Question

The number of grams of a dibasic acid (molecular weight 200 ) present is \(100 \mathrm{~mL}\) of its aqueous solution to give decinormal strength is: (a) \(1 \mathrm{~g}\) (b) \(2 \mathrm{~g}\) (c) \(3 \mathrm{~g}\) (d) \(4 \mathrm{~g}\)

Short answer

The correct answer is (a) 1 g.

Step-by-step solution

Understand the Problem

We need to find out how many grams of a dibasic acid with a molecular weight of 200 are required to create a decinormal (0.1 N) solution in 100 mL of water.

Determine Normality Formula

The normality (N) formula is given by \( N = \frac{\text{weight of solute in grams}}{\text{equivalent weight of solute} \times \text{volume in liters}} \). We aim to find the weight of the solute.

Calculate Equivalent Weight

For a dibasic acid, the equivalent weight equals the molecular weight divided by the number of replaceable hydrogen ions. Here, it is \( \frac{200}{2} = 100 \).

Set Up the Equation

Given that the desired solution has a normality of 0.1 and a volume of 0.1 L (100 mL), we substitute into the normality equation: \[ 0.1 = \frac{\text{weight of solute}}{100 \times 0.1} \].

Solve for Weight of Solute

Rearrange the equation to find the weight of the solute: \[ \text{weight of solute} = 0.1 \times 100 \times 0.1 = 1 \text{ gram} \]. This means 1 gram of the dibasic acid is needed.

Key concepts

Normality

In chemistry, normality is a way of expressing the concentration of a solution. It is particularly useful when dealing with reactions involving acids and bases because it accounts for the reactive capacity of different solutions. Normality, symbolized as "N," is defined as the number of gram equivalents of a substance dissolved in a liter of solution.

• Normality takes into account not just the amount of substance, but also its chemical nature.
• The formula used to calculate normality is: \( N = \frac{\text{weight of solute in grams}}{\text{equivalent weight of solute} \times \text{volume in liters}} \).

An essential point to note is that normality varies with the type of reaction because it considers the reactiveness or active moles of the compound in the solution.

Dibasic Acid

A dibasic acid is one that has two replaceable hydrogen ions, which can participate in chemical reactions. This characteristic is important because it affects the equivalent weight calculation. Here’s how it works:

• Examples of dibasic acids include sulfuric acid \( (H_2SO_4) \) and carbonic acid \( (H_2CO_3) \).
• In every chemical reaction, a dibasic acid will donate two hydrogen ions.

This property means that the molecular weight of the acid is divided by two when calculating the equivalent weight, thereby affecting the calculation of normality for solutions.

Equivalent Weight

Equivalent weight is a concept integral to understanding normality and is pivotal in titration calculations. It is defined as the mass of a compound that supplies or reacts with one mole of chemical equivalents. For dibasic acids, this calculation is simple and significant.

• To find the equivalent weight of a dibasic acid, divide its molecular weight by two, due to its ability to donate two protons.
• For example, if the molecular weight is 200, the equivalent weight is \( \frac{200}{2} = 100 \).

In essence, the equivalent weight helps determine how much acid is required to achieve a specific chemical change or reaction when it's involved in solutions.

Molar Calculations

Molar calculations are essential in finding out the quantities of substances used in chemical reactions. They help students and chemists alike to understand how compounds interact in reactions by relating moles, weight, and volume. In terms of the concept:

• Moles relate to the molecular weight of a substance and Avogadro's number \( (6.022 \times 10^{23}) \).
• These calculations offer a way to convert between mass and the amount of substance, allowing us to measure how much of a compound is needed for a reaction with a given concentration.

When dealing with acids like dibasic acids, these calculations involve the normality equation and the concept of equivalent weight to determine how many grams are needed to achieve the desired concentration.