Question

\(0.5 \mathrm{M}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is diluted from 1 litre to 10 litre, normaliy of the resulting solution is (a) \(1 \mathrm{~N}\) (b) \(10 \mathrm{~N}\) (c) \(11 \mathrm{~N}\) (d) \(0.1 \mathrm{~N}\)

Short answer

The normality of the resulting solution is \(0.1\, \mathrm{N}\).

Step-by-step solution

Understand the Given Information

We have an initial solution of sulfuric acid \( (\mathrm{H}_2\mathrm{SO}_4) \) with a molarity of \(0.5\, \mathrm{M}\) and a volume of \(1\, \mathrm{L}\). This solution is diluted to a total volume of \(10\, \mathrm{L}\). We need to find the normality of the resulting solution.

Remember Relationship Between Molarity and Normality

Normality (\(N\)) for sulfuric acid is calculated as \(N = n \times M\), where \(n\) is the number of equivalents per mole of solute. For \(\mathrm{H}_2\mathrm{SO}_4\), \(n = 2\) because it can donate two protons (\(H^+\)) per molecule.

Calculate Initial Normality of the Acid Solution

Initially, the molarity \( (M) \) of \(\mathrm{H}_2\mathrm{SO}_4\) is \(0.5\, \mathrm{M}\). Thus, the initial normality is calculated as: \(N = 2 \times 0.5 = 1\, \mathrm{N}\).

Determine Normality After Dilution

Upon dilution, the number of equivalents of solute remains the same, but the concentration changes. Since normality is proportional to molarity and inversely proportional to total volume, the new normality can be calculated by: \(\text{New Normality} = \frac{\text{Initial Normality} \times \text{Initial Volume}}{\text{Final Volume}} = \frac{1 \times 1}{10} = 0.1 \mathrm{~N}\).

Key concepts

Dilution

When a solution is diluted, it is essentially mixed with more solvent. This process decreases the concentration of the solute in the solution, meaning that the solute is less concentrated in the larger volume. For example, if you have one liter of a 0.5 M sulfuric acid solution and you dilute it to ten liters, you are spreading the same amount of sulfuric acid over a much larger volume. This results in a decreased concentration.
You can calculate the new concentration after dilution using the formula:

• \[\text{New Normality} = \frac{\text{Initial Normality} \times \text{Initial Volume}}{\text{Final Volume}}\]

Even though the concentration changes, the total number of moles of solute remains the same because dilution does not add more solute, just more solvent. This is a key concept in understanding the changes that occur in dilute solutions.

Sulfuric Acid

Sulfuric acid, with a chemical formula of \(\text{H}_2\text{SO}_4\), is a strong, diprotic acid. This means it can donate two protons or hydrogen ions (\(H^+\)) per molecule. Because of this, it behaves differently in normality calculations compared to monoprotic acids like hydrochloric acid, which can donate just one proton.
Sulfuric acid is widely used in industries and laboratories due to its strong acidic nature and ability to participate in a variety of chemical reactions. Its diprotic nature requires special consideration in calculations related to concentration:

• Normality \((N)\) is given by: \(N = n \times M\), where \(n\) is the number of equivalents per mole of acid.

For sulfuric acid, \(n = 2\) because it can release two protons, making it twice as effective in reactions that require available hydrogen ions. This property significantly impacts its concentration calculations and uses in reactions based on molarity and normality.

Molarity

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of a solute per liter of solution and is expressed in units of mol/L, or \(\text{M}\). Molarity is crucial in chemistry because it allows chemists to know how much of a substance is present in a given volume of solution.
The formula for molarity is:

• \[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]

In the context of sulfuric acid, knowing the molarity is the first step to understanding its reactivity and how it will affect a given reaction. A solution with a high molarity means a higher concentration of solute and greater reactivity. When solutions are diluted, the molarity decreases, but the total amount of solute remains unchanged. This concept is central to both preparing consistent chemical solutions and understanding reactions in laboratory settings.