The molecular mass of sulphuric acid is 98. If 49 g of the acid is dissolved in water to make one litre of solution, what will be the strength of the acid?
One normal
Calculating the Normality (Strength) of Sulphuric Acid Solution
The question asks for the strength of a sulphuric acid solution in terms of normality (N). Normality is a measure of concentration defined as the number of gram equivalents of solute per litre of solution.
To calculate the normality of the sulphuric acid (H<sub>2</sub>SO<sub>4</sub>) solution, we need to follow these steps:
- Determine the equivalent mass of sulphuric acid.
- Calculate the number of gram equivalents of sulphuric acid dissolved.
- Calculate the normality using the number of gram equivalents and the volume of the solution.
Step 1: Determine the Equivalent Mass of Sulphuric Acid
For an acid, the equivalent mass is calculated by dividing its molecular mass by its basicity (the number of replaceable hydrogen ions).
- Molecular mass of sulphuric acid (H<sub>2</sub>SO<sub>4</sub>) is given as 98 g/mol.
- Sulphuric acid (H<sub>2</sub>SO<sub>4</sub>) is a diprotic acid, meaning it can donate two protons (H<sup>+</sup>) in a reaction. Therefore, its basicity is 2.
The equivalent mass of H<sub>2</sub>SO<sub>4</sub> is:
\[\text{Equivalent Mass} = \frac{\text{Molecular Mass}}{\text{Basicity}}\] \[\text{Equivalent Mass} = \frac{98 \text{ g/mol}}{2 \text{ equivalents/mol}}\] \[\text{Equivalent Mass} = 49 \text{ g/equivalent}\]
Step 2: Calculate the Number of Gram Equivalents
The number of gram equivalents of the solute is calculated by dividing the given mass of the solute by its equivalent mass.
- Mass of sulphuric acid dissolved = 49 g.
- Equivalent mass of sulphuric acid = 49 g/equivalent.
The number of gram equivalents of 49 g of H<sub>2</sub>SO<sub>4</sub> is:
\[\text{Number of Gram Equivalents} = \frac{\text{Mass of Solute}}{\text{Equivalent Mass}}\] \[\text{Number of Gram Equivalents} = \frac{49 \text{ g}}{49 \text{ g/equivalent}}\] \[\text{Number of Gram Equivalents} = 1 \text{ equivalent}\]
Step 3: Calculate the Normality
Normality (N) is defined as the number of gram equivalents of solute per litre of solution.
- Number of gram equivalents of solute = 1 equivalent.
- Volume of the solution = 1 litre.
The normality of the solution is:
\[\text{Normality (N)} = \frac{\text{Number of Gram Equivalents}}{\text{Volume of Solution (in Litres)}}\] \[\text{Normality (N)} = \frac{1 \text{ equivalent}}{1 \text{ litre}}\] \[\text{Normality (N)} = 1 \text{ N}\]
Therefore, the strength of the sulphuric acid solution is 1 normal.
Revision Table: Sulphuric Acid Normality Calculation
| Parameter | Value/Formula | Explanation |
|---|---|---|
| Molecular Mass (H<sub>2</sub>SO<sub>4</sub>) | 98 g/mol | Given in the question. |
| Basicity (H<sub>2</sub>SO<sub>4</sub>) | 2 | Number of replaceable H<sup>+</sup> ions. |
| Equivalent Mass | \[\frac{\text{Molecular Mass}}{\text{Basicity}}\] = \[\frac{98}{2}\] = 49 g/equivalent | Mass per equivalent of reactive species. |
| Mass of Solute (H<sub>2</sub>SO<sub>4</sub>) | 49 g | Amount of acid dissolved. |
| Number of Gram Equivalents | \[\frac{\text{Mass of Solute}}{\text{Equivalent Mass}}\] = \[\frac{49}{49}\] = 1 equivalent | Total reactive equivalents. |
| Volume of Solution | 1 Litre | Final volume of the solution. |
| Normality (N) | \[\frac{\text{Gram Equivalents}}{\text{Volume (L)}}\] = \[\frac{1}{1}\] = 1 N | Concentration in equivalents per litre. |
Additional Information: Understanding Solution Strength
Solution strength can be expressed in various concentration units, each serving a specific purpose in chemistry. Besides normality, other common units include Molarity, Molality, Mass Percentage, and Parts Per Million (PPM).
- Molarity (M): Moles of solute per litre of solution. It is temperature-dependent because volume changes with temperature.
- Molality (m): Moles of solute per kilogram of solvent. It is temperature-independent as it is based on mass.
- Normality (N): Gram equivalents of solute per litre of solution. It is often used in acid-base titrations and redox reactions because it relates directly to the reactive capacity of the solute. The equivalent mass depends on the specific reaction the substance undergoes. For acids, it's based on basicity; for bases, on acidity; for salts, on the total positive or negative charge; and for redox agents, on the number of electrons transferred.
In the case of sulphuric acid (H<sub>2</sub>SO<sub>4</sub>), its normality can be related to its molarity. Since its basicity is 2, 1 mole of H<sub>2</sub>SO<sub>4</sub> contains 2 equivalents. Therefore, for H<sub>2</sub>SO<sub>4</sub>, Normality (N) = 2 \(\times\) Molarity (M).
In this specific problem, we calculated 1 equivalent of H<sub>2</sub>SO<sub>4</sub> in 1 litre, giving 1 N. The mass 49 g corresponds to 0.5 moles (49 g / 98 g/mol). So, the molarity would be 0.5 M. As expected, 1 N = 2 \(\times\) 0.5 M.
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