Problem 15 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(\mathrm{p} K_{\mathrm{a}}\) of a weak acid (HA) and \(\mathrm{p} K_{\mathrm{b}}\) of a weak base \((\mathrm{BOH})\) are \(3.2\) and 3.4, respectively. The \(\mathrm{pH}\) of their salt (AB) solution is (a) \(7.2\) (b) \(6.9\) (c) \(7.0\) (d) \(1.0\)

Step-by-Step Solution

Answer

The pH of the salt solution is 6.9.

Step 1: Understand the Relationship between pKa and pKb

For a salt formed by a weak acid and a weak base, the relation between \( {pH}\), \( {pK}_{ ext{a}} \) and \( {pK}_{ ext{b}} \) is given by the formula: \[ \text{pH} = 7 + \frac{1}{2}(\text{pK}_{\text{a}} - \text{pK}_{\text{b}}) \]

Step 2: Substitute Values into the Formula

Plug in the given values of \( {pK}_{ ext{a}} \) = 3.2 and \( {pK}_{ ext{b}} \) = 3.4 into the formula: \[ \text{pH} = 7 + \frac{1}{2}(3.2 - 3.4) \]

Step 3: Calculate the Difference Between pKa and pKb

Subtract the value of \( {pK}_{ ext{b}} \) from \( {pK}_{ ext{a}} \): \( 3.2 - 3.4 = -0.2 \).

Step 4: Compute the pH

Substitute the difference into the formula: \[ \text{pH} = 7 + \frac{1}{2}(-0.2) \] which simplifies to \[ \text{pH} = 7 - 0.1 = 6.9 \].

Step 5: Conclude the Solution

The calculated \( {pH} \) of the salt solution is \( 6.9 \), matching option (b).

Key Concepts

Weak Acid

A weak acid is a type of acid that doesn't completely dissociate in water. This means that only a small fraction of its molecules release hydrogen ions (H⁺) into the solution. Because of this partial dissociation, weak acids do not donate protons (H⁺) as readily as strong acids do.
Examples of weak acids include:

• Acetic acid (CH₃COOH)
• Formic acid (HCOOH)
• Hydrofluoric acid (HF)

Understanding weak acids is important in pH calculation because they influence the concentration of hydrogen ions in a solution, affecting acidity. When considering a weak acid in conjunction with a weak base, their equilibria play a significant role in determining pH of the solution.

Weak Base

Just like a weak acid, a weak base partially dissociates in solution. A weak base will accept only some hydrogen ions, which makes it less effective in increasing the OH⁻ concentration than a strong base. Weak bases are typically organic compounds or metal hydroxides that do not fully ionize in water.
Common examples of weak bases include:

• Ammonia (NH₃)
• Methylamine (CH₃NH₂)
• Pyridine (C₅H₅N)

This partial ionization behavior of weak bases must be considered, just like with weak acids, when calculating the pH of solutions involving salts derived from weak acids and bases. The interactions between the weak acid and base's properties define the resulting pH balance in a mixture or solution.

pKa

The term pKa refers to the negative logarithm of the acid dissociation constant (Ka) of a solution. It measures the strength of an acid in a solution. The smaller the pKa value, the stronger the acid is, as it indicates a greater tendency to release hydrogen ions.

The formula used to calculate pKa is:

\[pKa = -\log_{10}(Ka)\]

This concept is crucial when analyzing weak acids because it helps illustrate how readily an acid donates its hydrogen ions. In the context of the pH of a salt formed by a weak acid and a weak base, pKa helps dictate how the salt's cation will behave in water, impacting the overall pH.

pKb

Much like pKa for acids, pKb is the logarithmic measure associated with a base's strength, specifically of a weak base. pKb is calculated from the base ionization constant (Kb), and it provides insight into how readily a base can accept protons.

The formula for calculating pKb is:

\[pKb = -\log_{10}(Kb)\]

The lower the pKb, the stronger the base, indicating a higher tendency to accept protons. When a weak acid and weak base form a salt, the pKb plays a pivotal role alongside pKa to determine the pH of the solution. The interplay between these values allows chemists to predict the resulting pH in cases where both components are involved in salt formation.