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Chapter 12: Problem 410

Find \(\mathrm{K}_{\mathrm{b}}\) and \(\mathrm{pK}_{\mathrm{b}}\) for the acetate ion \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), The ionization constant of \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(\mathrm{K}_{\mathrm{a}}=1.75 \times 10^{-5} ; \mathrm{K}_{\mathrm{W}}=1.00 \times 10^{-14}\)

Short Answer

Expert verified
The ionization constant (\(K_b\)) for the acetate ion (CH3COO-) can be found using the formula \(K_b = \frac{K_w}{K_a}\), where \(K_w = 1.00 \times 10^{-14}\) and \(K_a = 1.75 \times 10^{-5}\). Calculating \(K_b\) gives us \(5.71 \times 10^{-10}\). To find the pKb value, we use the formula pKb = \(-log_{10}(K_b)\), which results in a pKb value of 9.24.

Step by step solution

01

Identify given values

We are given the following values: Ka (ionization constant of acetic acid) = \(1.75 \times 10^{-5}\) Kw (ionization constant of water) = \(1.00 \times 10^{-14}\)
02

Apply the relationship between Ka, Kb, and Kw

Now, we will use the formula Kb = Kw/Ka to find the ionization constant of the acetate ion (Kb): Kb = \(\frac{1.00 \times 10^{-14}}{1.75 \times 10^{-5}}\)
03

Calculate Kb

We'll calculate Kb as follows: Kb = \(\frac{1.00 \times 10^{-14}}{1.75 \times 10^{-5}} = 5.71 \times 10^{-10}\)
04

Calculate pKb

Now, we'll calculate pKb using the formula pKb = -log10(Kb): pKb = \(-log_{10}(5.71 \times 10^{-10})\)
05

Calculate the final pKb value

Finally, we get the pKb as follows: pKb = \(-log_{10}(5.71 \times 10^{-10}) = 9.24\) So, the ionization constant (Kb) for the acetate ion (CH3COO-) is \(5.71 \times 10^{-10}\) and its pKb value is 9.24.

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Key Concepts

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

Acetic Acid
Acetic acid, known chemically as \(\text{CH}_3\text{COOH}\), is a weak acid commonly found in vinegar. It's an essential chemical compound in the food and chemical industries. Acetic acid partially dissociates in water, a characteristic of weak acids, producing hydrogen ions \(\text{H}^+\) and acetate ions \(\text{CH}_3\text{COO}^-\).
This partial dissociation is quantified by the ionization constant \(K_a\), which gives the strength of the acid. A lower \(K_a\) value indicates a weaker acid. For acetic acid, \(K_a = 1.75 \times 10^{-5}\), signifying its relatively weak acidic property.
Ionization Constant of Water
Water naturally undergoes slight ionization into hydroxide \(\text{OH}^-\) and hydronium ions \(\text{H}_3\text{O}^+\). This equilibrium is described by the ionization constant of water, represented as \(K_w\).
The value of \(K_w\) at 25掳C is \(1.00 \times 10^{-14}\), a fundamental constant in aqueous chemistry. This small value highlights the minimal extent of water's self-ionization, yet it plays a significant role in calculations involving acids and bases.
Understanding \(K_w\) is crucial as it helps relate the ionization constants of acids and their conjugate bases.
Kb Calculation
To determine the base ionization constant \(K_b\) of the acetate ion \(\text{CH}_3\text{COO}^-\), we use its relationship with \(K_a\) and \(K_w\). The equation is given by:
  • \( K_b = \frac{K_w}{K_a} \)

In the case of acetate, where \(K_w = 1.00 \times 10^{-14}\) and \(K_a = 1.75 \times 10^{-5}\), we have:
  • \( K_b = \frac{1.00 \times 10^{-14}}{1.75 \times 10^{-5}}\)

This results in \(K_b = 5.71 \times 10^{-10}\). This value describes the extent to which the acetate ion accepts protons in a solution, demonstrating its role as a weak base.
pKb Value Calculation
After finding \(K_b\), we calculate \(pK_b\), a more convenient way to represent base strength on a logarithmic scale. Similar to \(pH\) and \(pK_a\), \(pK_b\) is derived from \(K_b\) using the formula:
  • \( pK_b = -\log_{10}(K_b) \)

For \(K_b = 5.71 \times 10^{-10}\), calculating \(pK_b\) results in:
  • \( pK_b = -\log_{10}(5.71 \times 10^{-10}) = 9.24 \)

Thus, a \(pK_b\) of 9.24 indicates a weak base, as larger \(pK_b\) values correspond to weaker basicity. This calculation helps in comparing the basic strength of different conjugate bases in a consistent manner.

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Most popular questions from this chapter

\(100 \mathrm{ml}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\) is mixed with \(100 \mathrm{ml}\) of \(0.15 \mathrm{M}\) HOAC. \(K_{a}=1.75 \times 10^{-5}\) for acetic acid, what is the \(\mathrm{pH}\) of this mixture?

Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of sodium acetate, \(\mathrm{NaOAc}\). The equilibrium constant for the dissociation of acetic acid is \(1.8 \times 10^{-5}\).

A solution of hydrogen peroxide is \(30 \%\) by weight \(\mathrm{H}_{2} \mathrm{O}_{2}\). Assuming a density of \(1.11 \mathrm{~g} / \mathrm{cm}^{3}\) and a dissociation constant of \(1.0 \times 10^{-12}\) for \(\mathrm{H}_{2} \mathrm{O}_{2}\), what is the \(\mathrm{pH}\) of the solution?
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