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(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 \(C H_{3} C O O^{-}\) and \(C 1 O^{-}.\)

Short Answer

Expert verified
(a) Acetic acid is the stronger acid. (b) The hypochlorite ion is the stronger base. (c) The \(K_b\) values for the acetate ion and the hypochlorite ion are \(5.6 \times 10^{-10}\) and \(3.3 \times 10^{-7}\), respectively.

Step by step solution

01

Understanding the relation between \(K_a\) and acid strength

Acids that have a larger \(K_a\) are considered stronger acids since they dissociate more into their ions when dissolved in water. Therefore, we can compare \(K_a\) values to determine which acid is stronger.
02

Comparing the given \(K_a\) values

We are given that \(K_a\) for acetic acid is \(1.8 \times 10^{-5}\) and \(K_a\) for hypochlorous acid is \(3.0 \times 10^{-8}\). Since \(1.8 \times 10^{-5} > 3.0 \times 10^{-8}\), acetic acid is the stronger acid. #b) Comparing base strength#
03

Understanding the relation between conjugate acid and base strength

The stronger the acid, the weaker its conjugate base. Therefore, to determine the stronger base, we have to look for the weaker acid.
04

Identifying the stronger base

Based on our comparison in part (a), acetic acid is the stronger acid. This means that its conjugate base, the acetate ion (\(CH_3COO^-\)), is weaker than the conjugate base of hypochlorous acid, the hypochlorite ion (\(ClO^-\)). Thus, the hypochlorite ion is the stronger base. #c) Calculating \(K_b\) values#
05

Understanding the relation between \(K_a\), \(K_b\) and \(K_w\)

For a conjugate acid-base pair, the product of their dissociation constants \(K_a\) and \(K_b\) is equal to the ion product constant of water, \(K_w\), i.e., \(K_a \times K_b = K_w\). The value of \(K_w\) at 25°C is \(1.0 \times 10^{-14}\).
06

Calculating \(K_b\) for acetate ion (\(CH_3COO^-\))

We know \(K_a\) for acetic acid, so we can calculate \(K_b\) for its conjugate base using the formula: \[K_b = \frac{K_w}{K_a}\] For the acetate ion, we have: \[K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}\]
07

Calculating \(K_b\) for hypochlorite ion (\(ClO^-\))

Similarly, we can calculate \(K_b\) for the hypochlorite ion: \[K_b = \frac{1.0 \times 10^{-14}}{3.0 \times 10^{-8}} = 3.3 \times 10^{-7}\] In summary, acetic acid is the stronger acid, the hypochlorite ion is the stronger base, and their respective \(K_b\) values are \(5.6 \times 10^{-10}\) and \(3.3 \times 10^{-7}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Strength of Acids and Bases
Understanding the strength of acids and bases is crucial for predicting the behavior of substances in chemical reactions. An acid's strength is measured by its acid dissociation constant, denoted as Ka. The higher the Ka value, the more an acid dissociates into protons (H+) and its conjugate base in a solution, thus making it a stronger acid.

In the given exercise, acetic acid has a Ka of 1.8 x 10^-5, which is larger than the Ka of hypochlorous acid, 3.0 x 10^-8. This clearly indicates that acetic acid is the stronger acid because it donates protons more readily in solution. Conversely, the strength of a base is often inferred from the weakness of its conjugate acid. In this context, since hypochlorous acid is the weaker acid, hypochlorite ion, its conjugate base, will be the stronger base.
Conjugate Acid-Base Pairs
A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton. In the context of our exercise, acetic acid (CH3COOH) and acetate ion (CH3COO-) form one such pair, while hypochlorous acid (HClO) and hypochlorite ion (ClO-) form another. The interesting aspect of these pairs is that the strength of the conjugate base is inversely related to the strength of its acid. When an acid donates a proton, it forms a conjugate base that is less likely to re-accept a proton.

As illustrated in the solution, because acetic acid is stronger than hypochlorous acid, acetate ion is a weaker base in comparison to the hypochlorite ion. Remember, when assessing the strength of bases, we often look to the corresponding acid's strength for insight. The weaker the acid, the more robust its conjugate base.
Ion Product Constant of Water
The ion product constant of water (Kw) is a vital concept in understanding acid-base chemistry. It represents the equilibrium constant for the self-ionization of water and is always 1.0 x 10^-14 at 25°C. In an aqueous solution, water dissociates into hydroxide (OH-) and hydronium ions (H3O+) at this constant rate.

The relationship between Ka and Kb for a conjugate acid-base pair is directly linked to Kw. The formula \(\text{Ka} \times \text{Kb} = \text{Kw}\) reflects that the product of the dissociation constants of a conjugate acid and its base always equals Kw. In the course of solving the exercise, we apply this formula to determine the Kb values of the conjugate bases. For instance, the Kb for the acetate ion is computed by dividing Kw by the Ka of acetic acid, yielding a Kb equal to 5.6 x 10^-10. Such calculations are essential for thoroughly understanding acid-base equilibrium in solutions.

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Most popular questions from this chapter

The acid-dissociation constant for chlorous acid \(\left(\mathrm{HClO}_{2}\right)\) is \(1.1 \times 10^{-2} .\) Calculate the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{ClO}_{2}^{-}\) and \(\mathrm{HClO}_{2}\) at equilibrium if the initial concentration of \(\mathrm{HClO}_{2}\) is 0.0125 \(\mathrm{M} .\)

Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the formula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid, or a species with negligible acidity: \((\mathbf{a})\mathrm{CH}_{3} \mathrm{COO}^{-},(\mathbf{b}) \mathrm{HCO}_{3}^{-},(\mathbf{c}) \mathrm{O}^{2-},(\mathbf{d}) \mathrm{Cl}^{-},(\mathbf{e}) \mathrm{NH}_{3}\)

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(0.0167 M \mathrm{HNO}_{3},(\mathbf{b}) 0.225 \mathrm{g}\) of \(\mathrm{HClO}_{3}\) in 2.00 \(\mathrm{L}\) of solution, \((\mathbf{c}) 15.00 \mathrm{mL}\) of 1.00 \(\mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{L},\) (d) a mixture formed by adding 50.0 \(\mathrm{mL}\) of 0.020 \(\mathrm{MHCl}\) to 125 \(\mathrm{mL}\) of 0.010 \(\mathrm{M} \mathrm{HI} .\)

Identify the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also identify the conjugate acid and conjugate base of each on the right side: (a) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q)\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (c)\(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)\)

(a) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l) .\) (c) What is the conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q) ?\) What is its conjugate base?

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