Question

a. Calculate the \(\mathrm{pH}\) of a buffered solution that is \(0.100 \mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5}\) ) and \(0.100 \mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\) b. Calculate the \(\mathrm{pH}\) after \(20.0 \%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) $$ to calculate the \(\mathrm{pH}\). c. Do the same as in part \(\mathrm{b}\), but use the following equilibrium to calculate the \(\mathrm{pH}\) : $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q) $$ d. Do your answers in parts \(\mathrm{b}\) and \(\mathrm{c}\) agree? Explain.

Short answer

In summary, the initial pH of the buffered solution is approximately 4.19. After 20% of the benzoic acid is converted to benzoate anion, the pH is approximately 4.37, calculated using both dissociation and hydrolysis equilibrium methods. The agreement between these methods confirms the validity of the calculations.

Step-by-step solution

Use the Henderson-Hasselbalch equation

To calculate the pH of the buffered solution, we can use the Henderson-Hasselbalch equation: $$ pH = pK_a + \log \frac{[\mathrm{C} _6 \mathrm{H} _5 \mathrm{CO}_{2} ^{-}]}{[\mathrm{C} _6 \mathrm{H} _5 \mathrm{CO}_{2} \mathrm{H}]} $$ Where \(pK_a = -\log K_a\) and the given \(K_a = 6.4 \times 10^{-5}\).

Calculate the pH

Now substitute the given values and concentrations into the equation: $$ pH = -\log(6.4 \times 10^{-5}) + \log \frac{0.100}{0.100} $$ The ratio inside the logarithm is equal to 1, so the equation simplifies to: $$ pH = -\log(6.4 \times 10^{-5}) $$ Calculating the final pH value: $$ pH \approx 4.19 $$ b. Calculate the pH after 20% of the benzoic acid is converted to benzoate anion.

Calculate new concentrations

We know that 20% of benzoic acid is converted to benzoate anion, so calculate the new concentrations: $$ [\mathrm{C} _6 \mathrm{H} _5 \mathrm{CO}_{2} \mathrm{H}] = 0.100 \times 0.80 = 0.080\ \mathrm{M} $$ $$ [\mathrm{C} _6 \mathrm{H} _5 \mathrm{CO}_{2} ^{-}] = 0.100 + 0.100 \times 0.20 = 0.120\ \mathrm{M} $$

Use the Henderson-Hasselbalch equation with new concentrations

Substitute the new concentrations into the Henderson-Hasselbalch equation: $$ pH = -\log(6.4 \times 10^{-5}) + \log \frac{0.120}{0.080} $$

Calculate the new pH

Evaluate the equation and calculate the new pH: $$ pH \approx 4.37 $$ c. Calculate the pH using the hydrolysis equilibrium.

Write the Kb expression from Ka

First, we need to find the \(K_b\) value for the benzoate anion using the given \(K_a\) value and the \(K_w\) value, which is equal to \(1.0 \times 10^{-14}\). $$ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.4 \times 10^{-5}} \approx 1.56 \times 10^{-10} $$

Write the hydrolysis equilibrium expression

Write the equilibrium expression for the hydrolysis of benzoate anion (C): $$ K_b = \frac{[\mathrm{C} _6 \mathrm{H} _5 \mathrm{CO}_{2} \mathrm{H}][\mathrm{OH}^{-}]}{[\mathrm{C} _6 \mathrm{H} _5 \mathrm{CO}_{2} ^{-}]} $$

Calculate the hydroxide ion concentration

Rearrange the equation to find the hydroxide ion concentration: $$ [\mathrm{OH}^{-}] = \frac{K_b [\mathrm{C} _6 \mathrm{H} _5 \mathrm{CO}_{2} ^{-}]}{[\mathrm{C} _6 \mathrm{H} _5 \mathrm{CO}_{2} \mathrm{H}]} = \frac{(1.56 \times 10^{-10})(0.120)}{0.080} $$ Calculate the hydroxide ion concentration: $$ [\mathrm{OH}^{-}] \approx 2.34 \times 10^{-10} $$

Calculate the new pH

Now, find the pOH: $$ pOH = -\log(2.34 \times 10^{-10}) \approx 9.63 $$ Finally, we can find the pH: $$ pH = 14 - pOH \approx 4.37 $$ d. Compare the answers in parts b and c. The answers in parts b and c are indeed the same, showing that the Henderson-Hasselbalch equation and the hydrolysis equilibrium give the same final pH value. This is expected, as both methods are essentially describing the same acidity-basicity balance within the buffered solution, but from different perspectives. The agreement between the two approaches confirms the validity of our calculations.

Key concepts

Henderson-Hasselbalch Equation

Understanding the pH of a buffer solution is crucial for many chemical and biological processes. A key tool for this is the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to its acid dissociation constant \(K_a\) and the concentrations of the acid and its conjugate base. The equation is expressed as:
\[ pH = pK_a + \log\frac{[\text{A}^-]}{[\text{HA}]} \]
In this equation, \( \text{A}^- \) represents the conjugate base of the acid \( \text{HA} \), and \( pK_a \) is the negative logarithm of the acid dissociation constant \(K_a\). The equation assumes that the concentrations of the acid and base are those that prevail at equilibrium. For a buffer system like benzoic acid and sodium benzoate, where the concentrations of acid and base are equal, the \( \log\frac{[\text{A}^-]}{[\text{HA}]} \) term becomes zero, simplifying the calculation of pH to just \( pK_a \). The power of this equation lies in its ability to predict the pH of a buffer system upon addition of small amounts of acids or bases, which is essential for maintaining pH stability in many practical applications.

Acid-Base Equilibrium

At the heart of buffer system functionality is the concept of acid-base equilibrium. It is the delicate balance that exists between an acid and its conjugate base in a solution. In the equilibrium process, benzoic acid (\(\text{HA}\)) dissociates into its conjugate base, the benzoate ion (\(\text{A}^-\)), and a proton (\(\text{H}^+\)):
\[ \text{HA} \rightleftharpoons \text{A}^- + \text{H}^+ \]
The equilibrium constant for this reaction, known as the acid dissociation constant (\(K_a\)), quantifies the extent of the dissociation at equilibrium. In buffer solutions, this equilibrium is crucial because it allows the solution to resist changes in pH when acids or bases are added. This functions through the buffer's ability to 'absorb' the extra hydrogen ions (or donate them) without a significant change in pH, which is particularly important in biological systems where enzymes are sensitive to pH changes.

Hydrolysis of Benzoate Anion

In the context of buffer solutions, it is also essential to understand the hydrolysis of the benzoate anion. Hydrolysis is a chemical reaction involving the breaking of a bond in a molecule using water. For the benzoate ion (\(\text{A}^-\)), the hydrolysis reaction proceeds as follows:
\[ \text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^- \]
Here, benzoate ions react with water to form benzoic acid and hydroxide ions. The equilibrium constant for this reaction is represented as the base dissociation constant (\(K_b\)), highlighting the reverse process of acid dissociation, resulting in formation of hydroxide ions instead of hydrogen ions. Balancing hydrolysis is an integral part of managing the buffer's pH, particularly when the buffer system comes into contact with bases, as the generation of hydroxide ions could increase the pH were it not for the buffering action of the acid-base pair.

Benzoic Acid and Sodium Benzoate Buffer System

The benzoic acid and sodium benzoate combination constitutes an effective buffer system that stabilizes pH against the introduction of external acids or bases. Benzoic acid (\( \text{C}_6\text{H}_5\text{CO}_2\text{H} \)) serves as the weak acid, and sodium benzoate (\( \text{C}_6\text{H}_5\text{CO}_2\text{Na} \)) acts as its conjugate base due to the deprotonation of the acid upon dissolution.

When a strong base is added to this buffer system, the benzoic acid donates a proton to neutralize hydroxide ions, forming the benzoate anion and water. Conversely, upon the addition of a strong acid, the benzoate anion can accept a proton, thus forming benzoic acid and mitigating the change in pH. This balance between the acid and its conjugate base enables the buffer to maintain a relatively constant pH, making this combination useful in various applications, from food preservation to enzymatic control in biochemical processes.