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?

Prepare for the Environmental Compliance Inspector Grade 1 Exam. Access comprehensive flashcards and multiple choice questions, with detailed hints and explanations for each. Ace your exam with confidence!

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

Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy