Question

A buffer contains 0.30 mol of propionic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)\) and \(0.25 \mathrm{~mol}\) of potassium propionate \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOK}\right)\) in 1.80 L. (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of \(0.10 \mathrm{~mol}\) of \(\mathrm{NaOH}\) ? \((\mathbf{c})\) What is the \(\mathrm{pH}\) of the buffer after the addition of \(0.10 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

Short answer

The pH of the buffer initially is approximately 4.68. After the addition of 0.10 mol of NaOH, the pH increases to about 5.23. After the addition of 0.10 mol of HCl, the pH decreases to about 4.34.

Step-by-step solution

(Step 1: Identify the conjugate acid-base pair and write their dissociation constant (Ka) expression)

(In this exercise, the propionic acid (C₂H₅COOH) and potassium propionate (C₂H₅COOK) are the conjugate acid-base pair. We need to find the Ka expression, knowing that the dissociation of propionic acid (HA) is written as: HA ⇌ H⁺ + A⁻ The Ka expression is: Ka = [H⁺][A⁻]/[HA])

(Step 2: Find initial concentrations of each component in the buffer)

(Given the moles and volume of the buffer, we can find the initial concentrations: Initial concentration of C₂H₅COOH: 0.30 mol / 1.80 L = 0.167 M Initial concentration of C₂H₅COOK: 0.25 mol / 1.80 L = 0.139 M)

(Step 3: Calculate the pH of the initial buffer solution)

(Now we can use the Henderson-Hasselbalch equation to find the pH of the initial buffer: pH = pKa + log([A⁻]/[HA]) First, we will need the pKa value, which can be found in a textbook or online source. For propionic acid, the pKa value is 4.88. Substitute the molarities into the equation: pH = 4.88 + log(0.139/0.167) pH ≈ 4.68)

(Step 4: Calculate the pH after the addition of NaOH)

(When we add 0.10 mol of NaOH to the buffer solution, it will react with the propionic acid according to: C₂H₅COOH + OH⁻ → C₂H₅COO⁻ + H₂O Calculate the new moles of propionic acid and potassium propionate after the reaction: Moles of propionic acid = 0.30 mol - 0.10 mol = 0.20 mol Moles of potassium propionate = 0.25 mol + 0.10 mol = 0.35 mol Now, find the new concentrations of each component in the buffer: New concentration of C₂H₅COOH: 0.20 mol / 1.80 L = 0.111 M New concentration of C₂H₅COOK: 0.35 mol / 1.80 L = 0.194 M Find the new pH using the Henderson-Hasselbalch equation: pH = 4.88 + log(0.194/0.111) pH ≈ 5.23)

(Step 5: Calculate the pH after the addition of HCl)

(When we add 0.10 mol of HCl to the buffer solution, it will react with the potassium propionate: C₂H₅COO⁻+ H⁺ → C₂H₅COOH Calculate the new moles of propionic acid and potassium propionate after the reaction: Moles of propionic acid = 0.30 mol + 0.10 mol = 0.40 mol Moles of potassium propionate = 0.25 mol - 0.10 mol = 0.15 mol Now, find the new concentrations of each component: New concentration of C₂H₅COOH: 0.40 mol / 1.80 L = 0.222 M New concentration of C₂H₅COOK: 0.15 mol / 1.80 L = 0.083 M Find the new pH using the Henderson-Hasselbalch equation: pH = 4.88 + log(0.083/0.222) pH ≈ 4.34) As a summary: a) The initial pH of the buffer is approximately 4.68 b) The pH of the buffer after the addition of NaOH is approximately 5.23 c) The pH of the buffer after the addition of HCl is approximately 4.34

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation is an essential tool in chemistry for understanding buffer solutions. A buffer solution resists changes in pH when small amounts of acid or base are added. This equation provides a simple way to calculate the pH of a buffer solution using the concentrations of the acid and its conjugate base.

The equation is expressed as follows:

\[pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)\]

Here:

• \(pK_a\) is the negative logarithm of the acid dissociation constant \(K_a\).
• \([A^-]\) is the concentration of the conjugate base.
• \([HA]\) is the concentration of the acid.

The beauty of the Henderson-Hasselbalch Equation lies in its ability to connect the idea of pH with the concentrations of an acid and its conjugate base. This makes it particularly helpful in assessing the effective strength of a buffer solution.

Conjugate Acid-Base Pair

In the realm of buffer solutions, the concept of a conjugate acid-base pair is fundamental. A conjugate acid-base pair consists of two species that transform into each other by gaining or losing a proton \((H^+)\). In the given buffer solution exercise, propionic acid \((\mathrm{C}_2\mathrm{H}_5\mathrm{COOH})\) and potassium propionate \((\mathrm{C}_2\mathrm{H}_5\mathrm{COOK})\) are a conjugate acid-base pair.

A few key points about conjugate acid-base pairs include:

• The acid in the pair can donate a proton to become its conjugate base.
• The conjugate base can accept a proton to revert to the acid form.
• This reversible transformation plays a crucial role in buffer solutions, as it helps neutralize added acids or bases, thus stabilizing the pH.

Understanding how conjugate acid-base pairs function allows students to grasp the mechanics behind buffering action, helping maintain a stable pH despite the addition of external acids or bases.

pH Calculation

Calculating pH is a vital skill not only within buffer solution adjustments, such as in the given exercise, but also more broadly in chemistry and biology fields. The core concept involves determining the hydrogen ion concentration in a solution, as pH is defined as the negative base-10 logarithm of the \(H^+\) concentration:

\[pH = - \log[H^+]\]

In buffer systems, using the calculated pH allows scientists and students to verify the efficiency of a buffer. For example, before any adjustments, the calculated pH using the Henderson-Hasselbalch Equation gave insight into the state of the buffer solution.

After adding specific amounts of strong acids or bases, as seen in the exercise with \( \mathrm{NaOH} \) and \( \mathrm{HCl} \), the buffer's ability to resist pH shifts is observed, and pH is recalculated. The adjustment in concentrations of the acid and base due to such additions directly translates to shifts in pH, which can also be determined using the Henderson-Hasselbalch Equation effectively.