Question

The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation below. Suppose

Answer 1

Answer:

• 1.13M

Explanation:

The change of color indicates that the solution was neutralized.

1. Neutralization equation(given):

    $H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O$

2. Mole ratio:

• 2moles of KOH are neutralized by 1 mol of H₂SO₄

3. Number of  moles of H₂SO₄

• Molarity, M = 0.20M
• Volume, V = 0.017 liter

• M = n/V ⇒ n = M×V =0.20M × 0.017 liter = 0.034mol

4. Number of moles of KOH

       $0.034molH_2SO_4\times \dfrac{2molKOH}{1molH_2SO_4}=0.068molKOH$

5. Concnetration KOH

• n = 0.0680 mol

• V = 0.06 liter

• M = n/V = 0.068 mol / 0.06 liter = 1.13 M ← answer