Question

A 6.05-g sample of anhydrous sodium hydroxide, NaOH was dissolved in water and made up to 100.00 mL of solution. A 25.00-mL sample of the solution was then titrated with sulphuric acid, H₂50. (a) Find the molarity of the NaOH solution. [1.52] (b) Write a balanced chemical equation involved in the titration (c) How many moles of H.50, were required for the neutralisation? (0.0189] (d) Calculate the concentration of the sulphuric acid if 14.60 mL of the acid was required to reach the end point. [1.295​

Answer 1

Step-by-step explanation:

- The highlighted text describes an experiment where a 6.05-gram sample of anhydrous sodium hydroxide (NaOH) was dissolved in water to make a solution of 100.00 mL.

- A 25.00-mL sample of this solution was then titrated with sulphuric acid (H₂SO₄).

- The molarity of the NaOH solution was calculated to be 1.52.

- A balanced chemical equation was involved in the titration, which is NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

- The number of moles of H₂SO₄ required for neutralization was found to be 0.0189.

- The concentration of the sulphuric acid was calculated to be 1.295 M when 14.60 mL of the acid was required to reach the end point.

- This experiment is important because it helps determine the concentration of a solution and the amount of acid required to neutralize it.

- The molarity of a solution is defined as the number of moles of solute per liter of solution, and it is a measure of the concentration of a solution.

- Titration is a common laboratory technique used to determine the concentration of a solution by reacting it with a solution of known concentration.