Question

(a) Given that \(K_{a}\) for acetic acid is \(1.8 \times 10^{-5}\) and that for hypochlorous acid is \(3.0 \times 10^{-8}\), which is the stronger acid? (b) Which is the stronger base, the acetate ion or the hypochlorite ion? (c) Calculate \(K_{b}\) values for \(\mathrm{CH}_{3} \mathrm{COO}^{-}\)and \(\mathrm{ClO}^{-}\).

Short answer

(a) Acetic acid is the stronger acid because it has a higher \(K_{a}\) value. (b) The stronger base is the hypochlorite ion (\(\mathrm{ClO}^{-}\)). (c) The \(K_{b}\) values are: for the acetate ion, \(K_{b} = 5.56 \times 10^{-10}\), and for the hypochlorite ion, \(K_{b} = 3.33 \times 10^{-7}\).

Step-by-step solution

Determine the stronger acid

To determine the stronger acid, we need to compare the \(K_{a}\) values given for each acid. Acetic acid: \(K_{a} = 1.8 \times 10^{-5}\) Hypochlorous acid: \(K_{a} = 3.0 \times 10^{-8}\) Since the higher \(K_{a}\) value indicates a stronger acid, acetic acid is the stronger acid.

Determine the stronger base

As a general rule, the conjugate base of a weaker acid is stronger compared to the conjugate base of a stronger acid. We have already established that acetic acid is stronger than hypochlorous acid, therefore the conjugate base - the acetate ion - is weaker compared to the hypochlorite ion. So, the stronger base is the hypochlorite ion (\(\mathrm{ClO}^{-}\)).

Calculate \(K_{b}\) values for the acetate and hypochlorite ions

To calculate the \(K_{b}\) values for the given ions, we will use the relationship between \(K_{w}\), \(K_{a}\), and \(K_{b}\): \(K_{w} = K_{a} \cdot K_{b}\) The ion product of water \(K_{w}\) is \(1.0 \times 10^{-14}\) at \(25^{\circ} \mathrm{C}\). We can find \(K_{b}\) for each ion using the given \(K_{a}\) values and the formula above. For the acetate ion (\(\mathrm{CH}_{3} \mathrm{COO}^{-}\)): \(K_{b} = \frac{K_{w}}{K_{a}} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\) For the hypochlorite ion (\(\mathrm{ClO}^{-}\)): \(K_{b} = \frac{K_{w}}{K_{a}} = \frac{1.0 \times 10^{-14}}{3.0 \times 10^{-8}} = 3.33 \times 10^{-7}\) Therefore, the \(K_{b}\) values are: Acetate ion: \(K_{b} = 5.56 \times 10^{-10}\) Hypochlorite ion: \(K_{b} = 3.33 \times 10^{-7}\)

Key concepts

Acid Strength

Acid strength refers to the ability of an acid to donate a proton (H⁺) to a base. This characteristic is quantified by the acid dissociation constant, or \(K_a\). The larger the \(K_a\) value, the stronger the acid since it indicates a greater tendency to dissociate and release protons.
For example, acetic acid has a \(K_a\) value of \(1.8 \times 10^{-5}\), while hypochlorous acid has a \(K_a\) value of \(3.0 \times 10^{-8}\). Since acetic acid has a higher \(K_a\), it is the stronger acid because it more readily donates protons compared to hypochlorous acid.
When comparing acid strength:

• Higher \(K_a\) = Stronger acid
• Lower \(K_a\) = Weaker acid

Base Strength

Base strength is determined by the ability of a base to accept protons. It's related to the acid from which the base is derived - its conjugate acid. Generally, the conjugate base of a weak acid will be stronger than the conjugate base of a strong acid.
In the exercise, we see acetic acid is a stronger acid than hypochlorous acid due to its higher \(K_a\). Consequently, its conjugate base, the acetate ion (\(\mathrm{CH}_3\mathrm{COO}^-\)), is weaker because strong acids have weak conjugate bases.
On the other hand, hypochlorous acid's conjugate base, the hypochlorite ion (\(\mathrm{ClO}^-\)), is stronger since it comes from the weaker hypochlorous acid. Hence:

• Stronger acids = Weaker conjugate bases
• Weaker acids = Stronger conjugate bases

Ka and Kb Calculations

\(K_a\) and \(K_b\) are related quantities used to describe the strength of acids and bases. \(K_a\) is the acid dissociation constant, while \(K_b\) is the base dissociation constant. These values are linked by the ion product of water \(K_w\), which is \(1.0 \times 10^{-14}\) at \(25^\circ C\).
To calculate \(K_b\) for a conjugate base from a given \(K_a\):

• Use the formula \(K_w = K_a \times K_b\)
• Rearrange to find \(K_b\): \(K_b = \frac{K_w}{K_a}\)

For the acetate ion, given \(K_a = 1.8 \times 10^{-5}\):
\[K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\]
For the hypochlorite ion, given \(K_a = 3.0 \times 10^{-8}\):
\[K_b = \frac{1.0 \times 10^{-14}}{3.0 \times 10^{-8}} = 3.33 \times 10^{-7}\]
Therefore, calculating \(K_b\) helps us understand the base strength and how it relates to its conjugate acid.