Question

\(0.3 \mathrm{~g}\) of an acid is neutralized by \(40 \mathrm{~cm}^{3}\) of \(0.125 \mathrm{~N}\) \(\mathrm{NaOH}\). Equivalent mass of the acid is a. 20 b. 60 c. 30 d. 46

Short answer

The equivalent mass of the acid is 60.

Step-by-step solution

Understand the Relation Between Normality and Equivalent Mass

The normality of a solution is defined as the number of gram equivalents of solute per liter of solution. When an acid is neutralized by a base, the relation can be expressed as \( \text{Equivalent Mass of Acid} = \frac{\text{Mass of Acid (g)}}{\text{Volume of Base (L)} \times \text{Normality of Base}} \).

Convert Units for Calculation

First, convert the volume of the sodium hydroxide solution from cubic centimeters to liters. Given that \( 40 \mathrm{~cm}^{3} = 0.040 \mathrm{~L} \), use this value in the calculations.

Calculate the Equivalent Mass of the Acid

Substitute the known values into the formula: Equivalent Mass of Acid = \( \frac{0.3 \mathrm{~g}}{0.040 \mathrm{~L} \times 0.125 \mathrm{~N}} \). Calculate this to determine the equivalent mass of the acid.

Perform the Calculation

Calculate by solving: \( \frac{0.3}{0.040 \times 0.125} = \frac{0.3}{0.005} = 60 \).

Key concepts

Equivalent Mass

The concept of equivalent mass is essential in chemistry, particularly when dealing with neutralization reactions. Equivalent mass refers to the mass of a substance that will react with or displace one mole of hydrogen ions (\(1\,\text{mol}\,\text{H}^+\)) in a chemical reaction.

In the context of acids and bases, the equivalent mass of an acid can be found by using the formula: \[\text{Equivalent Mass of Acid} = \frac{\text{Mass of Acid (g)}}{\text{Volume of Base (L)} \times \text{Normality of Base}}\]This formula helps determine how much of an acid is needed to react completely with a base, allowing for precise calculations in laboratory settings. Understanding this helps in analyzing and solving chemical problems effectively. Make sure to convert all your measurements into consistent units to avoid calculation errors.

When the given problem states that 0.3 g of acid is neutralized by the sodium hydroxide solution, this formula is applied to find the acid's equivalent mass, which in this case is calculated as 60.

Normality

Normality is a way of expressing the concentration of a solution, often used in acid-base chemistry and titrations. It represents the gram equivalent weight of a solute per liter of solution. The normality (\(N\)) of a solution is related to the concept of equivalents; it indicates how many equivalents of the solute are available per liter.

Normality can be particularly useful in neutralization reactions. For example, in acid-base reactions, where substances combine based on their equivalent amounts according to their reaction stoichiometry, understanding normality helps you determine how much of one reactant will react with a known quantity of another.

In the given exercise, the normality of the sodium hydroxide solution is 0.125 N. This indicates that there are 0.125 equivalents of sodium hydroxide per liter of solution. When calculating equivalent masses, normality can directly influence the accuracy of your results.

Stoichiometry

Stoichiometry fundamentally deals with the calculation of relative quantities of reactants and products in chemical reactions. It allows chemists to predict how much product will form from certain amounts of reactants, or conversely, how much of the reactants you need to form a desired amount of product.

In neutralization reactions, stoichiometry is crucial to balance the equation based on equivalent quantities of acids and bases. This balance is achieved using the relationship between moles, mass, and equivalent masses, aiding in precise quantitative understanding in chemical analysis.

With the original exercise problem, stoichiometry is used through the equivalence equation that balances the acid's mass, the solution's volume, and the normality of the base. Understanding stoichiometry ensures that you are correctly predicting and quantifying chemical interactions, leading to successful experiment outcomes.