Problem 273
Question
The pKa of a weak acid, HA is \(4.80\). The \(\mathrm{pK}_{\mathrm{b}}\) of a weak base, \(\mathrm{BOH}\) is \(4.78\). The \(\mathrm{pH}\) of an aqueous solution of the corresponding salt, BA will be \([2008]\) (a) \(9.58\) (b) \(4.79\) (c) \(7.01\) (d) \(9.22\)
Step-by-Step Solution
Verified Answer
The pH of the salt solution BA is 4.79, which is option (b).
1Step 1: Understanding the Problem
To find the pH of the salt solution formed from the weak acid HA and the weak base BOH, we need to calculate the hydrolysis of the salt BA formed by the acid-base reaction.
2Step 2: Calculate the pKw
The sum of the \( \mathrm{pK}_\mathrm{a} \) of the acid and \( \mathrm{pK}_\mathrm{b} \) of the base is related to the water's ion product constant. So, \( \mathrm{pK}_\mathrm{w} = \mathrm{pK}_\mathrm{a} + \mathrm{pK}_\mathrm{b} = 4.80 + 4.78 = 9.58 \).
3Step 3: Use Hydrolysis Equation for Salt
The pH of a solution of the salt from a weak acid and a weak base is given by the equation: \( \text{pH} = \frac{1}{2}(\text{pK}_\text{w}) \. \) Using \( \text{pK}_\text{w} = 9.58 \), we calculate \( \text{pH} = \frac{1}{2}(9.58) = 4.79 \. \)
4Step 4: Compare with Options
The calculated pH value of 4.79 matches with option (b).
Key Concepts
Hydrolysis of SaltsWeak Acid and Weak BaseIon Product Constant of Water (pKw)
Hydrolysis of Salts
When a salt like BA, formed from a weak acid HA and a weak base BOH, is dissolved in water, it undergoes a reaction known as hydrolysis.
Hydrolysis occurs when ions in the salt react with water to form the original weak acid and weak base. This can slightly change the pH of the solution.
This process involves the breaking apart of the salt into its constituent ions. The balancing act between these ions and water can lead to either an acidic or basic solution, depending on the relative strength of the acid and base. For the salt BA, formed from a weak acid and base:
This process involves the breaking apart of the salt into its constituent ions. The balancing act between these ions and water can lead to either an acidic or basic solution, depending on the relative strength of the acid and base. For the salt BA, formed from a weak acid and base:
- The weak acid's conjugate base (A-) can react with water to produce OH⁻ ions.
- The weak base's conjugate acid (B+) can react with water to produce H⁺ ions.
Weak Acid and Weak Base
A weak acid, like HA, and a weak base, like BOH, only partially dissociate in water.
This partial dissociation results in less free ions available in solution, affecting the overall acidity or basicity of the solution.
Because neither the acid nor the base completely dissociates, the resulting salt solution can have a close to neutral pH. However, due to hydrolysis, the pH will not be exactly neutral.
Because neither the acid nor the base completely dissociates, the resulting salt solution can have a close to neutral pH. However, due to hydrolysis, the pH will not be exactly neutral.
- The extent of dissociation of the weak acid is defined by its dissociation constant, Ka, which is derived from its pKa value.
- Similarly, the dissociation of the weak base is defined by its Kb, derived from its pKb value.
Ion Product Constant of Water (pKw)
The ion product constant of water, denoted as Kw, is a fundamental parameter in understanding acidity, basicity, and the neutral point of water solutions.
In any aqueous solution, water can self-ionize to form H⁺ and OH⁻ ions. The equilibrium constant for this process is represented by Kw, which is always equal to the product of the concentrations of these ions: \[ K_w = [H^+][OH^-] \]
The logarithmic form of this equilibrium constant, pKw, is often used in pH calculations:\[ \text{pK}_w = \text{pK}_a + \text{pK}_b \]
For the given problem, when HA and BOH form the salt BA, we use the pKa of HA and pKb of BOH to determine the pKw. With their sum, we find pKw = 9.58, which facilitates calculating the pH of the resultant solution. This value is indicative of the medium in which the salt is placed and determines how close to neutral the solution is.
In any aqueous solution, water can self-ionize to form H⁺ and OH⁻ ions. The equilibrium constant for this process is represented by Kw, which is always equal to the product of the concentrations of these ions: \[ K_w = [H^+][OH^-] \]
The logarithmic form of this equilibrium constant, pKw, is often used in pH calculations:\[ \text{pK}_w = \text{pK}_a + \text{pK}_b \]
For the given problem, when HA and BOH form the salt BA, we use the pKa of HA and pKb of BOH to determine the pKw. With their sum, we find pKw = 9.58, which facilitates calculating the pH of the resultant solution. This value is indicative of the medium in which the salt is placed and determines how close to neutral the solution is.
Other exercises in this chapter
Problem 270
The solubility product of a salt having general formula \(\mathrm{MX}_{2}\) in water is, \(4 \times 10^{-12}\) \([2005]\) The concentration of \(\mathrm{M}^{2+}
View solution Problem 272
Four species are listed below \([\mathbf{2 0 0 8}]\) (I) \(\mathrm{HCO}_{3}^{-}\) (II) \(\mathrm{H}_{3} \mathrm{O}^{+}\) (III) \(\mathrm{HSO}_{4}^{-}\) (IV) \(\
View solution Problem 274
Solid \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) is gradually dissolved in a \(1.0 \times 10^{-4}\) \(\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) solution
View solution Problem 275
Solubility product of silver bromide is \(5.0 \times 10^{-13} .\) The quantity of potassium bromide (molar mass taken as \(120 \mathrm{~g} \mathrm{~mol}^{-1}\)
View solution