What is the normality of a sodium hydroxide solution of 25 ml that is neutralized by 100 ml of a 0.01 N sulfuric acid solution?

What is the normality of a sodium hydroxide solution of 25 ml that is neutralized by 100 ml of a 0.01 N sulfuric acid solution?

• A. 0.001 N
• B. 0.0025 N
• C. 0.005 N
• D. 0.01 N

Answer: (B)

To determine the normality of the sodium hydroxide (NaOH) solution, it's essential to analyze the neutralization reaction between sodium hydroxide and sulfuric acid (H₂SO₄). Normality, in this case, refers to the amount of equivalent solute per liter of solution. Sulfuric acid is a diprotic acid, meaning each molecule can donate two hydrogen ions (H⁺). Thus, when sulfuric acid is in a 0.01 N concentration, it provides 0.01 equivalents of H⁺ per liter. During the neutralization, the moles of NaOH will equal the moles of H⁺ provided by sulfuric acid. The equation for neutralization between NaOH and H₂SO₄ can be considered: \[ \text{H₂SO₄} + 2 \text{NaOH} \rightarrow \text{Na₂SO₄} + 2 \text{H₂O} \] The 100 ml of 0.01 N sulfuric acid contributes: \[ \text{Equivalents of H⁺} = \text{Volume (L)} \times \text{Normality} = 0.100

To determine the normality of the sodium hydroxide (NaOH) solution, it's essential to analyze the neutralization reaction between sodium hydroxide and sulfuric acid (H₂SO₄). Normality, in this case, refers to the amount of equivalent solute per liter of solution.

Sulfuric acid is a diprotic acid, meaning each molecule can donate two hydrogen ions (H⁺). Thus, when sulfuric acid is in a 0.01 N concentration, it provides 0.01 equivalents of H⁺ per liter.

During the neutralization, the moles of NaOH will equal the moles of H⁺ provided by sulfuric acid. The equation for neutralization between NaOH and H₂SO₄ can be considered:

[

\text{H₂SO₄} + 2 \text{NaOH} \rightarrow \text{Na₂SO₄} + 2 \text{H₂O}

]

The 100 ml of 0.01 N sulfuric acid contributes:

[

\text{Equivalents of H⁺} = \text{Volume (L)} \times \text{Normality} = 0.100