Problem 40 - Chemistry: Structure and Properties

Question

Calculate the pH of the solution that results from each mixture. a. 150.0 mL of 0.25 M HF with 225.0 mL of 0.30 M NaF b. 175.0 mL of 0.10 M C2H5NH2 with 275.0 mL of 0.20 M C2H5NH3Cl

Step-by-Step Solution

Answer

The pH for the mixture in part a is 3.4253, and the final pH for part b will be calculated by first determining the pOH using the Henderson-Hasselbalch equation then converting the pOH to pH by subtracting pOH from 14.

Step 1: Understanding the Buffer Systems

Both mixtures represent buffer systems. For part a, HF is a weak acid and NaF its conjugate base. For part b, C2H5NH2 is a weak base and C2H5NH3Cl its conjugate acid. Buffers resist changes in pH when small amounts of acid or base are added.

Step 2: Calculate the Moles

First calculate the initial moles of HF and NaF before mixing by using the formula: Moles = Concentration (M) * Volume (L). For HF, that's 0.25 M * 0.150 L = 0.0375 moles. For NaF, it's 0.30 M * 0.225 L = 0.0675 moles.

Step 3: Use the Henderson-Hasselbalch Equation for Part a

Since we have a mixture of a weak acid and its conjugate base, use the Henderson-Hasselbalch equation to find the pH: pH = pKa + log([A-]/[HA]). For HF, we need the pKa, which is typically around 3.17. We then plug in the values: pH = 3.17 + log(0.0675/0.0375).

Step 4: Calculate the pH for Part a

Using the values from step 3, the pH for part a is calculated to be pH = 3.17 + log(1.8), which approximately equals to pH = 3.17 + 0.2553 = 3.4253.

Step 5: Calculate the Moles for Part b

Again calculate the moles of C2H5NH2 and C2H5NH3Cl before mixing. For C2H5NH2, that's 0.10 M * 0.175 L = 0.0175 moles. For C2H5NH3Cl, it's 0.20 M * 0.275 L = 0.055 moles.

Step 6: Use the Henderson-Hasselbalch Equation for Part b

This time, since we have a weak base and its conjugate acid, we adapt the Henderson-Hasselbalch equation to find the pOH first: pOH = pKb + log([B+]/[B]). Then we convert pOH to pH. First we need to know the pKa for the conjugate acid C2H5NH3+, which can then be used to find pKb (pKb = 14 - pKa).

Step 7: Calculate the pH for Part b

Assuming pKa for C2H5NH3+ is approximately 10.7, the pKb = 14 - 10.7 = 3.3. Using the adapted Henderson-Hasselbalch equation: pOH = 3.3 + log(0.055/0.0175). Calculate the pOH, then subtract it from 14 to find the pH. The pH = 14 - pOH.

Key Concepts

Henderson-Hasselbalch equation

Understanding the Henderson-Hasselbalch equation is crucial for calculating the pH of buffer solutions. It is an expression derived from the acid dissociation constant (Ka) that provides insight into the pH of a solution containing a weak acid and its conjugate base, or a weak base and its conjugate acid.

For weak acid and conjugate base systems, the equation is written as:
\[ pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right) \]
where \( pKa \) is the negative logarithm of the acid dissociation constant, \( [A^-] \) is the concentration of the conjugate base, and \( [HA] \) is the concentration of the acid.

For weak base systems, the equation is adapted by converting pKa to pKb (since pKb = 14 - pKa for water at 25 °C) and then calculating pOH instead of pH:
\[ pOH = pKb + \log\left(\frac{[B^+]}{[B]}\right) \]
Finally, the pH is obtained by subtracting pOH from 14:
\[ pH = 14 - pOH \]
The elegance of this equation lies in its simplicity, allowing students to readily determine the pH of a buffer system if they have knowledge of the concentrations and dissociation constants involved.

Weak acid and conjugate base

In buffer solutions, the pair of a weak acid and its conjugate base works in tandem to neutralize added acids or bases. A weak acid (HA) only partially dissociates in solution, releasing hydrogen ions (H+) and conjugate base ions (A-).

Let's break down these components:

• A weak acid is an acid that does not completely dissociate in aqueous solution.
• The conjugate base (A-) is what the acid becomes after donating a hydrogen ion.

Together, this weak acid-conjugate base pair can resist drastic changes in pH. This is because the weak acid can neutralize added bases while the conjugate base can neutralize added acids. The ability to maintain pH makes this combination ideal for buffer systems in various applications, including biological fluids and industrial processes.

Weak base and conjugate acid

Similar to the weak acid-conjugate base, there's also a weak base and its conjugate acid which constitute a buffer system. A weak base (B), like its acid counterpart, does not fully dissociate in water and thus provides buffering action by absorbing excess H+ ions.

The key points to remember are:

• A weak base partially accepts protons (H+), typically forming a partially charged or neutral species.
• The conjugate acid (B+) is the species formed when a base gains a proton.

When an acid is added to such a system, the weak base will neutralize some of the added H+ ions, while the conjugate acid will release H+ ions if a base is added to the system. This interplay allows for the regulation of pH within a narrow range, which is why these pairs are essential components in many biological and chemical buffer solutions.

Buffer systems

Buffer systems are the unsung heroes when it comes to maintaining a stable pH environment. They are composed of a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid. These systems have the remarkable ability to mitigate the effects of acids or bases, thus keeping the pH of a solution relatively constant even when perturbed by external factors.

Some characteristics and uses of buffer systems include:

• Buffer systems are essential in biological processes, maintaining the pH necessary for enzymatic activities within cells.
• They are also used in laboratory settings, where reactions often require a stable pH.
• Industrial applications such as dyeing fabrics, fermentation processes, and pharmaceutical production rely on buffer systems.

By using the principles of how buffer systems function, students and professionals can predict and control the pH of solutions important in both academic and practical applications.