The weight of oxalic acid completely neutrilising 100ml of N/2 alkali will be??

To determine the weight of oxalic acid needed to completely neutralize 100 ml of N/2 alkali, we first need to understand the relationship between the acid and the alkali in terms of their neutralization reaction. Oxalic acid (H2C2O4) is a diprotic acid, meaning it can donate two protons (H+ ions) in a reaction. The alkali in this case is a base, which we can assume to be sodium hydroxide (NaOH) for this explanation.

Understanding the Neutralization Reaction

The neutralization reaction between oxalic acid and sodium hydroxide can be represented as follows:

• H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O

This equation shows that one mole of oxalic acid reacts with two moles of sodium hydroxide. Since we are dealing with a normality of N/2 (which is equivalent to 0.5 N), we can calculate how much oxalic acid is required to neutralize the given volume of alkali.

Calculating the Moles of Alkali

First, we need to find out how many equivalents of the alkali are present in 100 ml of N/2 solution:

• Normality (N) = 0.5 N
• Volume (V) = 100 ml = 0.1 L

The number of equivalents of the alkali can be calculated using the formula:

Equivalents = Normality × Volume

Substituting the values:

Equivalents = 0.5 N × 0.1 L = 0.05 equivalents

Relating Equivalents of Acid to Alkali

Since one mole of oxalic acid provides two equivalents of H+, we can determine how many moles of oxalic acid are needed to neutralize 0.05 equivalents of the alkali:

Moles of oxalic acid = Equivalents of alkali / 2

Thus:

Moles of oxalic acid = 0.05 / 2 = 0.025 moles

Calculating the Mass of Oxalic Acid

Next, we need to find the mass of oxalic acid required. The molar mass of oxalic acid (H2C2O4) is approximately 90 g/mol. Therefore, we can calculate the mass:

Mass = Moles × Molar Mass

Substituting the values:

Mass = 0.025 moles × 90 g/mol = 2.25 grams

Final Answer

To completely neutralize 100 ml of N/2 alkali, you would need approximately 2.25 grams of oxalic acid. This calculation illustrates the stoichiometric relationships in acid-base reactions and how to apply them to practical scenarios.