Question

A student makes a standard solution of sulfuric acid by taking 100 ml of a super concentrated stock solution and diluting it to 2.50 L. He then standardizes the diluted solution and diluting it to 40 ml with 28.58 ml of a 0.4050 M solution of KOH. What is the concentration of the standard solution and the stock solution.

Answer 1

Total volume of diluted solution = 40 mL

Volume of KOH (V₂) = 28.58 mL

Molarity of KOH (M₂) = 0.405 M

Thus, Volume of standardized acid (V₁) = 40 - 28.58 = 11.42 mL

Let M₁ be the concentration of standard solution

Now 2 mol of KOH will standardize 1 mol of H₂SO₄

From Molarity equation

2*M₁V₁ = M₂V₂

M₁ = M₂V₂ / V₁ = 0.405 M * 28.58 mL / (11.42 mL * 2)= 0.507 M

Thus, the concentration of standard solution = 0.507 M

Volume of Stock solution (V₂') = 100 mL

Volume of standard solution (V₁') = 2.50 L = 2500 mL

Molarity of standard solution (M₁') = 0.507 M

Let M₂' be the molarity of stock solution

Using molarity equation, we have:

M₁'V₁' = M₂' V₂'

M₂' = M₁'*V₁' / V₂' = 0.507 M * 2500 mL / 100 mL = 12.67 M

Thus, the concentration of stock solution = 12.67 M