The beaker on the right contains 0.1 Macetic acid solution with methyl orange as an indicator. The beaker on the left contains a mixture of 0.1\(M\) acetic acid and 0.1\(M\) sodium acetate with methyl orange. (a) Using Figures 16.8 and 16.9, which solution has a higher pH? (b) Which solution is better able to maintain its pH when small amounts of NaOH are added? Explain. [Sections 17.1 and 17.2]

In the beaker on the right, there is a 0.1M acetic acid solution with methyl orange as an indicator. Acetic acid is a weak acid, and its dissociation in water can be represented as: \[CH_3COOH_{(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}\] In the beaker on the left, there is a mixture of 0.1M acetic acid and 0.1M sodium acetate with methyl orange as an indicator. Sodium acetate is the sodium salt of the weak acid, acetic acid. The acetate ion can react with water as a base: \[CH_3COO^-_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3COOH_{(aq)} + OH^-_{(aq)}\] This solution is an example of a buffer solution, which is a mixture of a weak acid and its conjugate base.

In order to find the pH of the two solutions, we need to identify the dissociation constants (\(K_a\)) of acetic acid and the acetate ion. The \(K_a\) for acetic acid is \(1.8 \times 10^{-5}\). We can use the pH equation to find the pH of the acetic acid solution in the beaker on the right: \[pH = -\log[H^+]\] Since we are given the molarity of the acetic acid solution, we can calculate the [H+] concentration using the equilibrium expression: \[K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}\] For the buffer solution in the beaker on the left, we can use the Henderson-Hasselbalch equation to find the pH: \[pH = pK_a + \log\frac{[A^-]}{[HA]}\] where \(pK_a\) is the negative logarithm of \(K_a\), \([A^-]\) is the concentration of the acetate ion, and \([HA]\) is the concentration of acetic acid. Since both acetic acid and sodium acetate have the same concentration (0.1 M), we can calculate the pH of the buffer solution.

Let’s first find the pH of the acetic acid solution. Using the equilibrium expression, we can estimate [H+] then calculate the pH. Now find the pH of the buffer solution, by plugging in the values into the Henderson-Hasselbalch equation. Finally, compare both pH values and identify which solution has a higher pH.

When NaOH (a strong base) is added to the solutions, it reacts with any available hydrogen ions present. In the case of the acetic acid solution, the addition of NaOH will neutralize free hydrogen ions creating water and acetate ions which will push the reaction further to the right, decreasing the pH value. In a buffer solution, the addition of NaOH will react with the weak acid (acetic acid) initially, causing the production of its conjugate base (acetate ions). The acetate ions will then react with water as shown in the second reaction, producing hydroxide ions and acetic acid. The presence of both the weak acid and its conjugate base will resist changes in the pH when small amounts of NaOH are added. The solution that will maintain its pH better when small amounts of NaOH are added is the buffer solution (the mixture of acetic acid and sodium acetate) because both the weak acid and its conjugate base are present.