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 \(0.2\) 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 POH of the solution is approximately 1.3, which is closest to the option (d) 4.2.

Step 1: Calculate Initial Moles of HCl and NaOH

Firstly, calculate the initial moles of each reactant before mixing. \( Moles = Concentration \times Volume \). For \( HCl \): \( 0.1 \text{ M} \times 50 \text{ L} = 5 \text{ moles} \). For \( NaOH \): \( 0.2 \text{ M} \times 50 \text{ L} = 10 \text{ moles} \).

Step 2: Determine the Reaction Outcome

Write the neutralization reaction: \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \). HCl will react fully due to the lower initial moles. After the reaction, \( 5 \text{ moles} \) of NaOH remain since it is in excess.

Step 3: Calculate the Concentration of Excess NaOH

Calculate the concentration of the leftover NaOH in the solution. The total volume is \( 100 \text{ L} \) after mixing. Thus, \( \text{Concentration of NaOH} = \frac{5 \text{ moles}}{100 \text{ L}} = 0.05 \text{ M} \).

Step 4: Calculate the pOH

The pOH of the solution can be calculated using the formula \( \text{pOH} = -\log_{10}[\text{OH}^-] \). Substitute \( [\text{OH}^-] = 0.05 \text{ M} \): \( \text{pOH} = -\log_{10}(0.05) \approx 1.3 \).

Step 5: Revising Final Calculation

Double-check the log calculation. \( \text{pOH} \approx 1.3 \), suggestion from previous steps confirms processed calculations are correct for resulting figures.

Key Concepts

Neutralization Reaction

A neutralization reaction is a chemical process where an acid and a base react to form water and a salt. This reaction typically involves a proton (\(\text{H}^+\)) from the acid combining with the hydroxide ion (\(\text{OH}^-\)) from the base. The general formula of a neutralization reaction can be expressed as:

• \(\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}\)

In the given exercise, hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to produce sodium chloride (NaCl) and water (H\(_2\)O). This specific reaction can be written as:\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]When performing a neutralization reaction in a lab or an exercise, it is crucial to consider the quantities needed for complete neutralization. In this case, 5 moles of hydrochloric acid completely react with 5 moles of sodium hydroxide (from the excess 10 moles), leaving 5 moles of sodium hydroxide free to affect the concentration of the final solution.

Concentration Calculation

When it comes to concentration calculation, determining the concentration of remaining reactants or products after a reaction is vital. Concentration is essentially the quantity of solute present in a given volume of solution, and is expressed using molarity (\(\text{M}\)).For this exercise, it's crucial to calculate the concentration of sodium hydroxide that remains unreacted in the mixture. Initially, you start with the following quantities:

• 50 litres of 0.1 M hydrochloric acid, resulting in 5 moles.
• 50 litres of 0.2 M sodium hydroxide, which gives you 10 moles.

After the reaction, there are 5 moles of leftover sodium hydroxide, since only the 5 moles of hydrochloric acid will react completely. To find the concentration of this excess NaOH in a new total volume of 100 litres, apply the formula:\[ \text{Concentration} = \frac{\text{Moles of solute}}{\text{Total Volume of solution}} \]Substituting the values, you receive a concentration for NaOH of 0.05 M, which is used later in pH and pOH calculations to determine the nature of the final solution.

pH and pOH Calculations

Understanding pH and pOH is fundamental for assessing the acidity or basicity of a solution. The pH scale ranges from 0 to 14 where:

• 0-7 indicates acidic conditions.
• 7 is neutral.
• 7-14 signifies basic conditions.

Meanwhile, pOH provides details on the concentration of hydroxide ions (\(\text{OH}^-\)) in a solution, complementing the pH. The relationship between pH and pOH is given by the equation:\[ \text{pH} + \text{pOH} = 14 \]In the solution provided, we focus on calculating the pOH since we have the concentration of the excess hydroxide ions, 0.05 M. To obtain the pOH, use the equation:\[ \text{pOH} = -\log_{10}[\text{OH}^-] \]Substituting the concentration value:\[ \text{pOH} = -\log_{10}(0.05) \approx 1.3 \]You can cross-verify these calculations to ensure they're accurate. If needed, the pH can also be easily determined using its relationship with pOH (i.e., pH = 14 - 1.3 = 12.7), confirming the solution remains basic.