Problem 82 - Chemistry An Atoms First Approach

Question

a. Calculate the pH of a buffered solution that is 0.100 \(M\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5}\) ) and \(0.100 M\) 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 $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q)$$ to calculate the pH. c. Do the same as in part b, but use the following equilibrium to calculate the 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 \(b\) and c agree? Explain.

Step-by-Step Solution

Answer

The pH of the buffered solution is 4.19. After converting 20% of benzoic acid to benzoate anion, the pH is 4.37 using the dissociation equilibrium. Using the given equilibrium involving benzoate anion and water, the pH is 4.97. The disagreement between the results is due to the assumption that the concentration of \(H_3O^+\) and \(OH^-\) has negligible effects on the pH in the Henderson-Hasselbalch equation.

Step 1: Use the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \[pH = pK_a + log\frac{[A^-]}{[HA]}\] where, \(pH\) - pH of the solution \(pK_a\) - \(-log(K_a)\) \([A^-]\) - molar concentration of benzoate anion (\(C_6H_5CO_2^-\)) \([HA]\) - molar concentration of benzoic acid (\(C_6H_5CO_2H\))

Step 2: Calculate \(pK_a\)

Given \(K_a = 6.4 \times 10^{-5}\), we can find \(pK_a\) using the formula \(pK_a = -log(K_a)\): \(pK_a = -log(6.4 \times 10^{-5}) = 4.19\)

Step 3: Calculate the pH

Since the concentration of benzoic acid and benzoate ion are both 0.100 \(M\), we can use the Henderson-Hasselbalch equation to find the pH. \(pH = 4.19+ log\frac{0.100}{0.100} = 4.19+0 = 4.19\) So, the pH of the buffered solution is 4.19. b. Calculate the pH after 20% of benzoic acid is converted to benzoate anion

Step 1: Update the concentrations

When 20% of benzoic acid is converted to benzoate anion, the concentrations change: \([C_6H_5CO_2H] = 0.100 \times 0.8 = 0.080\, M\) \([C_6H_5CO_2^-] = 0.100 + 0.020 = 0.120\, M\)

Step 2: Calculate the new pH

Now we will use the new concentrations of benzoic acid and benzoate anion to find the new pH using the Henderson-Hasselbalch equation. \(pH = 4.19 + log\frac{0.120}{0.080} = 4.19 + 0.18 = 4.37\) So, the pH after converting 20% of benzoic acid to benzoate anion is 4.37. c. Calculate the pH using the given equilibrium involving benzoate anion and water

Step 1: Write the equilibrium expression

The equilibrium expression for the given reaction is: \(K_b = \frac{[C_6H_5CO_2H][OH^-]}{[C_6H_5CO_2^-]}\)

Step 2: Calculate the \(K_b\)

Using the relationship \(K_a \times K_b = K_w\), we can find the \(K_b\) for benzoate anion: \(K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.4 \times 10^{-5}} = 1.56 \times 10^{-10}\)

Step 3: Calculate \([OH^-]\) using the equilibrium expression

Assuming the initial concentration of hydroxide is 0, \(K_b = \frac{(0.020)(x)}{(0.120)}\) Solving for x, we get: \[x = [OH^-] = 9.36 \times 10^{-10}\]

Step 4: Calculate new pH

Finally, we can calculate the new pH using the relationship \(pH = 14 - pOH\): \(pOH = -log[OH^-] = -log(9.36 \times 10^{-10}) = 9.03\) \(pH = 14 - pOH = 14 - 9.03 = 4.97\) So, the pH calculated using this equilibrium is 4.97. d. Comparing the results The pH calculated using the dissociation equilibrium (part b, pH = 4.37) and the pH calculated using the equilibrium involving benzoate anion and water (part c, pH = 4.97) are not equal. This disagreement is because the Henderson-Hasselbalch equation assumes that the concentration of \(H_3O^+\) and \(OH^-\) has negligible effects on the pH. The presence of \(OH^-\) affects the concentrations of benzoic acid and benzoate anion and should not be neglected in a more accurate calculation.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental tool for calculating the pH of buffer solutions. It provides a simple relationship between the pH, the pKa of the acid, and the ratio of the concentrations of the acid and its conjugate base.
The equation is expressed as: \[ pH = pK_a + \log\frac{[A^-]}{[HA]} \]Where:

• \(pH\) is the measure of acidity or alkalinity.
• \(pK_a\) 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.

This equation helps predict how the pH will change if the concentrations of acid or base in the solution are altered. It's especially useful for designing buffer solutions that resist drastic changes in pH.

Benzoic Acid

Benzoic acid, with the formula \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\), is a simple aromatic carboxylic acid. It's commonly used in the food industry as a preservative due to its antibacterial properties.

In the context of buffer solutions, benzoic acid acts as the weak acid component. It can donate protons (\(\mathrm{H}^+\)) to maintain pH levels when bases are added. Its conjugate base, benzoate ion \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^-\), stabilizes the solution by accepting protons if an acid is introduced.

Understanding benzoic acid's role in equilibria is essential for manipulating and predicting the pH of solutions it is involved in.

pH Calculation

Calculating pH is crucial to understanding the acidity or basicity of a solution. A buffer solution resists changes in pH when small amounts of acids or bases are added.

In the original problem, the pH calculation involves the use of the Henderson-Hasselbalch equation, where the concentrations of benzoic acid and its conjugate base, benzoate, are known:

\[pH = 4.19 + \log\frac{0.100}{0.100} = 4.19 \]

Even after altering concentrations due to the conversion of some of the benzoic acid into benzoate, the equation remains applicable. Understanding how to rearrange and adjust values in the equation is fundamental for accurate pH predictions.

Dissociation Equilibrium

Dissociation equilibrium involves the exchange of ions between molecules in solution. For benzoic acid, the dissociation into its ions is represented by the equation:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(aq) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^-(aq) + \mathrm{H}^+(aq)\).
This process defines how readily the benzoic acid will donate protons in aqueous solutions, affecting the pH.

• \(K_a\) describes the equilibrium constant, showing the balance between reactants and products.
• A higher \(K_a\) value indicates a stronger acid, meaning more dissociation occurs.

In buffers, analyzing dissociation helps predict how the pH will shift when acid or base components are added. Recognizing the equilibrium dynamics ensures buffered solutions function effectively.