What volume of 0.2 M KOH is required to fully neutralize 20 ml of 0.1 M H₂SO₄?
-
20 ml
-
10 ml
-
40 ml
-
5 ml
📝 Detailed explanation
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O.
n(H₂SO₄) = 0.02 L * 0.1 M = 0.002 mol.
n(KOH) needed = 0.002 * 2 = 0.004 mol.
Volume KOH = 0.004 / 0.2 = 0.02 L = 20 ml.
← Previous
→ Next
Space — Show answer
🎯 Take a test on the topic «CAU Admission Test Sample Questions 10»
This topic contains 30 questions.
Test your knowledge!
🔐 Log in to take the test
📚 All tests by subject «Bio+Chem»
View all collections and topics in Bio+Chem
Go to subject →
