Question

Calculate the \(\mathrm{pH}\) of a buffer that is (a) \(0.012 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\left(K_{\mathrm{a}}=6.3 \times 10^{-5}\right)\) and 0.033 \(\mathrm{M} \mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{COO}\) (b) \(0.408 \mathrm{M} \mathrm{NH}_{3}\) and \(0.153 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\)

Short answer

The pH of a buffer solution for part (a) is 5.02, and for part (b) is 3.93.

Step-by-step solution

Identify the buffer components and understand the Henderson-Hasselbalch equation

The buffer system contains a weak acid C6H5COOH and its conjugate base NaC6H5COO in part (a), and a weak base NH3 and its conjugate acid NH4Cl in part (b). The Henderson–Hasselbalch equation is as follows: \[pH = pKa + log_{10} \left(\frac{[A^-]}{[HA]}\right)\] for the acid and its conjugate base, and \[pOH = pKb + log_{10} \left(\frac{[B]}{[HB^+]}\right)\] for the base and its conjugate acid, where [A^-] represents the concentration of the base, [HA] the concentration of the acid, [B] the concentration of the base, and [HB^+] the concentration of the conjugate acid. We will determine pH using these equations.

Calculate the pH for part (a)

Substitute the given values into the Henderson-Hasselbalch equation for the acid and its conjugate base: \[ Ka = 6.3 × 10^{−5}\], \[ [A^-] = [NaC6H5COO] = 0.033 M\], and \[ [HA] = [C6H5COOH] = 0.012 M\]. First, calculate pKa = -log(Ka) = -log(6.3 × 10^{−5}) = 4.2, then substitute these values into the equation, we get: pH = pKa + log_{10} \left(\frac{[A^-]}{[HA]}\right) = 4.2 + log_{10} \left(\frac{0.033}{0.012}\right) = 4.2 + 0.82 = 5.02.

Calculate the pH for part (b)

Now use the Henderson-Hasselbalch equation for a base and its conjugate acid. The base is NH3 with a given Kb of 1.8 × 10^{-5}. For NH3 and NH4Cl, \[ [B] = [NH3] = 0.408 M\] and \[ [HB^+] = [NH4Cl] = 0.153 M\]. Start by converting Kb to Ka using the equation \[ Ka = \frac{Kw}{Kb} = \frac{1.0 × 10^{−14}}{1.8 × 10^{−5}} = 5.56 × 10^{−10}\]. Then, calculate pKa = -log(Ka) = -log(5.56 × 10^{−10}) = 9.25. Now, find pOH using the Henderson-Hasselbalch equation: \[pOH = pKa + log_{10} \left(\frac{[B]}{[HB^+]}\right) = 9.25 + log_{10} \left(\frac{0.408}{0.153}\right) = 9.25 + 0.82 = 10.07\]. Convert pOH to pH using the equation \[pH = 14 - pOH = 14 - 10.07 = 3.93.\]

Interpret the results

The pH of a buffer solution for part (a) is 5.02, and for part (b) is 3.93. These measurements indicate that both buffers are acidic since their pH values are less than 7. pH of part (b) is lower than that of part (a) which means it is more acidic.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a mathematical formula that simplifies the calculation of the pH of buffer solutions. It is derived from the acid dissociation constant expression and provides a relationship between pH, pKa (the negative logarithm of the acid dissociation constant, Ka), and the ratio of the concentration of the conjugate base to the weak acid.

For a buffered solution containing a weak acid (HA) and its conjugate base (A⁻), the equation is given by:
\[ pH = pKa + \log_{10} \left(\frac{[A^-]}{[HA]}\right) \] This equation allows you to calculate the pH by knowing:

• The pKa of the acid, which is a fixed value.
• The concentration of the acid ([HA]) and the base ([A⁻]).

Similarly, for a buffer containing a weak base (B) and its conjugate acid (HB⁺), the equation can be adapted to calculate pOH:

\[ pOH = pKb + \log_{10} \left(\frac{[B]}{[HB^+]}\right) \] To find pH in such a case, you first calculate pOH and then convert it using the identity \[ pH = 14 - pOH \].

This equation is particularly useful in biochemistry and chemistry, where buffer solutions are needed to maintain a stable pH in reactions or solutions.

weak acid and conjugate base

In the realm of chemistry, a weak acid is an acid that does not completely dissociate into its ions when dissolved in water. A conjugate base, on the other hand, is what remains after a weak acid has donated a hydrogen ion. Together, they form a buffer system that can resist drastic pH changes.

For example, in part (a) of the exercise, benzoic acid (\(\mathrm{C_6H_5COOH}\)) acts as the weak acid and sodium benzoate (\(\mathrm{NaC_6H_5COO}\)) is the conjugate base. They work together to maintain the buffer’s pH.

• The weak acid provides or removes \(\mathrm{H^+}\) ions to maintain equilibrium.
• The conjugate base can neutralize added acids by capturing \(\mathrm{H^+}\) ions.

This capacity to moderate pH is crucial for many chemical and biological processes, as maintaining a specific pH range is necessary for enzyme activity, cellular functions, and chemical reactions.
Buffer solutions are designed by choosing combinations of weak acids and their conjugate bases so that the desired pH is preserved, forming an essential part of chemical and biological buffer systems.

pH calculation of buffers

Calculating the pH of a buffer solution involves utilizing the components of the buffer to stabilize pH against the addition of moderate amounts of acids or bases. As shown in the exercise, the process can be detailed into a few straightforward steps.

In part (a), with the components benzoic acid and sodium benzoate:

• You first determine the pKa of the weak acid, which is derived from the acid dissociation constant, Ka, using the formula: \( pKa = -\log_{10}(Ka) \).
• Then, apply the Henderson-Hasselbalch equation: \( pH = pKa + \log_{10}\left(\frac{[base]}{[acid]}\right) \) to obtain the pH value, which in this example is 5.02.

In part (b), with ammonia and ammonium chloride:

• First, derive Ka from Kb using \( Ka = \frac{Kw}{Kb} \), where Kw is the ion product of water.
• Convert this to pKa and apply the Henderson-Hasselbalch for bases to find pOH.
• Finally, convert pOH to pH using \( pH = 14 - pOH \), resulting in a buffer pH of 3.93.

These methods illustrate how buffer solutions work efficiently in buffering systems and help provide stable environments necessary for various biochemical and industrial processes.