Question

A buffer is prepared using the butyric acid/butyrate \(\left(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2} / \mathrm{C}_{4} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\right)\) acid-base pair. The ratio of acid to base is 2.2 and \(K_{\mathrm{a}}\) for butyric acid is \(1.54 \times 10^{-5}\). (a) What is the \(\mathrm{pH}\) of this buffer? (b) Enough strong base is added to convert \(15 \%\) of butyric acid to the butyrate ion. What is the \(\mathrm{pH}\) of the resulting solution? (c) Strong acid is added to the buffer to increase its \(\mathrm{pH}\). What must the acid/base ratio be so that the pH increases by exactly one unit (e.g., from 2 to 3 ) from the answer in (a)?

Short answer

(b) What is the new pH after 15% of butyric acid is converted to the butyrate ion? (c) What is the acid/base ratio required for a 1-unit increase in pH? Solution: (a) To find the pH of the buffer solution, use the Henderson-Hasselbalch equation: $$\mathrm{pH} = -\log_{10} \left(1.54 \times 10^{-5}\right) + \log_{10} \left(\frac{1}{2.2}\right)$$ Calculate the pH. (b) To find the new pH after 15% of butyric acid is converted to the butyrate ion, use the new acid/base ratio in the Henderson-Hasselbalch equation: $$\mathrm{pH} = -\log_{10} \left(1.54 \times 10^{-5}\right) + \log_{10} \left(\frac{1.15}{1.87}\right)$$ Calculate the new pH. (c) To find the new acid/base ratio that increases the pH by exactly 1 unit, rearrange the Henderson-Hasselbalch equation and calculate the new ratio: $$\frac{\text{Concentration of base}}{\text{Concentration of acid}} = 10^{[\text{pH(a)} + 1] - \log_{10}(1.54 \times 10^{-5})}$$ Calculate the new acid/base ratio required for a 1-unit increase in pH.

Step-by-step solution

(a) Finding the pH of the buffer

To find the pH of the buffer solution, we can use the Henderson-Hasselbalch equation: $$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log_{10} \left(\frac{\text{Concentration of base}}{\text{Concentration of acid}}\right)$$ Given that the ratio of acid to base is 2.2, and \(K_a = 1.54 \times 10^{-5}\): $$\mathrm{pH} = -\log_{10} \left(1.54 \times 10^{-5}\right) + \log_{10} \left(\frac{1}{2.2}\right)$$ Now, calculate the pH.

(b) Finding the new pH after adding strong base

After converting 15% of butyric acid to the butyrate ion, we need to find the new acid/base ratio and then use the Henderson-Hasselbalch equation again to find the new pH. If the original ratio was 2.2, the acid fraction decreases by 15% and the base fraction increases by 15%: $$\text{New acid fraction} = 0.85(2.2) = 1.87$$ $$\text{New base fraction} = 1 + 0.15 = 1.15$$ $$\text{New ratio} = \frac{1.15}{1.87}$$ Now, use the new ratio in the Henderson-Hasselbalch equation: $$\mathrm{pH} = -\log_{10} \left(1.54 \times 10^{-5}\right) + \log_{10} \left(\frac{1.15}{1.87}\right)$$ Calculate the new pH.

(c) Finding the acid/base ratio for a 1-unit increase in pH

To find the new acid/base ratio that increases the pH by exactly 1 unit from the answer in (a), rearrange the Henderson-Hasselbalch equation to solve for the ratio: $$\frac{\text{Concentration of base}}{\text{Concentration of acid}} = 10^{\text{pH} - \mathrm{p}K_{\mathrm{a}}}$$ Assuming the pH increases by 1 unit, $$\frac{\text{Concentration of base}}{\text{Concentration of acid}} = 10^{[\text{pH(a)} + 1] - \log_{10}(1.54 \times 10^{-5})}$$ Calculate the new acid/base ratio required for a 1-unit increase in pH.

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a mathematical expression that relates the pH of a buffer solution to its acid-base composition. A buffer solution is designed to maintain a stable pH upon the addition of small amounts of acids or bases. The equation is written as:
\[ \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log_{10} \left(\frac{\text{Concentration of base}}{\text{Concentration of acid}}\right) \]
In this formula, \( \mathrm{p}K_\mathrm{a} \) is the negative logarithm (base 10) of the acid dissociation constant \( K_\mathrm{a} \), a measure of the strength of an acid in solution. The equation is derived from the acid dissociation equilibrium and has wide applications in biochemistry and analytical chemistry, especially for calculating the pH of blood and other biological fluids.

Acid-Base Equilibrium

The balance between acids and bases in solution is described as acid-base equilibrium. This equilibrium is central to understanding how buffer solutions work. An acid is a substance that can donate protons \( (H^+) \), while a base can accept them. When an acid is dissolved in water, it dissociates to form its conjugate base and a hydrogen ion:
\[ \text{HA} \rightleftharpoons \text{A}^- + \text{H}^+ \]
Here, \( \text{HA} \) represents the acid and \( \text{A}^- \) the conjugate base. The dissociation constant \( K_\mathrm{a} \) quantifies the extent of this reaction and is influenced by the strength of the acid and its environment. In a buffer, the presence of both the acid and its conjugate base allows the solution to resist pH changes when acids or bases are added.

pKa and pH Relationship

The relationship between \( \mathrm{p}K_\mathrm{a} \) and pH is crucial for buffer solutions. The \( \mathrm{p}K_\mathrm{a} \) value is a property of the acid that indicates its acidity level—the lower the \( \mathrm{p}K_\mathrm{a} \), the stronger the acid. In the context of buffer solutions, when the pH of the solution equals the \( \mathrm{p}K_\mathrm{a} \) of the acid, there is an equal concentration of acid and conjugate base. It's at this point that a buffer has its maximum capacity to resist changes in pH. Thus, by knowing the \( \mathrm{p}K_\mathrm{a} \) of an acid, one can predict how the buffer solution will behave when small amounts of acid or base are added.

Chemical Buffers

Chemical buffers are solutions that minimize changes in pH when an acid or a base is added. They consist of a weak acid and its conjugate base (or a weak base and its conjugate acid) in similar concentrations. Buffers are vital in biological systems and laboratory settings because they maintain the pH at a nearly constant value, which is essential for the proper functioning of enzymes and other biochemical processes.
An example of a buffer system is the butyric acid/butyrate pair used in the exercise. When butyric acid (the acid component) is in solution with butyrate ion (the conjugate base), it forms a buffer that can absorb added hydrogen ions without a significant change in pH. This makes it an effective buffer, capable of withstanding external influences that otherwise would alter the pH dramatically.