Question

An aqueous solution of \(6.3 \mathrm{~g}\) oxalic acid dehydrate is made up to \(250 \mathrm{ml}\). The volume of \(0.1 \mathrm{~N} \mathrm{NaOH}\) required to completely neutralize \(10 \mathrm{~mL}\) of this solution is: (a) \(4 \mathrm{~mL}\) (b) \(20 \mathrm{~mL}\) (c) \(40 \mathrm{~mL}\) (d) \(60 \mathrm{~mL}\)

Short answer

The volume of NaOH required is 20 mL.

Step-by-step solution

Find the Number of Moles of Oxalic Acid

The molecular weight of oxalic acid dihydrate \( \text{C}_2\text{H}_2\text{O}_4 \cdot 2\text{H}_2\text{O} \) is \( 126 \text{ g/mol} \). Number of moles \( = \frac{6.3 \text{ g}}{126 \text{ g/mol}} = 0.05 \text{ moles} \).

Find the Concentration of Oxalic Acid Solution

Oxalic acid solution is made up to \(250 \text{ mL} = 0.25 \text{ L} \). The molarity \( \text{M} \) is given by \( \frac{\text{moles}}{\text{liters}} = \frac{0.05 \text{ moles}}{0.25 \text{ L}} = 0.2 \text{ M} \).

Calculate the Milliequivalents of Oxalic Acid in 10 mL

The normality \( \text{N} \) of the oxalic acid solution is equal to its molarity \( \text{M} \) because 1 mole of oxalic acid provides 2 equivalents. For 10 mL, the milliequivalents \( = \text{N} \times \text{volume (mL)} = 0.2 \times 10 = 2 \text{ meq} \).

Calculate the Volume of NaOH Solution Required

The normality of NaOH is \(0.1 \text{ N}\), so the volume needed is given by \( \frac{\text{milliequivalents of oxalic acid}}{\text{normality of NaOH}} = \frac{2 \text{ meq}}{0.1 \text{ N}} = 20 \text{ mL} \).

Key concepts

Molarity Calculation

Molarity is a fundamental concept in chemistry that denotes the concentration of a solution. It tells us how many moles of a substance are present per liter of solution. For our problem, we start by determining the number of moles of oxalic acid dihydrate, which has a molecular weight of 126 g/mol.
First, we find the moles by using the mass provided:

• Number of moles = \( \frac{6.3 \text{ g}}{126 \text{ g/mol}} = 0.05 \text{ moles} \)

Next, we calculate the molarity \( M \) of the solution by dividing the number of moles by the volume of the solution in liters.

• Molarity \( M = \frac{0.05 \text{ moles}}{0.25 \text{ L}} = 0.2 \text{ M} \)

This calculation gives us a concentration of 0.2 M, which is essential for the subsequent steps in the problem.

Normality and Equivalents

Normality is another way to express the concentration of a solution, especially in the context of acid-base reactions or redox reactions. It measures the equivalent concentration in a given volume. Since oxalic acid is a dibasic acid, each mole provides two equivalents in an acid-base reaction.
For our oxalic acid solution:

• The normality \( N \) of oxalic acid is the same as its molarity times the number of equivalents per mole.
• Therefore, \( N = 0.2 \text{ M} \times 2 = 0.4 \text{ N} \).

When we consider a specific volume, like the 10 mL of the solution in the exercise:

• The milliequivalents (meq) are calculated as: \( \text{meq} = \text{N} \times \text{volume in mL} = 0.2 \times 10 = 2 \text{ meq} \).

Understanding to calculate normality and equivalents helps to accurately determine how much base is needed for neutralization.

Volumetric Analysis

Volumetric analysis is a practical technique used to determine the concentration of a substance by reacting it with a standard solution. In our case, oxalic acid is titrated with sodium hydroxide (NaOH).The concept of milliequivalents is crucial here.

• In this scenario, the oxalic acid provides 2 meq in the 10 mL of solution mentioned.
• The normality of NaOH is given as 0.1 N.

To find out how much NaOH is required to completely neutralize the oxalic acid:

• We use the formula: \( \text{volume of NaOH} = \frac{\text{milliequivalents of acid}}{\text{normality of NaOH}} = \frac{2 \text{ meq}}{0.1 \text{ N}} = 20 \text{ mL} \).

This type of analysis is widely used in labs to achieve precise chemical measurements and is critical for industrial standards.