What is the normality of a 0.5 solution of H2SO4

To find the normality of a 0.5 M (molarity) solution of \( \text{H}_2\text{SO}_4 \) (sulfuric acid), you need to understand that normality (N) is related to molarity (M) by the equation:

\[
\text{Normality (N)} = \text{Molarity (M)} \times \text{Number of equivalents (n)}
\]

For sulfuric acid, \( \text{H}_2\text{SO}_4 \), the number of equivalents \( n \) depends on the number of protons \( (\text{H}^+) \) it can donate in a reaction. Since \( \text{H}_2\text{SO}_4 \) can donate 2 protons:

\[
n = 2
\]

Therefore, the normality (N) is calculated as:

\[
\text{N} = 0.5 \, \text{M} \times 2 = 1.0 \, \text{N}
\]

So, the normality of the 0.5 M \( \text{H}_2\text{SO}_4 \) solution is **1.0 N**.

Would you like more details or have any other questions?

Here are five related questions:
1. How does normality differ from molarity in acid-base titrations?
2. What is the normality of a 1 M solution of \( \text{H}_3\text{PO}_4 \)?
3. How would you calculate the normality of a base like \( \text{NaOH} \)?
4. What happens to the normality if you dilute the solution?
5. Can normality be used for redox reactions as well?

**Tip:** Normality is especially useful in titrations because it directly relates to the equivalent amount of reactants.

Learn how to calculate the normality of a 0.5 M sulfuric acid (H2SO4) solution step-by-step. Understand the relationship between molarity and normality, and how it applies to acid-base reactions. Suitable for advanced high school and college students.

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