Problem 162 - A Complete Resource Book in Chemistry for JEE Main

Question

The \(\mathrm{pH}\) of a solution containing \(0.1 \mathrm{~mol}\) of \(\mathrm{CH}_{3} \mathrm{COOH}\), \(0.2\) mol of \(\mathrm{CH}_{3} \mathrm{COONa}\) and \(0.05 \mathrm{~mol}\) of \(\mathrm{NaOH}\) in \(1 \mathrm{~L}\). \(\left(\mathrm{pK}_{\mathrm{a}}\right.\) of \(\mathrm{CH}_{3} \mathrm{COOH}=4.74\) and \(\left.\log 5=0.7\right)\) (a) \(4.56\) (b) \(5.44\) (c) \(5.04\) (d) \(3.74\)

Step-by-Step Solution

Answer

The pH of the solution is 5.44.

Step 1: Identify the Type of Solution

This is a buffer solution consisting of acetic acid (\(\mathrm{CH}_3\mathrm{COOH} \)) and its conjugate base acetate ion (from \(\mathrm{CH}_3\mathrm{COONa}\)). We further have the addition of \(\mathrm{NaOH}\), a strong base, which will react with \(\mathrm{CH}_3\mathrm{COOH}\) to form more acetate ions.

Step 2: Calculate the moles of acid and base after reaction

Initially, \(\mathrm{CH}_3\mathrm{COOH}\) = 0.1 mol, \(\mathrm{CH}_3\mathrm{COONa}\) = 0.2 mol, and \(\mathrm{NaOH}\) = 0.05 mol. \(\mathrm{NaOH}\) reacts with \(\mathrm{CH}_3\mathrm{COOH}\):\[\mathrm{NaOH} + \mathrm{CH}_3\mathrm{COOH} \rightarrow \mathrm{CH}_3\mathrm{COONa} + \mathrm{H}_2\mathrm{O} \]After the reaction:- \(\mathrm{CH}_3\mathrm{COOH}\) = 0.1 - 0.05 = 0.05 mol- \(\mathrm{CH}_3\mathrm{COONa}\) = 0.2 + 0.05 = 0.25 mol

Step 3: Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation for a buffer solution is:\[\mathrm{pH} = \mathrm{pK}_a + \log\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)\]Here, \(\mathrm{pK}_a = 4.74\), \([\mathrm{Base}] = 0.25\), \([\mathrm{Acid}] = 0.05\). Calculating the \(\log\) term:\[\log\left(\frac{0.25}{0.05}\right) = \log(5) = 0.7\]

Step 4: Calculate the pH

Substitute the values into the Henderson-Hasselbalch equation:\[\mathrm{pH} = 4.74 + 0.7 = 5.44\]

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is integral to understanding buffer solutions. It is particularly useful for calculating the pH of a buffer solution when you know the concentrations of the acid and its conjugate base. The equation is expressed as:\[\mathrm{pH} = \mathrm{pK}_a + \log\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)\]This formula is derived from the expression for the equilibrium constant of a weak acid. It allows us to understand how adding an acid or base to a solution will affect the pH. This is especially helpful in determining the pH of solutions like those in biological systems. The convenience of the Henderson-Hasselbalch equation is its ability to provide an estimate of pH changes without solving complex equilibrium equations. Always remember, it is crucial to know the pKa of the acid involved, as this constant is fundamental to the calculation.

pH calculation

When calculating the pH of a buffer solution, you can rely on the Henderson-Hasselbalch equation to simplify the process. Begin by identifying the concentrations of the acid and its conjugate base. In our example, the acetic acid (\(\text{CH}_3\text{COOH}\)) is the acid, and the acetate ion (from \(\text{CH}_3\text{COONa}\)) is the conjugate base. After accounting for any reactions, you use these concentrations in the equation.After identifying fully reacted states, place the concentrations into the Henderson-Hasselbalch equation to calculate the pH. In this problem, after reaction, you have:

• Acetic acid: 0.05 M
• Acetate ion: 0.25 M

Substitute these values into the equation, along with the pKa of 4.74, and solve for pH as shown: \(\mathrm{pH} = 4.74 + 0.7 = 5.44\). This process helps in predicting how acidic or basic a solution is, especially important in buffer solutions that are designed to resist changes in pH.

acetic acid buffer

An acetic acid buffer is a type of buffer solution that includes both acetic acid (\(\text{CH}_3\text{COOH}\)) and its conjugate base, the acetate ion (\(\text{CH}_3\text{COO}^-\)). In these solutions, the weak acid and its salt act together to resist changes in pH when small amounts of a strong acid or base are added.The presence of both the acid and its conjugate base allows the solution to neutralize added acids or bases. When a strong base like \(\text{NaOH}\) is added to an acetic acid buffer, it reacts with the acetic acid to form water and more of the acetate ion, thereby minimizing changes in pH. This characteristic is vital in many biological and chemical processes, where maintaining a stable pH is crucial for function and stability.A practical example of an acetic acid buffer can be found in vinegar solutions used in laboratories to maintain a steady pH in reactions.

conjugate acid-base pair

In a buffer solution, the term "conjugate acid-base pair" refers to a weak acid and its corresponding base that differ by just one proton. Understanding conjugate acid-base pairs is key to grasping how buffers work.When a weak acid dissociates, it loses a proton (\(\text{H}^+\)) to form its conjugate base. For acetic acid (\(\text{CH}_3\text{COOH}\)), the dissociated form, acetate ion (\(\text{CH}_3\text{COO}^-\)), serves as the conjugate base. Together, they balance changes in hydrogen ion concentration in a solution. For instance, if an acid is added to the buffer, the conjugate base (\(\text{CH}_3\text{COO}^-\)) will react with the additional \(\text{H}^+\) ions, thereby converting itself back to the weak acid form (\(\text{CH}_3\text{COOH}\)). This provides a stabilizing effect on the pH, showcasing the essential role of conjugate pairs in buffering systems. This mechanism is the foundation for maintaining the pH levels in natural systems, such as blood.