Question

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

Short answer

40 mL of 0.1 N NaOH is required to neutralize 10 mL of the oxalic acid solution.

Step-by-step solution

Determine the molarity of oxalic acid

First, calculate the molarity. The molecular weight of oxalic acid dihydrate is approximately 126 g/mol. The number of moles in 6.3 g of oxalic acid dihydrate is \( \frac{6.3}{126} = 0.05 \) moles. Since the solution is made up to 250 mL (0.25 L), the molarity is \( \frac{0.05}{0.25} = 0.2 \text{ M} \).

Calculate normality of oxalic acid

Since oxalic acid is a diprotic acid (it donates two protons per molecule), the normality is twice the molarity. Thus, the normality of the oxalic acid solution is \(0.2 \times 2 = 0.4 \text{ N} \).

Determine equivalent of oxalic acid in 10 mL

The normality of the oxalic acid solution is \(0.4 \text{ N}\). For 10 mL, the equivalent is \( \frac{0.4 \times 10}{1000} = 0.004 \text{ equivalents} \).

Calculate the volume of NaOH solution required

The NaOH solution is \(0.1 \text{ N}\). To find the volume required to neutralize 0.004 equivalents of oxalic acid, use the equation \( \text{Equivalents of acid} = \text{Equivalents of base} \), which gives \(0.004 = 0.1 \times V_{\text{NaOH}} \). Solving for \(V_{\text{NaOH}}\), we get \(V_{\text{NaOH}} = \frac{0.004}{0.1} = 0.04 \text{ L} \), or 40 mL.

Key concepts

Molarity Calculation

Molarity is an important concept in chemistry that refers to the concentration of a solution. It is expressed as the number of moles of solute per liter of solution. To calculate molarity, you need to know the amount of solute (in moles) and the total volume of the solution (in liters).
To calculate molarity for the given exercise with oxalic acid dihydrate:

• The mass of the oxalic acid dihydrate is given as 6.3 g.
• The molecular weight (molar mass) of oxalic acid dihydrate is 126 g/mol.
• Find the number of moles using the formula: \( ext{Moles of oxalic acid} = \frac{\text{mass}}{\text{molar mass}} \)\( = \frac{6.3}{126} = 0.05 \text{ moles} \).
• The total volume of the solution is 250 mL or 0.25 L.
• Calculate the molarity using the formula:\( \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \)\( = \frac{0.05}{0.25} = 0.2 \text{ M} \).

This tells us that the concentration of the solution is 0.2 M, meaning there are 0.2 moles of oxalic acid per liter of solution.

Normality Calculation

Normality is another way to express the concentration of a solution, specifically in terms of the equivalent concentration. It is defined as the number of gram equivalents of solute present in one liter of solution. When dealing with acids, normality can change depending on how many protons (H+) the acid can donate.
In this exercise, oxalic acid is a diprotic acid, meaning it can donate two protons per molecule. Thus, the normality of an acid is calculated as the molarity multiplied by the number of protons it can donate:

• Given that the oxalic acid solution has a molarity of 0.2 M.
• The normality is calculated as:\( \text{Normality} = \text{Molarity} \times n \)where \( n \) is the number of protons, or in this case, 2.\( 0.2 \times 2 = 0.4 \text{ N} \).

The normality of our oxalic acid solution is 0.4 N, indicating there are 0.4 equivalents of acid per liter of solution.

Neutralization Reaction

Neutralization reactions are fundamental in acid-base chemistry, where an acid reacts with a base to produce a salt and water. This process results in the neutralization of the acidic and basic properties of the reactants.
In the scenario of titrating oxalic acid with sodium hydroxide (NaOH):

• The oxalic acid solution has a normality of 0.4 N.
• We calculate the equivalent of oxalic acid present in 10 mL of solution:\( \text{Equivalents} = \frac{\text{Normality} \times \text{Volume (mL)}}{1000} \)\( = \frac{0.4 \times 10}{1000} = 0.004 \text{ equivalents} \).
• Neutralization implies the equivalents of acid equals the equivalents of base needed to react completely.
• The NaOH solution has a normality of 0.1 N, and the equation is set: \( 0.004 = 0.1 \times V_{\text{NaOH}} \).Solving gives \( V_{\text{NaOH}} = \frac{0.004}{0.1} = 0.04 \text{ L} \), or 40 mL.

Thus, 40 mL of NaOH is required for complete neutralization of the oxalic acid in the given solution.

Diprotic Acid Concept

A diprotic acid is capable of donating two protons (hydrogen ions) per molecule to a base during a chemical reaction. This is an important classification in acid-base chemistry, influencing calculations such as normality and equivalent weight. Oxalic acid is a classic example of a diprotic acid.
Key considerations of a diprotic acid:

• It reacts in two steps or stages, each involving the donation of one proton:
• The first dissociation produces a hydrogen ion and a hydrogen oxalate ion.
• The second dissociation releases another hydrogen ion and oxalate ion.
• For titrations, this means the normality is directly related to the number of protons it can release, which is important for calculating neutralization reactions and finding accurate normality values.

Understanding the behavior of diprotic acids helps clarify why we multiply molarity by the number of dissociable protons to find the normality. This ensures precise calculations in acid-base reactions.