Question Detail

CHM 101 CHM 101 Past Questions Objective Question
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Question

What volume of a 0.200 M solution of KOH is required to neutralize 50.0 mL of a 0.100 M solution of H₂SO₄?
Options
A 25.0 mL
B 50.0 mL
Correct Answer
C 100.0 mL
D 12.5 mL
Correct Answer

Option B is the correct answer.

Detailed Explanation

Reaction: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O. Moles H₂SO₄ = 0.0500 L × 0.100 M = 0.00500 mol. Moles KOH needed = 2 × 0.00500 = 0.0100 mol. Volume KOH = 0.0100 mol / 0.200 M = 0.0500 L = 50.0 mL.

Hint

Use the balanced equation to get the mole ratio.

Question Info

Course: CHM 101
Topic: CHM 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 1312 of 1872
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