Problem 176 - A Complete Resource Book in Chemistry for JEE Main

Question

A buffer solution is prepared by mixing \(20 \mathrm{ml}\) of \(0.1 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(40 \mathrm{ml}\) of \(0.5 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa}\) and then diluted by adding \(100 \mathrm{ml}\) of distilled water. The \(\mathrm{pH}\) of resulting buffer solution is (Given \(\mathrm{pKa} \mathrm{CH}_{3} \mathrm{COOH}=4.76\) ) (a) \(5.76\) (b) \(4.67\) (c) \(3.48\) (d) \(5.9\)

Step-by-Step Solution

Answer

The pH of the resulting buffer solution is 5.76 (option a).

Step 1: Calculate Moles of Acid and Base

First, find the moles of the weak acid (\( \mathrm{CH_3COOH} \)) and the conjugate base (\( \mathrm{CH_3COONa} \)). \For \( \mathrm{CH_3COOH} \):\[ \text{moles of } \mathrm{CH_3COOH} = \text{Volume} \times \text{Concentration} = 20\, \mathrm{ml} \times 0.1\, \mathrm{M} = 2\, \mathrm{mmol} \]For \( \mathrm{\ CH_3COONa} \):\[ \text{moles of } \mathrm{CH_3COONa} = \text{Volume} \times \text{Concentration} = 40\, \mathrm{ml} \times 0.5\, \mathrm{M} = 20\, \mathrm{mmol} \]

Step 2: Determine Total Volume of the Solution

The solution is diluted with 100 ml of distilled water after mixing the acid and base. Calculate the total volume \\[ \text{Total volume} = 20\, \mathrm{ml} + 40\, \mathrm{ml} + 100\, \mathrm{ml} = 160\, \mathrm{ml} \]

Step 3: Find Concentrations of Acid and Base in the Buffer

Calculate the concentrations of \( \mathrm{CH_3COOH} \) and \( \mathrm{CH_3COONa} \) in the final solution. For \( \mathrm{CH_3COOH} \):\[ \text{Concentration of } \mathrm{CH_3COOH} = \frac{2\, \mathrm{mmol}}{160\, \mathrm{ml}} = \frac{1}{80}\, \mathrm{M} \] \For \( \mathrm{CH_3COONa} \):\[ \text{Concentration of } \mathrm{CH_3COONa} = \frac{20\, \mathrm{mmol}}{160\, \mathrm{ml}} = \frac{1}{8}\, \mathrm{M} \]

Step 4: Apply Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates the pH to the \( \mathrm{pK_a} \) and the ratio of the concentrations of the base (\( \mathrm{CH_3COONa} \)) and acid (\( \mathrm{CH_3COOH} \)). Using the equation: \[ \mathrm{pH} = \mathrm{pKa} + \log\left(\frac{\text{concentration of base}}{\text{concentration of acid}}\right) \] \Substitute the values: \[ \mathrm{pH} = 4.76 + \log\left(\frac{1/8}{1/80}\right) \] \\[ \mathrm{pH} = 4.76 + \log(10) \] \\[ \mathrm{pH} = 4.76 + 1 = 5.76 \]

Step 5: Identify the Correct Option

The calculated pH is 5.76, which corresponds to option (a).

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is an extremely useful tool in chemistry for understanding the relationship between pH, pKa, and the concentrations of acid and base in a buffer solution. This equation is expressed as:
\[\mathrm{pH} = \mathrm{pKa} + \log\left(\frac{\text{concentration of base}}{\text{concentration of acid}}\right)\]It helps in estimating the pH of a buffer solution by considering the equilibrium between the weak acid and its conjugate base. The pKa, or the acid dissociation constant, is a value that indicates the acid's tendency to donate protons.

• A lower pKa value means a stronger acid.
• The converse is true; a higher pKa indicates a weaker acid.

By determining the ratio of the concentrations of the base (the salt of the weak acid) to the acid, the equation gives us a simple way to calculate the buffer's pH. In our example, the calculated pH using this equation indicates that the buffer solution has a pH of 5.76.

acetic acid and sodium acetate buffer

Acetic acid and sodium acetate are a classic pair for creating a buffer solution. This buffer system is useful because it operates well around the pH of 4.76, close to the pKa of acetic acid. A buffer solution resists changes in pH when small amounts of acids or bases are added, making it incredibly valuable in many chemical and biological processes.
Here's how this specific buffer pair works:

• **Acetic Acid**: A weak acid represented by \(\mathrm{CH_3COOH}\). It can donate protons to the solution.
• **Sodium Acetate**: The salt of acetic acid, represented by \(\mathrm{CH_3COONa}\). It acts as a source of acetate ions, the conjugate base that can accept protons.

Together, acetic acid and sodium acetate maintain a stable pH by neutralizing added acids or bases. The amount of these components can affect the buffer's capacity and pH range. In our exercise, we mixed specific volumes and concentrations of these compounds to create the buffer.

dilution effect on pH

When dealing with buffers, it's important to understand how dilution affects the pH. Typically, diluting a buffer involves adding more solvent to the solution, which in our exercise was water. An interesting property of buffers is that their pH stays relatively constant even when some amount of dilution occurs.
However, it's critical to remember that:

• While the overall concentration of ions decreases, the ratio of acid to base concentrations remains constant.
• This ratio is what primarily determines the buffer's pH, as shown in the Henderson-Hasselbalch equation.

Therefore, dilution doesn't have a significant effect on the pH of a well-prepared buffer within its capacity limits. In the given exercise, after adding 100 ml of distilled water, the buffer was diluted; however, the pH remained at 5.76, highlighting the buffer's effectiveness in maintaining its pH.