Problem 172 - A Complete Resource Book in Chemistry for JEE Main

Question

Calculate the ratio of \(\mathrm{pH}\) of a solution containing 1 mole of \(\mathrm{CH}_{3} \mathrm{COONa}+1\) mole of \(\mathrm{HCl}\) 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) \(\frac{3}{2}\)

Step-by-Step Solution

Answer

The correct answer is (c) \(\frac{2}{3}\).

Step 1: Identify Reactants in Solution 1

In the first solution, we have 1 mole of CH₃COONa and 1 mole of HCl per litre. CH₃COONa dissociates into CH₃COO⁻ and Na⁺ ions. HCl completely dissociates into H⁺ and Cl⁻ ions.

Step 2: Determine Reaction in Solution 1

H⁺ ions from HCl will react with CH₃COO⁻ ions to form CH₃COOH:\[\text{CH₃COO⁻} + \text{H⁺} \rightarrow \text{CH₃COOH}\]Since both H⁺ and CH₃COO⁻ are equimolar, they completely react, resulting in a 1 molar solution of CH₃COOH.

Step 3: Calculate pH of Solution 1

Solution 1 is now a pure 1 molar solution of CH₃COOH, a weak acid. The pH can be obtained from its dissociation constant, but since the whole concentration is converted to acid, we approximate it as pH = -log [H⁺]. For typical acetic acid, this simplifies to around pH = 2 assuming full conversion.

Step 4: Identify Composition in Solution 2

In the second solution, we have 1 mole of CH₃COONa and 1 mole of CH₃COOH per litre. These two form a buffer solution, where the concentrations of the acetate ion and acetic acid remain in a 1:1 ratio.

Step 5: Calculate pH of Solution 2 using Henderson-Hasselbalch Equation

For a buffer solution of acetic acid and sodium acetate:\[\text{pH} = \text{pK}_a + \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right)\]Since the base (CH₃COO⁻ from CH₃COONa) and the acid (CH₃COOH) have the same concentration, the ratio is 1 and \(\log(1) = 0\). Therefore, \(\text{pH} = \text{pK}_a\) of CH₃COOH, which is around 4.76.

Step 6: Calculate the Ratio of pH Values

The pH of solution 1 is about 2 and the pH of solution 2 is about 4.76. The ratio of their pH values is:\[\frac{2}{4.76} \approx \frac{2}{3}\]Thus, the ratio is closest to \(\frac{2}{3}\).

Key Concepts

pH Calculation

Calculating the pH of a solution is a fundamental concept in chemistry, as it gives us an idea of how acidic or basic the solution is. The "pH" stands for "potential of hydrogen" and is a measure of the concentration of hydrogen ions (\(\text{H}^+\)) in a solution. The formula for calculating pH is:
\[ \text{pH} = -\log[\text{H}^+] \]where [H⁺] is the molarity of hydrogen ions.

For example, if the concentration of hydrogen ions in a solution is 0.01 M, the pH would be calculated as:
\[ \text{pH} = -\log(0.01) = 2 \]
In this exercise, we calculated the pH of two solutions. The first solution was identified as having a lower pH because it became a strong acid solution, and the second solution had a higher pH as it formed a buffer. These calculations help in understanding the comparative acidity and basicity of different solutions.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation is an important tool for calculating the pH of buffer solutions. A buffer solution can resist changes in pH when acids or bases are added to it. It usually consists of a weak acid and its conjugate base.

The equation is expressed as:
\[\text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\]
where:

• \( \text{pK}_a \) is the negative logarithm of the acid dissociation constant, \( K_a \), of the weak acid.
• [Base] is the concentration of the conjugate base.
• [Acid] is the concentration of the weak acid.

In our exercise, the second solution formed a buffer with equal concentrations of acetic acid (CH₃COOH) and acetate ions (CH₃COO⁻), which simplified the expression because the log of 1 is 0. Thus, the pH remained equal to the \( \text{pK}_a \) of acetic acid, which is approximately 4.76.

Acid-Base Reactions

Acid-base reactions are essential in understanding how different chemical species interact in solution. A typical acid-base reaction involves the transfer of protons (\(\text{H}^+\)) from an acid to a base.

For instance, when hydrochloric acid (HCl), a strong acid, reacts with sodium acetate (CH₃COONa), a source of acetate ions, the \(\text{H}^+\) ions from HCl will combine with acetate ions (CH₃COO⁻) to form acetic acid (CH₃COOH):
\[\text{CH₃COO}^- + \text{H}^+ \rightarrow \text{CH₃COOH}\]
In this way, the strong acid completely neutralizes the base, converting it into a weak acid.

This is what occurred in the first solution of our exercise. In contrast, the second solution's acid and base were already in equimolar amounts, serving as a stabilizing buffer without any significant net reaction, which maintained its pH more stable. Understanding these reactions enables chemists to predict and control the outcomes in diverse chemical environments.