Problem 217 - A Complete Resource Book in Chemistry for JEE Main

Question

\(50 \mathrm{~mL}\) of \(0.2 \mathrm{M}\) aqueous \(\mathrm{CH}_{3} \mathrm{COOH}\) is mixed with \(50 \mathrm{~mL}\) of \(0.2 \mathrm{M}\) aqueous KOH solution. The \(\mathrm{pH}\) of resulting solution is \(\left(\mathrm{pK}_{2}\right.\) of acetic acid is \(4.7\) ) (a) \(7.0\) (b) \(9.35\) (c) \(8.85\) (d) \(6.05\)

Step-by-Step Solution

Answer

After mixing, the pH of the solution is approximately 9.35.

Step 1: Identify the Reactants and Their Concentrations

We are given two solutions: 50 mL of 0.2 M acetic acid (\(\mathrm{CH}_3\mathrm{COOH}\)) and 50 mL of 0.2 M potassium hydroxide (KOH). Both solutions have the same volume and molarity.

Step 2: Determine the Complete Reaction Equation

The reaction between acetic acid and KOH is a neutralization reaction: \[\mathrm{CH}_3\mathrm{COOH} + \mathrm{KOH} \rightarrow \mathrm{CH}_3\mathrm{COOK} + \mathrm{H}_2\mathrm{O}\]Acetic acid reacts with KOH to form potassium acetate and water.

Step 3: Calculate the Moles of Reactants

The moles of acetic acid and KOH are both calculated using the formula \( \text{moles} = \text{volume} \times \text{molarity} \).For both: \[0.050 \, \text{L} \times 0.2 \, \text{mol/L} = 0.01 \, \text{mol}\]Thus, we have 0.01 mol of \(\mathrm{CH}_3\mathrm{COOH}\) and 0.01 mol of KOH.

Step 4: Determine the Reaction Completion and Formed Product

Since both reactants are present in equal moles and completely react, the reaction will go to completion forming 0.01 mol of potassium acetate (\(\mathrm{CH}_3\mathrm{COOK}\)).

Step 5: Calculate the pH of the Resulting Solution

Potassium acetate (\(\mathrm{CH}_3\mathrm{COOK}\)) is a salt formed from a weak acid and a strong base. It can slightly hydrolyze in water:\[\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \leftrightarrow \mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^-\]This shifts the pH towards the basic side. The pKa of acetic acid is given as 4.7, so the \(\mathrm{pK}_b\) of acetate is: \[\mathrm{pK}_b = 14 - 4.7 = 9.3\]The approximate pH for the buffer can thus be estimated as higher than 7 due to its nature as a basic salt.

Key Concepts

Neutralization Reaction

In a neutralization reaction, an acid and a base react to form water and a salt. It is a type of double displacement reaction. When acetic acid (a weak acid) reacts with potassium hydroxide (a strong base), a neutralization reaction occurs. The formula for this reaction is:

• \[\mathrm{CH}_3\mathrm{COOH} + \mathrm{KOH} \rightarrow \mathrm{CH}_3\mathrm{COOK} + \mathrm{H}_2\mathrm{O}\]

The result is the formation of water and potassium acetate, a salt. Since both reactants are present in equal molarity and volume, the reaction completely converts them to the products. Neutralization is key in balancing opposing chemical properties. In this case, the acidic property of the acetic acid and the basic property of KOH cancel each other out, resulting in a mostly neutral product, apart from the basic nature of the salt formed.

Buffer Solutions

Buffer solutions resist drastic changes in pH upon the addition of small amounts of acid or base. In this exercise, we deal with potassium acetate, which forms when acetic acid reacts with KOH. This salt can play a role in forming a buffer system:

• Since acetic acid is a weak acid and its conjugate base (acetate) forms part of the products, it can contribute to a buffer system.
• In a buffer system, the acid and its conjugate base maintain the pH. The acetate ion from potassium acetate can react with water to slightly form acetic acid again, affecting the overall pH of the solution.

Buffer solutions are crucial because they maintain a stable pH in biological and chemical systems, allowing reactions to proceed under optimal conditions. In our example, even though potassium acetate isn't a complete buffer by itself, it shows buffer-like behavior because of its origin from a weak acid.

Acid-Base Reaction

An acid-base reaction involves the transfer of protons (hydrogen ions) between reactants. In our exercise, acetic acid donates a proton to the hydroxide ion from potassium hydroxide:

• Acetic acid (\[\mathrm{CH}_3\mathrm{COOH}\]) acts as the proton donor (acid).
• Potassium hydroxide (\[\mathrm{KOH}\]) provides the hydroxide ion, which acts as the proton acceptor (base).

The transfer of the proton facilitates the conversion into water and the salt, potassium acetate. This reaction is a quintessential example of how substances interact chemically via acid-base principles. Understanding these reactions is fundamental in both chemistry and biochemistry, as they occur frequently in nature, industry, and laboratories.

Acetic Acid

Acetic acid is an organic compound with the chemical formula \[\mathrm{CH}_3\mathrm{COOH}\]. It is known for being a weak acid, meaning it doesn't completely ionize in water. The ionization can be represented as:

• \[\mathrm{CH}_3\mathrm{COOH} \leftrightarrow \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+\]

The partial ionization is crucial in reactions like the one in the exercise because it determines the extent of the acid's capability to donate protons. This affects how it neutralizes bases such as KOH. Acetic acid is commonly known as vinegar's active ingredient, and it plays a significant role in food preservation, flavoring, and as a chemical reagent. Understanding its behavior and role in reactions deepens knowledge of acidity and its practical applications.

Potassium Acetate

Potassium acetate (\[\mathrm{CH}_3\mathrm{COOK}\]) is a salt that results from the neutralization of acetic acid by potassium hydroxide. It's noteworthy because it participates in hydrolysis reactions in aqueous solutions:

• When dissolved, the acetate ion (\[\mathrm{CH}_3\mathrm{COO}^-\]) can react with water, converting slightly back to acetic acid and hydroxide ions: \[\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \leftrightarrow \mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^-\]
• This raises the pH of the solution, making it slightly basic.

In practical applications, potassium acetate is used as a preservative and buffering agent. Its ability to affect the pH of solutions is leveraged in various chemical processes and formulations.