Problem 216 - A Complete Resource Book in Chemistry for JEE Main

Question

\(50 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) aqueous \(\mathrm{NaOH}\) is mixed with \(10 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) aqueous acetic acid. The conc. \(\left(\mathrm{OH}^{-}\right)\)of resulting solution is (a) \(0.04 \mathrm{M}\) (b) \(0.080 \mathrm{M}\) (c) \(0.0607 \mathrm{M}\) (d) \(0.0667 \mathrm{M}\)

Step-by-Step Solution

Answer

The concentration of \( \mathrm{OH}^- \) is \( 0.0667 \mathrm{M} \) (option d).

Step 1: Calculate Initial Moles of Each Reactant

Calculate the moles of \ \( \mathrm{NaOH} \ \) and acetic acid used in the reaction. \( \mathrm{NaOH} \) has a concentration of \( 0.1 \mathrm{M} \) and a volume of \( 50 \mathrm{~mL} \), making the moles \( 0.1 \times 0.050 = 0.005 \). For the acetic acid, with the same concentration but a volume of \( 10 \mathrm{~mL} \), the moles are \( 0.1 \times 0.010 = 0.001 \).

Step 2: Determine the Reaction Between Reactants

\( \mathrm{NaOH} \) is a strong base and completely dissociates as \( \mathrm{OH}^- \). Acetic acid is a weak acid. When mixed, they undergo a neutralization reaction: \[ \mathrm{OH}^- + \mathrm{CH}_3\mathrm{COOH} \rightarrow \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \] Determine the limiting reactant. Since acetic acid \( (0.001) \) is less than the \( \mathrm{NaOH} \) \( (0.005) \), it will be completely neutralized.

Step 3: Calculate Remaining Moles of \( \mathrm{OH}^- \)

After the neutralization, all 0.001 moles of acetic acid have reacted with an equivalent number of moles of \( \mathrm{OH}^- \). Thus, the remaining moles of \( \mathrm{OH}^- \) will be \( 0.005 - 0.001 = 0.004 \) moles.

Step 4: Calculate Total Volume of the Solution

The total volume of the mixed solution is the sum of the volumes of the reactants. This is \( 50 \mathrm{~mL} + 10 \mathrm{~mL} = 60 \mathrm{~mL} \), or \( 0.060 \mathrm{~L} \).

Step 5: Calculate Final \( \mathrm{OH}^- \) Concentration

With \( 0.004 \) moles of \( \mathrm{OH}^- \) remaining and a total solution volume of \( 0.060 \mathrm{~L} \), the concentration is: \[ \text{Concentration of } \mathrm{OH}^- = \frac{0.004}{0.060} = 0.0667 \mathrm{M} \]

Key Concepts

Understanding Acid-Base Reactions

Acid-base reactions are foundational concepts in chemistry where acids and bases react to form water and salt. Acids are substances that can donate a proton (\( ext{H}^+\)), while bases are substances that can accept a proton or donate a hydroxide ion (\( ext{OH}^-\)). In a typical acid-base reaction, an acid will react with a base to neutralize each other, often resulting in the formation of water.
In the reaction between sodium hydroxide (\( ext{NaOH}\)) and acetic acid (\( ext{CH}_3 ext{COOH}\)), \( ext{NaOH}\) acts as a strong base, meaning it dissociates completely in water to produce \( ext{OH}^-\) ions. Acetic acid, being a weak acid, only partially dissociates and is generally given the formula \( ext{HA}\) when writing reactions.

• Understanding the strength of acids and bases helps predict the extent of reactions and whether they'll go to completion.
• Knowledge of how acids and bases react is crucial for tasks like titration or determining \( ext{pH}\) in solutions.

Molarity Calculations Explained

Molarity is a measurement of the concentration of a solution. It is defined as the number of moles of a solute per liter of solution, commonly denoted as \( ext{M}\). Knowing how to perform molarity calculations is essential for interpreting chemical reactions and mixtures.
To determine molarity, use the formula:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]In the example exercise, we calculated the moles of \( ext{NaOH}\) and \( ext{CH}_3 ext{COOH}\) separately using their given volumes and concentrations. For \( ext{NaOH}\), having 0.005 moles in a 0.050 L solution, the molarity initially was \(0.1 ext{ M}\).

• Calculating molarity helps ensure precise formulations in reactions.
• Molarity is used not only to describe how much solute is dissolved but also predicts the outcome of reactions.

The Process of Neutralization Reactions

Neutralization reactions occur when an acid and a base react to form water and a salt. In these reactions, the acidic properties of the acid and the basic properties of the base are effectively cancelled out, leading to neutral substances.
In the case of mixing 50 mL of \(0.1 ext{ M} ext{ NaOH}\) with 10 mL of \(0.1 ext{ M} ext{ acetic acid}\), the acetic acid is completely neutralized by the \( ext{NaOH}\). This is a perfect illustration of a neutralization reaction where the strong base \( ext{NaOH}\) reacts with the weaker acid \( ext{CH}_3 ext{COOH}\).

• The moles of \( ext{OH}^-\) remaining after neutralization show how much base is left unreacted.
• Such reactions help understand the \( ext{pH}\) control in solutions and find practical uses in fields such as medicine and water purification.