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 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 \(\mathrm{pH} :\) $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q) $$ d. Do your answers in parts \(b\) and \(c\) agree? Explain.

Short answer

In summary, the initial pH of the buffered solution is 4.20. After 20% of benzoic acid is converted to benzoate anion, the pH increases to 4.46 using both dissociation equilibria methods. Both results agree, indicating that the dissociation equilibria can be used to solve this problem.

Step-by-step solution

Part a: Calculate the initial pH of the buffered solution

We will use the Henderson-Hasselbalch equation to calculate the pH of the buffered solution consisting of benzoic acid and its sodium salt, sodium benzoate. $$ \mathrm{pH}=\mathrm{p}K_{a}+\log \frac{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\mathrm{H}]} $$ Given that both benzoic acid and sodium benzoate have the same concentration of 0.1 M and \(K_{a}=6.4 \times 10^{-5}\), we can plug these numbers into the equation. $$ \mathrm{pH}=\mathrm{-\log}(6.4 \times 10^{-5})+\log \frac{0.100}{0.100} $$ Solving this equation, we get the initial pH value.

Part b: Calculate the pH after 20.0% of benzoic acid is converted to benzoate anion

Adding a strong base will convert 20% of benzoic acid to benzoate anion. Let's first find the new concentration of both benzoic acid and benzoate anion. New concentration of benzoic acid: 0.100 - 0.100x0.20 = 0.080 M New concentration of benzoate anion: 0.100 + 0.100x0.20 = 0.120 M Now, using the Henderson-Hasselbalch equation again, $$ \mathrm{pH}=\mathrm{-\log}(6.4 \times 10^{-5})+\log \frac{0.120}{0.080} $$ Solving this equation, we get the pH value after the conversion.

Part c: Calculate the pH using a different dissociation equilibrium

We can also use the dissociation equilibrium between benzoate anion and water to calculate the pH. $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q) $$ Here, the equilibrium constant, \(K_b\), can be calculated as follows: $$ K_b = \frac{K_w}{K_a}= \frac{10^{-14}}{6.4 \times 10^{-5}} $$ Using the above equilibrium expression and \(K_b\), we can write: $$ K_b = \frac{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\mathrm{H}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}]} $$ Given the new concentrations of benzoic acid (0.080 M) and benzoate anion (0.120 M), we can solve for the concentration of \(\mathrm{OH}^{-}\). After finding the concentration of \(\mathrm{OH}^{-}\), we can calculate the pOH by taking the negative logarithm of \(\mathrm{OH}^{-}\) concentration and then find the pH using: $$ \mathrm{pH} = 14 - \mathrm{pOH} $$

Part d: Comparing the answers from parts b and c

Compare the pH values obtained in parts b and c. They should be approximately equal, which confirms that both dissociation equilibria can be used to solve this problem.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a powerful tool for calculating the pH of a buffer solution. A buffer contains a weak acid and its conjugate base (or vice versa) that resists changes in pH upon the addition of small amounts of acid or base. The equation is given as: \[\mathrm{pH} = \mathrm{p}K_{a} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)\]In this formula, \(\mathrm{p}K_a\) is the negative logarithm of the acid dissociation constant \(K_a\). The ratio \([\text{base}]/[\text{acid}]\) is crucial since it indicates the proportion of the conjugate base and acid in the solution. By using this equation, you can calculate the initial pH of a buffer solution when both components are at equal concentrations.

Benzoic acid dissociation

Benzoic acid is a weak acid that partially dissociates in water. Its dissociation can be represented by the following equation:\[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}(aq) \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}(aq) + \mathrm{H}^{+}(aq)\]The benzoate ion (\(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}\)) is the conjugate base of benzoic acid. The extent of this dissociation is measured by the equilibrium constant \(K_a\), which is specific to benzoic acid. Working with benzoic acid involves considering its weak acid property, where a certain percentage can be converted into the benzoate ion by reacting with a base. This key concept is important when working with equilibrium calculations, and it influences the pH of the solution due to the association-disassociation dynamics.

Equilibrium constant calculation

Equilibrium constants help us understand the extent of a reaction. In our scenario with benzoic acid, knowing \(K_a\) allows us to comprehend how much of the acid dissociates into ions. When calculating the pH after some of the benzoic acid is converted to its conjugate base, we rely on a new equilibrium.Besides the original dissociation, an alternative equilibrium involves the reaction of the benzoate ion with water:\[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}(aq) + \mathrm{OH}^{-}(aq)\]To find the equilibrium constant \(K_b\), we use the relationship: \[K_b = \frac{K_w}{K_a}\]This relationship uses the water ion product \(K_w\) and the acid dissociation constant \(K_a\), allowing us to then solve for the concentration of \(\mathrm{OH}^{-}\) ions and ultimately calculate pH from pOH.

pH and pOH relationship

pH and pOH provide insight into the acidity and alkalinity of solutions, respectively. They are inversely related through the ion product of water:\[\mathrm{pH} + \mathrm{pOH} = 14\]This relationship helps in determining one value if the other is known. For instance, if we calculate the concentration of \(\mathrm{OH}^{-}\) ions (from the alternate equilibrium reaction discussed earlier), we can determine pOH using:\[\mathrm{pOH} = -\log [\mathrm{OH}^{-}]\]Subsequently, knowing the pOH lets us calculate the pH by using the above relationship. Understanding how pH and pOH are interconnected is fundamental for balancing equations and predicting the behavior of solutions in different chemical contexts.