Problem 136 - A Complete Resource Book in Chemistry for JEE Main

Question

50 litres of \(0.1 \mathrm{M} \mathrm{HCl}\) are mixed with 50 litres of \(0.2\) \(\mathrm{M} \mathrm{NaOH}\). The POH of the resulting solution is (a) \(12.70\) (b) \(12.34\) (c) \(8.7\) (d) \(4.2\)

Step-by-Step Solution

Answer

The correct pOH is actually 1.30, not matching any listed option.

Step 1: Calculate moles of HCl

To find the number of moles of HCl in the mixture, use the formula for moles: \( ext{moles} = ext{concentration} \times ext{volume} \). For HCl, the concentration is 0.1 M and the volume is 50 L, so the moles of HCl is \( 0.1 \times 50 = 5 \) moles.

Step 2: Calculate moles of NaOH

Similarly, for NaOH, the concentration is 0.2 M, and the volume is 50 L. Thus, the moles of NaOH is \( 0.2 \times 50 = 10 \) moles.

Step 3: Determine the excess reactant

In the reaction between HCl and NaOH, they react in a 1:1 ratio. Since there are fewer moles of HCl (5 moles) than NaOH (10 moles), HCl is the limiting reactant. After reacting, the excess moles of NaOH are \( 10 - 5 = 5 \) moles.

Step 4: Calculate the concentration of excess OH⁻ ions

After the reaction has occurred, the remaining NaOH is dissolved in the total solution volume (100 L). The concentration of OH⁻ in the solution is given by \( \frac{5 \, \text{moles}}{100 \, \text{L}} = 0.05 \mathrm{M} \).

Step 5: Calculate the pOH of the solution

The pOH can be calculated using the formula \( \text{pOH} = -\log[ ext{OH}^-] \), where \([ ext{OH}^-] = 0.05\). Therefore, \( \text{pOH} = -\log(0.05) = 1.30 \).

Step 6: Final calculation: Check against options

The pOH calculated for the resulting solution is 1.30. Since the problem asks for pOH options, note that this seems to be a mistake with the available options. The correct pOH is indeed 1.30 and should be compared to the potential options provided.

Key Concepts

HCl and NaOH Reaction

When hydrochloric acid (HCl) and sodium hydroxide (NaOH) are mixed in a solution, a reaction occurs that is both fundamental and well-understood in chemistry. HCl is a strong acid, and NaOH is a strong base. They undergo a classic acid-base neutralization reaction, resulting in the formation of water (H₂O) and sodium chloride (NaCl).

The balanced chemical equation for the reaction is:

• HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

This equation shows a 1:1 reaction ratio, meaning one mole of HCl reacts with one mole of NaOH. Such reactions are common in chemistry, allowing chemists to calculate and predict the outcome of chemical processes. It is this reaction's simplicity and predictability that makes it a perfect model for understanding acid-base interactions.

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely consumed first, thus determining the extent to which the reaction proceeds. In our exercise, HCl is the limiting reactant because it has fewer moles compared to NaOH. Even though both reactants have an equal volume, their molarity differs, resulting in different amounts of moles.

Here, we used the mole calculation to find that HCl had 5 moles, while NaOH had 10 moles. Since they react in a 1:1 ratio, after the complete consumption of HCl, 5 moles of NaOH remain unreacted. These leftover NaOH molecules are crucial as they contribute to the solution’s basicity, affecting the pOH calculation.

Mole Calculation

The concept of moles is a cornerstone in chemistry, as it allows chemists to count the number of particles participating in a reaction. Moles represent a quantity that lets us move from a microscopic to a macroscopic view. In the provided exercise, calculating moles involved using the formula:

• Moles = Concentration (M) × Volume (L)

For HCl, with a concentration of 0.1 M and a volume of 50 L, the calculation gives us 5 moles. Similarly, for NaOH with a concentration of 0.2 M and a volume of 50 L, we find 10 moles.

Understanding mole calculations is essential for accurately determining the limiting reactant and the amount of reactants remaining after a reaction.

Acid-Base Reaction

An acid-base reaction, typically referred to as neutralization, involves an acid and a base producing a salt and water. These reactions are characterized by the transfer of protons ( H⁺ ) from the acid to the base, resulting in the formation of water. In our exercise, HCl donates a proton to OH⁻ provided by the dissociation of NaOH, forming water.

These reactions are important in many fields, from industrial processes to biological systems, and understanding them is crucial for a wide range of scientific applications. The product of the reaction, in our case, was a basic solution due to the excess of NaOH after the HCl had been used up. This leftover base allowed us to calculate the solution's pOH, emphasizing the importance of knowing how reactions proceed past neutralization in complete solution analysis.