderson-Hasselbalch equation

For each buffered solution, use the concentrations of the weak acid and weak base, along with the calculated \(\mathrm{p}K_\mathrm{a}\) value, in the Henderson-Hasselbalch equation to find the \(\mathrm{pH}\): \[\mathrm{pH}=\mathrm{p}K_\mathrm{a}+\log\frac{[\mathrm{base}]}{[\mathrm{acid}]}\]

Calculate the \(\mathrm{pH}\) values for each buffered solution

a. For the buffered solution with \(0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2} / 0.25\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl}\), plug in the concentrations of the weak base and weak acid, and the calculated \(\mathrm{p}K_\mathrm{a}\) value, into the Henderson-Hasselbalch equation to find the \(\mathrm{pH}\). b. For the buffered solution with \(0.25\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{NH}_{2} / 0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl}\), plug in the concentrations of the weak base and weak acid, and the calculated \(\mathrm{p}K_\mathrm{a}\) value, into the Henderson-Hasselbalch equation to find the \(\mathrm{pH}\). c. For the buffered solution with \(0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2} / 0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl}\), plug in the concentrations of the weak base and weak acid, and the calculated \(\mathrm{p}K_\mathrm{a}\) value, into the Henderson-Hasselbalch equation to find the \(\mathrm{pH}\). After completing these steps, the \(\mathrm{pH}\) values of the given buffered solutions will be calculated.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a crucial formula for calculating the pH of buffered solutions, where both a weak acid and its conjugate base or a weak base and its conjugate acid are present. It is given by:\[\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \frac{[\mathrm{base}]}{[\mathrm{acid}]}\]Here:

• \(\mathrm{pH}\) represents the acidity or basicity of the solution.
• \(\mathrm{p}K_a\) is the negative logarithm of the acid dissociation constant, \(K_a\), indicating the strength of the weak acid.
• \([\mathrm{base}]\) and \([\mathrm{acid}]\) are the molar concentrations of the weak base and weak acid in the solution.

Breaking it down simply, the equation helps us find out the resultant pH when we know the concentrations and the strength of the weak components. This equation is especially valuable because it connects the chemical equilibrium to the measurable property of pH.

weak acid and weak base

Understanding weak acids and bases is essential in chemistry, especially for buffered solutions. A weak acid, like ethylamine hydrochloride \((\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl})\), only partially dissociates in water. This means not all the acid molecules donate protons to the solution.On the flip side, a weak base, such as ethylamine \((\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2})\), does not fully accept protons or dissociate completely in water. This partial dissociation results in the equilibrium between the undissociated and dissociated forms of the acid or base in the solution.Buffered solutions become effective at resisting pH changes because they contain these partially dissociated acids and bases. The combination of a weak acid with its conjugate base or a weak base with its conjugate acid helps maintain a stable pH, even when small amounts of acids or bases are added.

pH calculation

Calculating pH is an important skill in chemistry. For buffered solutions, the Henderson-Hasselbalch equation simplifies this process. First, identify the weak acid and the weak base in your solution. Then, determine their molar concentrations.To perform the calculation:

• Use the \(\mathrm{p}K_a\) value of the weak acid.
• Use the Henderson-Hasselbalch equation: \(\mathrm{pH} = \mathrm{p}K_a + \log \frac{[\mathrm{base}]}{[\mathrm{acid}]}\).
• Substitute the values into the equation, ensuring the concentrations are in molarity (M).

This calculation gives the pH, providing insight into the chemical nature of the solution. Using this equation can predict how the solution will behave when faced with additional acids or bases.

pKa and pKb

\(\mathrm{p}K_a\) and \(\mathrm{p}K_b\) are measures of the strengths of acids and bases, respectively. They are related to the acid dissociation constant \(\mathrm{K}_a\) and base dissociation constant \(\mathrm{K}_b\), using the negative logarithm:\[\mathrm{p}K_a = -\log(\mathrm{K}_a)\]\[\mathrm{p}K_b = -\log(\mathrm{K}_b)\]These values express how easily an acid donates protons or a base accepts them.Importantly, \(\mathrm{p}K_a\) and \(\mathrm{p}K_b\) are interconnected through the ion product of water \(\mathrm{K}_w\):\[\mathrm{K}_a \times \mathrm{K}_b = \mathrm{K}_w\]\[\mathrm{p}K_a + \mathrm{p}K_b = 14\]These relationships assert that stronger acids have lower \(\mathrm{p}K_a\) values and stronger bases have lower \(\mathrm{p}K_b\) values. In buffered solutions, using \(\mathrm{p}K_a\) and \(\mathrm{p}K_b\) ensures accurate pH calculations.

ethylamine

Ethylamine \((\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2})\) is a primary aliphatic amine, functioning as a weak base in chemical solutions. It partially reacts with water to produce hydroxide ions and its conjugate acid, ethylammonium ion \((\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}^+)\).In buffered solutions, ethylamine is often paired with ethylamine hydrochloride \((\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl})\), forming a conjugate acid-base pair. This weak base has a base dissociation constant \(\mathrm{K}_b = 5.6 \times 10^{-4}\), allowing calculation of other related values such as \(\mathrm{p}K_a\) using the inverse relationship between \(\mathrm{K}_a\), \(\mathrm{K}_b\), and \(\mathrm{K}_w\).Ethylamine's significance in buffered solutions lies in its ability to stabilize pH by absorbing or donating protons, enhancing the resilience of the solution against pH shifts.