Problem 30 - Chemistry in 30 days

Question

An aqueous solution of \(6.3 \mathrm{~g}\) of oxalic acid dihydrate 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) \(40 \mathrm{~mL}\) (b) \(20 \mathrm{~mL}\) (c) \(10 \mathrm{~mL}\) (d) \(4 \mathrm{~mL}\)

Step-by-Step Solution

Answer

The volume of NaOH required is 40 mL.

Step 1: Determine the molar mass of oxalic acid dihydrate

Oxalic acid dihydrate, \ ext{H}_2 ext{C}_2 ext{O}_4 ext{.}2 ext{H}_2 ext{O}\, has a molar mass calculated as follows: \ \(\text{H}_2: 2 \times 1 = 2\, ext{g/mol}\), \(\text{C}_2: 2 \times 12 = 24\, ext{g/mol}\), \(\text{O}_4: 4 \times 16 = 64\, ext{g/mol}\), \(2\text{H}_2 2\text{O}: 4 \times 1 + 2 \times 16 = 36\, ext{g/mol}\) \Adding all these gives \(126\, ext{g/mol}\).

Step 2: Calculate moles of oxalic acid in 250 mL

Given \(6.3 \text{ g}\) of oxalic acid dihydrate in \(250 \text{ mL}\), calculate the moles: \[\text{Moles of } \text{H}_2 \text{C}_2 \text{O}_4 \text{.}2\text{H}_2\text{O} = \frac{6.3 \text{ g}}{126 \text{ g/mol}} = 0.05 \text{ mol}\] This amount is in the entire \(250 \text{ mL}\) solution.

Step 3: Calculate moles of oxalic acid in 10 mL

Since the total moles in \(250 \text{ mL}\) is \(0.05 \text{ mol}\), the moles in \(10 \text{ mL}\) are: \[\text{Moles in } 10 \text{ mL} = \frac{0.05 \times 10}{250} = 0.002 \text{ mol}\]

Step 4: Write the balanced neutralization reaction equation

The reaction between oxalic acid and NaOH is: \\(\text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O} \)This shows that \(1\) mole of oxalic acid reacts with \(2\) moles of NaOH.

Step 5: Calculate the moles of NaOH needed

Using the stoichiometry of the reaction from the previous step, as \( 1 \) mole of \(\text{H}_2\text{C}_2\text{O}_4\) reacts with \(2\) moles of \(\text{NaOH}\), the required moles of \(\text{NaOH}\) are:\[0.002 \text{ mol of } \text{H}_2\text{C}_2\text{O}_4 \times 2 = 0.004 \text{ mol of } \text{NaOH}\]

Step 6: Determine volume of 0.1 N NaOH required

Using the normality equation \(\text{Volume}\, \text{N} = \text{Moles}\), for \(\text{NaOH}\) which is \(0.1\, \text{N}\), solve for the volume:\[\text{Volume of NaOH} = \frac{0.004 \text{ mol}}{0.1 \text{ mol/L}} = 0.04 \text{ L} = 40 \text{ mL}\]

Step 7: Conclusion

Based on the calculations, the volume of \(0.1 \text{ N NaOH}\) required is \(40 \text{ mL}\).

Key Concepts

Oxalic Acid Dihydrate

Oxalic Acid Dihydrate is a crystalline solid that you might encounter under its chemical formula \( \text{H}_2 \text{C}_2 \text{O}_4 \cdot 2 \text{H}_2\text{O} \). This compound is notable for including both oxalic acid and water molecules within its structure. The presence of water molecules makes it a hydrate, which is common in various chemical substances. The water molecules in the dihydrate form are not merely trapped but are part of the crystalline structure, influencing properties like molar mass. To use oxalic acid dihydrate in reactions, it is critical to consider its full molecular weight, which includes both the oxalic acid and the water of hydration. Understanding this composition is crucial for calculating the amounts needed in chemical reactions like neutralization.

Molar Mass Calculation

Calculating molar mass is key in chemistry for determining the quantity of a substance. Oxalic acid dihydrate's molar mass is calculated by summing the atomic masses of its components:

• \( \text{H}_2: \ 2 \times 1 \ \text{g/mol} = 2 \ \text{g/mol} \)
• \( \text{C}_2: \ 2 \times 12 \ \text{g/mol} = 24 \ \text{g/mol} \)
• \( \text{O}_4: \ 4 \times 16 \ \text{g/mol} = 64 \ \text{g/mol} \)
• \( 2 \text{H}_2 \text{O}: \ 2 \times (2 \times 1 \ \text{g/mol} + 16 \ \text{g/mol}) = 36 \ \text{g/mol} \)

Adding these up gives a total molar mass of \(126 \ \text{g/mol} \). This calculation is fundamental for figuring out how much oxalic acid dihydrate you need when preparing solutions or for reactions.

Stoichiometry

Stoichiometry is all about the quantitative relationships in chemical reactions. In our context, it involves using the balanced chemical equation for the neutralization reaction between oxalic acid and sodium hydroxide (NaOH):When oxalic acid (\( \text{H}_2\text{C}_2\text{O}_4 \)) neutralizes NaOH, they react in a 1:2 ratio. This means one mole of oxalic acid requires two moles of NaOH:\[ \text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O} \]Stoichiometric calculations ensure that you add just enough NaOH to completely react with the given amount of oxalic acid. Using stoichiometry helps you predict the outcomes of reactions, prevent wastage of chemicals, and maintain efficiency.

Normality Calculation

Normality is a measure of concentration equivalent to molarity for reactions involving acids and bases. It accounts for the reactive capacity of a given solution. Normality becomes especially useful when dealing with titration and neutralization reactions. For oxalic acid in this reaction, \( 0.1 \text{ N NaOH} \) is used as a standard solution. The normality of a solution tells you how many equivalents of substance react. Given \( 0.002 \text{ mol} \) of oxalic acid in a 10 mL solution from the initial 250 mL, we need \( 0.004 \text{ mol} \) of NaOH for complete reaction, following the equation \( \text{N} \text{V} = \text{moles} \). Thus, \( 0.1 \text{ N} \) NaOH requires 40 mL to react fully with the oxalic acid present. This precise normality calculation ensures the accuracy and success of the neutralization reaction.