Answer:
0.478 M
Explanation:
Let's consider the neutralization reaction between KOH and H₂SO₄.
2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O
12.7 mL of 1.50 M H₂SO₄ react. The reacting moles of H₂SO₄ are:
0.0127 L × 1.50 mol/L = 0.0191 mol
The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of KOH are 2 × 0.0191 mol = 0.0382 mol
0.0382 moles of KOH are in 80.0 mL. The molarity of KOH is:
M = 0.0382 mol/0.0800 L = 0.478 M
