Question

55.2 mL of 0.500 M potassium hydroxide is used to neutralize 27.4 mL of sulfuric

Answer 1

Answer:

0.504 M

Explanation:

Step 1: Write the balanced neutralization reaction

2 KOH + H₂SO₄ ⇒ K₂SO₄ + 2 H₂O

Step 2: Calculate the reacting moles of KOH

55.2 mL (0.0552 L) of 0.500 M KOH react. The reacting moles of KOH are:

0.0552 L × 0.500 mol/L = 0.0276 mol

Step 3: Calculate the moles of H₂SO₄ that reacted with 0.0276 moles of KOH

The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of H₂SO₄ are 1/2 × 0.0276 mol = 0.0138 mol

Step 4: Calculate the concentration of H₂SO₄

0.0138 moles of H₂SO₄ are in 27.4 mL (0.0274 L). The molarity of H₂SO₄ is:

[H₂SO₄] = 0.0138 mol/0.0274 L = 0.504 M