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The pKa of a weak acid, HA is \(4.80\). The \(\mathrm{pK}_{\mathrm{b}}\) of a weak base, BOH is \(4.78\). The \(\mathrm{pH}\) of an aqueous solution of the corresponding salt, BA will be a. \(9.58\) b. \(4.79\) c. \(7.01\) d. \(9.22\)

Short Answer

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9.58

Step by step solution

01

Identify the Relationship between pKa and pKb

We start by recalling the relationship between the \( ext{pKa}\) and \( ext{pKb}\) of a conjugate acid-base pair. It is given by the formula: \[ ext{pKa} + ext{pKb} = 14\] Given \( ext{pKa} = 4.80\) and \( ext{pKb} = 4.78\), use this information for further calculation.
02

Calculate the pKw

Since \( ext{pKa} + ext{pKb} = 14\), we can calculate for the ion product of water (\( ext{pKw}\)) at 25掳C: \(4.80 + 4.78 = 9.58\)This means that \( ext{pKw} = 14\).
03

Find the pH Using pKa and pKb

The \( ext{pH}\) of the salt solution can be determined from the average of \( ext{pKa}\) and \( ext{pKw}\): \[ \text{pH} = \frac{1}{2} \left(\text{pKa} + \text{pKw} \right) \] Substituting the known values gives: \[ \text{pH} = \frac{1}{2} \left(4.80 + 4.78 \right) = 9.58 \]
04

Conclude the Correct pH Value

Since the calculated \( ext{pH}\) is 9.58, the correct answer is therefore option (a) 9.58. This indicates a basic solution as expected for a salt formed from a weak acid and a weak base where \( ext{pKb}< ext{pKa}\).

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Key Concepts

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

pKa and pKb relationship
The relationship between pKa and pKb is a core concept in understanding the behavior of acids, bases, and their salts in aqueous solutions. It is based on the principle that the sum of the \(\text{pKa}\) of an acid and the \(\text{pKb}\) of its conjugate base is equal to 14 at 25掳C. This is expressed in the formula: \[\text{pKa} + \text{pKb} = 14\].
This formula arises from the ion product for water (\(\text{pKw}\)), which is 14 at this temperature. It encapsulates the equilibrium nature of weak acids and bases in water. This relationship helps predict whether a salt will form an acidic, neutral, or basic solution when dissolved in water.
For example, if a weak acid has a \(\text{pKa}\) of 4.80 and a weak base has a \(\text{pKb}\) of 4.78, their respective relationship through the formula still holds true, allowing us to proceed with calculations related to salt formation and its pH value in water.
weak acid and weak base
Understanding weak acids and weak bases is essential in chemistry, particularly in the context of salt solutions and their resulting pH. A weak acid partially dissociates in water, releasing fewer hydrogen ions compared to a strong acid. Similarly, a weak base partially accepts hydrogen ions, generating fewer OH鈦 ions than a strong base. Because of this partial dissociation, weak acids and bases have corresponding equilibrium constants: Ka for acids and Kb for bases.
The resulting salts from such weak acids and bases can exhibit either acidic or basic properties depending on the relative strengths (Ka and Kb). When a salt is made from a weak acid and a weak base, its pH depends on the relative magnitudes of these strengths.
For example, if \(\text{pKb}\) of the base is less than the \(\text{pKa}\) of the acid, like in our exercise with \(\text{pKa} = 4.80\) and \(\text{pKb} = 4.78\), the resulting solution tends to be basic. This is due to the stronger tendency of the base part of the salt to accept protons compared to the acid's strength in donating them.
aqueous solution chemistry
Aqueous solution chemistry plays a crucial role in understanding how substances behave when dissolved in water. The properties and interactions of solutes, like salts, in water greatly determine the chemical behavior of solutions.
When a salt derived from a weak acid and weak base is dissolved in water, the resultant solution's pH is influenced by the degree of dissociation. The relative acid and base ionization constants (Ka and Kb) dictate whether the solution will be acidic, neutral, or basic.
In an aqueous solution, the cation resulting from a weak base tends to lower the pH by donating H鈦 ions, while the anion from a weak acid can increase the pH by accepting H鈦 ions. The balance between these two influences鈥攖he Ka of the acid and the Kb of the base鈥攄etermines whether the salt significantly alters the pH from neutrality.
pKw at 25掳C
pKw represents the ion product constant of water, and it is essential for calculations involving acid-base equilibria in water. At 25掳C, \(\text{pKw}\) is always 14, representing the sum of the \(\text{pKa}\) and \(\text{pKb}\) for any acid-base pair. This constant is derived from the equilibrium expression of water dissociation: \[\text{H}_2\text{O} \leftrightarrows \text{H}^+ + \text{OH}^-\].
The importance of knowing \(\text{pKw}\) at 25掳C lies in its use in equilibrium calculations, enabling predictions about the behavior and nature of solutions. This concept becomes particularly valuable when analyzing salts formed from weak acids and bases. By using the relationship \[\text{pKa} + \text{pKb} = \text{pKw}\], we can determine the expected pH level of such solutions.
Hence, understanding \(\text{pKw}\) informs us that even though the products of a weak acid and weak base may not fully dissociate, their influence on solution pH is quantifiable and predictable.

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Most popular questions from this chapter

Cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) undergoes a molecular rearrangement in the presence of \(\mathrm{AlCl}_{3}\) to form methylcyclopentane \(\left(\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right)\), according to the equation \(\mathrm{C}_{6} \mathrm{H}_{12} \rightleftharpoons \mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\) If \(\mathrm{K}_{\mathrm{C}}=0.143\) at \(25^{\circ} \mathrm{C}\) for this reaction, find the equilibrium concentrations of \(\mathrm{C}_{6} \mathrm{H}_{12}\) and \(\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\) if the initial concentrations are \(0.200 \mathrm{M}\) and \(0.100\) M respectively a. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.286\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.016 \mathrm{M}\) b. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.262\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.038 \mathrm{M}\) c. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.186\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.162 \mathrm{M}\) d. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.164\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.621 \mathrm{M}\)

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