Question

A 25.00 ml solution of sulfuric acid H2SO4 is titrated to phenolphthalein end point with 27.00 ml of 1.700 M KOH

Answer 1

Answer:

0.918 M

Explanation:

Assuming the question requires we calculate the Molarity of sulfuric acid:

We are given:

• Volume of the acid, H₂SO₄ = 25.00 ml
• Volume of the base, KOH = 27.00 mL
• Molarity of the base, KOH is 1.70 M

We can calculate the molarity of the acid using the following steps;

Step 1: Write the chemical equation for the reaction.

The reaction is an example of a neutralization reaction where a base reacts with an acid to form salt and water.

Therefore, the balanced equation will be;

H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)

Step 2: Determine the moles of the base, KOH

When given molarity and the volume of a solution, the number of moles can be calculated by multiplying molarity with volume.

Number of moles = Molarity × Volume

                             = 1.700 M × 0.027 L

                              = 0.0459 moles

Thus, moles of KOH used is 0.0459 moles

Step 3: Determine the number of moles of the Acid, H₂SO₄

From the reaction, 1 mole of the acid reacts with 2 moles of KOH

Therefore, the mole ratio of H₂SO₄ to KOH is 1 : 2

Thus, moles of H₂SO₄ = Moles of KOH ÷ 2

                                     = 0.0459 moles ÷ 2

                                     = 0.02295 moles

Step 4: Calculate the molarity of the Acid

Molarity is the concentration of a solution in moles per liter

Molarity = Moles ÷ Volume

Molarity of the acid = 0.02295 moles ÷ 0.025 L

                                = 0.918 M

Thus, the molarity of the acid, H₂SO₄ is 0.918 M