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Chapter 8: Problem 126

At \(25^{\circ} \mathrm{C}\) the dissociation constant of a base, \(\mathrm{BOH}\) is \(1.0 \times 10^{-12}\), the concentration of hydroxyl ions \(0.01 \mathrm{M}\) aqueous solution of the base would become (a) \(2.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(1.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}\) (d) \(1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)

Short Answer

Expert verified
The concentration of hydroxyl ions is \(1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\), which is option (d).

Step by step solution

01

Write the Dissociation Equation

The dissociation of a base BOH should be written in the form:\[ \mathrm{BOH} \rightleftharpoons \mathrm{B}^+ + \mathrm{OH}^- \]The dissociation constant \(K_b\) of the base is given as \(1.0 \times 10^{-12}\).
02

Establish Expression for Kb

The expression for the base dissociation constant \(K_b\) is written as:\[ K_b = \frac{[\mathrm{B}^+][\mathrm{OH}^-]}{[\mathrm{BOH}]} \]Assuming the degree of dissociation \(x\), at equilibrium, \([\mathrm{B}^+] = [\mathrm{OH}^-] = x\), and \([\mathrm{BOH}] = 0.01 - x\).
03

Simplifying Assumptions

Since \(K_b\) is very small, we can assume that \(x \ll 0.01\). Thus, \([\mathrm{BOH}] \approx 0.01\).
04

Solve for [OH-]

Substitute into the \(K_b\) expression:\[ 1.0 \times 10^{-12} = \frac{x^2}{0.01} \]So, \(x^2 = 1.0 \times 10^{-14}\), and solving for \(x\):\[ x = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \]Thus, \([\mathrm{OH}^-] = 1.0 \times 10^{-7}\, \mathrm{mol}\, \mathrm{L}^{-1}\).
05

Verify the Answer

The calculated concentration of hydroxyl ions \([\mathrm{OH}^-]\) is \(1.0 \times 10^{-7}\, \mathrm{mol}\, \mathrm{L}^{-1}\). This matches option \((d)\).

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Key Concepts

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

Dissociation of a Base
When a base dissociates, it separates into its constituent ions when placed in water. Consider an example with a generic base, denoted as BOH. When BOH is dissolved in water, it will dissociate to form one hydroxide ion (OH鈦�) and one positive ion (B鈦�). Here's what this looks like:
  • BOH (aqueous) 鈬� B鈦� (aqueous) + OH鈦� (aqueous)
This reaction is reversible and can reach what is known as equilibrium. At equilibrium, the rate of dissociation of the base is equal to the rate of recombination of the ions. The extent of this dissociation is quantified by the base dissociation constant, known as Kb.
The value of Kb is a measure of how strongly a base, like BOH, dissociates in water. A small Kb value implies that the base is weak, meaning it does not dissociate much, keeping the solution less basic.
Concentration of Hydroxyl Ions
The concentration of hydroxyl ions, [OH鈦籡, in a solution of a base is crucial for understanding the solution's alkaline nature. When a base like BOH is placed in water and dissociates, it liberates hydroxide ions. These ions increase the basicity of the solution.
To determine the concentration of OH鈦�, one must know how much the base has dissociated. In chemical equilibrium scenarios, such as with weak bases, we assume simple dissociation to calculate the hydroxide concentration. This involves solving mathematical equations involving Kb. For our exercise, with a given Kb of 1.0 脳 10鈦宦孤�, by establishing a balance between dissociated ions and undissociated base, we found the concentration of OH鈦� to be 1.0 脳 10鈦烩伔 mol/L. This implies that the base is weakly dissociated.
Degree of Dissociation
The degree of dissociation \( x \) represents the fraction of the base that has dissociated into ions in solution. It helps us understand how much of the original compound remains un-ionized. For weak bases, the degree of dissociation is generally small.
In mathematical terms, the degree of dissociation \( x \) is the change in concentration of the species dissociated divided by the initial concentration of the species. The initial concentration of BOH was set at 0.01 M. Applying the value of Kb to the equation \[ K_b = \frac{x^2}{0.01} \], we calculated \( x \) to be 1.0 脳 10鈦烩伔.
  • This \( x \) value represents the concentration of OH鈦� ions at equilibrium.
  • It confirms that the base weakly dissociates.
In our example, x is also approximately equal to [OH鈦籡, underscoring that only a tiny portion of the base dissociates in water.

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Most popular questions from this chapter

Calculate the ratio of \(\mathrm{pH}\) of a solution containing 1 mole of \(\mathrm{CH}_{3} \mathrm{COONa}+1\) mole of HCl per litre to that of a solution containing 1 mole of \(\mathrm{CH}_{3} \mathrm{COONa}+1\) mole of \(\mathrm{CH}_{3} \mathrm{COOH}\) per litre. (a) \(\frac{2}{1}\) (b) \(\frac{1}{2}\) (c) \(\frac{2}{3}\) (d) \(\frac{3}{2}\)

If \(\mathrm{Ksp}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(1.0 \times 10^{-15} \cdot \mathrm{M}\). Find at what \(\mathrm{pH}\) does \(1.0 \times 10^{-3} \cdot \mathrm{M} \mathrm{Al}^{3+}\) precipitate on the addition of buffer of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{NH}_{4} \mathrm{OH}\) solution. (a) 10 (b) \(10.5\) (c) 11 (d) 12

\(500 \mathrm{ml}\) of \(0.2 \mathrm{M} \mathrm{HCl}\) is mixed with \(500 \mathrm{ml}\) of \(0.2 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH} .25 \mathrm{ml}\) of the mixture is titrated with \(0.1\) M \(\mathrm{NaOH}\) solution. By how many units does the \(\mathrm{pH}\) change from the start to the stage when \(\mathrm{HCl}\) is just completely neutralized. \(\mathrm{K}_{\text {, for acetic acid }}=2.0 \times 10^{-5}\). (a) \(3.7\) (b) \(4.4\) (c) \(2.0\) (d) \(3.0\)

The Ka value of formic acid and acetic acid are respectively \(1.77 \times 10^{-4}\) and \(1.75 \times 10^{-5}\). the ratio of the acid strength of \(0.1 \mathrm{~N}\) acids is (a) \(0.1\) (b) \(0.3\) (c) \(3.178\) (d) 100

Find the molar solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in a buffer solution that \(0.10 \mathrm{M}\) in \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(0.10 \mathrm{M}\) in \(\mathrm{NH}_{3} .\) If \(\mathrm{K}_{\mathrm{b}}\) \(\left(\mathrm{NH}_{3}\right)=1.8 \times 10^{-5}\) and \(\mathrm{Ksp}\left[\mathrm{Fe}(\mathrm{OH})_{3}\right]=2.6 \times 10^{-39}\) (a) \(4.458 \times 10^{-25} \mathrm{M}\) (b) \(3.458 \times 10^{-25} \mathrm{M}\) (c) \(2.229 \times 10^{-24} \mathrm{M}\) (d) \(4.458 \times 10^{-22} \mathrm{M}\)
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