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.

Step by step solution

01

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.

02

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.

03

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: $$ \mathrm{pH} = 14 - \mathrm{pOH} $$

04

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.

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