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Chapter 15: Problem 84

The \(K_{2}\) value for oxalic acid (HOOCCOOH) is \(5.9 \times 10^{-2},\) and the \(K_{2,2}\) value is \(6.4 \times 10^{-5} .\) What are the values of \(K_{\mathrm{b}_{1}}\) and \(K_{\mathrm{b}_{2}}\) of the oxalate anion \(\left(^{-} \mathrm{OOCCOO}^{-}\right) ?\)

Short Answer

Expert verified
Answer: The values for Kb1 and Kb2 of the oxalate anion are 1.69 脳 10鈦宦孤� and 1.56 脳 10鈦宦光伆, respectively.

Step by step solution

01

Write down the ionization reactions for oxalic acid

Oxalic acid is a diprotic acid which means it can lose two hydrogen ions (protons) in two different reactions. The ionization reactions for oxalic acid are: 1) \( HOOCCOOH \rightleftharpoons H^+ + ^{-}OOCCOOH \) 2) \(^{-}OOCCOOH \rightleftharpoons H^+ + ^{-}OOCCOO^{-}\)
02

Write the expressions for the acid ionization constants K2 and K2,2

Now let's write the expressions for the acidic ionization constants K2 and K2,2: $$K_{2} = \frac{[H^+][^{-}OOCCOOH]}{[HOOCCOOH]}$$ $$K_{2,2} = \frac{[H^+][^{-}OOCCOO^{-}]}{[^{-}OOCCOOH]}$$
03

Write the ionization reactions and expressions for Kb1 and Kb2

Since we are asked to find Kb1 and Kb2 for the oxalate anion, we first need to write the base ionization reactions for the conjugate bases formed after each ionization step: 1) \(^{-}OOCCOOH + H_{2}O \rightleftharpoons HOOCCOOH + OH^{-}\) 2) \(^{-}OOCCOO^{-} + H_{2}O \rightleftharpoons ^{-}OOCCOOH + OH^{-}\) Now, we can write the expressions for Kb1 and Kb2: $$K_{\mathrm{b}_{1}} = \frac{[HOOCCOOH][OH^{-}]}{[^{-}OOCCOOH]}$$ $$K_{\mathrm{b}_{2}} = \frac{[^{-}OOCCOOH][OH^{-}]}{[^{-}OOCCOO^{-}]}$$
04

Use the relationship between Ka and Kb to find Kb1 and Kb2

The relationship between Ka and Kb is given by the following equation: $$K_a \times K_b = K_w$$ where \(K_w\) is the ion product of water (\(1.0 \times 10^{-14}\) at 25掳C) For Kb1, we use the relationship with K2: $$K_{2} \times K_{\mathrm{b}_{1}} = K_w$$ $$K_{\mathrm{b}_{1}} = \frac{K_w}{K_{2}} = \frac{1.0 \times 10^{-14}}{5.9 \times 10^{-2}} = 1.69 \times 10^{-13}$$ For Kb2, we use the relationship with K2,2: $$K_{2,2} \times K_{\mathrm{b}_{2}} = K_w$$ $$K_{\mathrm{b}_{2}} = \frac{K_w}{K_{2,2}} = \frac{1.0 \times 10^{-14}}{6.4 \times 10^{-5}} = 1.56 \times 10^{-10}$$ So, the values for \(K_{\mathrm{b}_{1}}\) and \(K_{\mathrm{b}_{2}}\) of the oxalate anion are \(1.69 \times 10^{-13}\) and \(1.56 \times 10^{-10}\) respectively.

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Key Concepts

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

Acid-Base Equilibria
Understanding acid-base equilibria involves analyzing the balance between hydrogen ions and their conjugate partners in solution. With oxalic acid, this balance is particularly interesting because it's a diprotic acid鈥攁 substance that can donate two protons (H鈦�). When studying its ionization in water, we observe two separate equilibria: one for each proton lost.

This leads to two different sets of equilibrium reactions and constants. These equilibria are crucial because they determine the strength and behavior of the acid in solution. Understanding these principles is vital to grasp how compounds interact in aqueous environments.
Ionization Constants
Ionization constants, denoted as Ka, help us understand how completely an acid donates its protons in a solution. For oxalic acid, which is diprotic, we have two constants: \(K_2\) and \(K_{2,2}\). These values describe the ionization steps where oxalic acid loses its first and second protons, respectively.

\(K_2 = 5.9 \times 10^{-2}\) and \(K_{2,2} = 6.4 \times 10^{-5}\), meaning the first step ionizes more readily than the second. These constants reflect the acid鈥檚 strength in each ionization stage and are pivotal to understanding its acid-base behavior.
  • A higher Ka value indicates a stronger acid, meaning it dissociates more in water.
  • These constants are crucial for calculating pH and understanding reaction dynamics in solutions.
Diprotic Acids
Diprotic acids like oxalic acid can donate two hydrogen ions. This means they have two ionization stages, each with its specific equilibrium constant. In oxalic acid, the first proton is easier to lose than the second due to the molecule's structure.

Because there are two protons, two conjugate bases form. The progression of the ionization can be seen in the two steps:
  • First: \(HOOCCOOH \rightarrow H^+ + ^{-}OOCCOOH\)
  • Second: \(^{-}OOCCOOH \rightarrow H^+ + ^{-}OOCCOO^{-}\)
Each step becomes progressively harder, which reflects in the decreasing ionization constants values. Understanding diprotic acids helps us predict their behavior in biological and chemical systems.
Conjugate Base
When an acid loses a proton, it forms a conjugate base. In the case of oxalic acid, the conjugate bases are \(^{-}OOCCOOH\) and \(^{-}OOCCOO^{-}\).

These species can accept protons, undergoing a reverse reaction from the ionization. The ability of a conjugate base to attract protons relates to its Kb, the base ionization constant:
  • Higher Kb values indicate stronger bases, meaning they are more likely to accept protons.
  • The relationship between Ka of an acid and the Kb of its conjugate base is \(Ka \times Kb = Kw\), where \(Kw\) is the ion product of water \((1.0 \times 10^{-14})\).
This balance underpins much of our understanding of chemical reactions in solutions where acids and bases are involved.

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Most popular questions from this chapter

A \(1.0 M\) aqueous solution of \(\mathrm{HNO}_{3}\) is a much better conductor of electricity than is a \(1.0 M\) solution of \(\mathrm{HNO}_{2}\) Explain why.

When \(1,\) 2-diaminoethane, \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2},\) dissolves in water, the resulting solution is basic. Write the formula of the ionic compound that is formed when hydrochloric acid is added to a solution of 1,2 -diaminoethane.

Painkillers Morphine is an effective painkiller but is also highly addictive. Codeine is a popular prescription painkiller because it is much less addictive than morphine. Codeine contains a basic nitrogen atom that can be protonated to give the conjugate acid of codeine. a. Calculate the \(\mathrm{pH}\) of a \(1.8 \times 10^{-3} M\) solution of morphine if its \(\mathrm{p} K_{\mathrm{b}}=5.79\) b. Calculate the pH of a \(2.7 \times 10^{-4} M\) solution of codeine if the \(\mathrm{p} K_{\mathrm{a}}\) of the conjugate acid is 8.21

The \(K_{\mathrm{a}}\) values of weak acids depend on the solvent in which they dissolve. For example, the \(K_{\mathrm{a}}\) of alanine in aqueous ethanol is less than its \(K_{\mathrm{a}}\) in water. a. In which solvent does alanine ionize more? b. Which is the stronger Bronsted-Lowry base: water or ethanol?

The awful odor of dead fish is due mostly to trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N},\) one of three compounds related to ammonia in which methyl groups replace \(1,2,\) or all 3 of the H atoms in ammonia. a. The \(K_{b}\) of trimethylamine \(\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\right]\) is \(6.5 \times 10^{-5}\) at \(25^{\circ} \mathrm{C} .\) Calculate the \(\mathrm{pH}\) of a \(3.00 \times 10^{-4} M\) solution of trimethylamine. b. The \(K_{b}\) of methylamine \(\left[\left(\mathrm{CH}_{3}\right) \mathrm{NH}_{2}\right]\) is \(4.4 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) Calculate the \(\mathrm{pH}\) of a \(2.88 \times 10^{-3} \mathrm{M}\) solution of methylamine. *c. The \(K_{\mathrm{b}}\) of dimethylamine \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\right]\) is \(5.9 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) What concentration of dimethylamine is needed for the solution to have the same \(\mathrm{pH}\) as the solution in part (b)?
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