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. Calculate the molarity of the acid solution? H2SO4(aq) + 2KOH(aq) o K2SO4(aq) + 2H2O(l)

Answer 1

0.9180 M

Step-by-step explanation:

Step 1: Write the balanced equation

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

Step 2: Calculate the reacting moles of KOH

27.00 mL of 1.700 M KOH react. The reacting moles of KOH are:

`0.02700L * (1.700mol)/(L) = 0.04590mol`

Step 3: Calculate the reacting moles of H₂SO₄

The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 0.04590 mol = 0.02295 mol.

Step 4: Calculate the molarity of H₂SO₄

0.02295 moles of H₂SO₄ are in 25.00 mL of solution. The molarity of the acid solution is:

`M = (0.02295 mol)/(0.02500) = 0.9180 M`