To reach the equivalence point in the titration of 25.0 mL of 0.100 M NaOH with 0.100 M HCl, what volume of HCl is required?

Study for the Chemistry 1LD Test with comprehensive flashcards and multiple choice questions. Each question is accompanied by hints and detailed explanations to ensure your success. Prepare thoroughly for your exam!

Multiple Choice

To reach the equivalence point in the titration of 25.0 mL of 0.100 M NaOH with 0.100 M HCl, what volume of HCl is required?

Explanation:
At the equivalence point for a strong base–strong acid titration, moles of base equal moles of acid. The NaOH solution provides 25.0 mL × 0.100 M = 2.50 × 10^-3 moles of OH-. To neutralize that amount, you must add the same number of moles of H+, and with 0.100 M HCl, the required volume is (2.50 × 10^-3 mol) / (0.100 mol/L) = 0.0250 L = 25.0 mL. So the necessary volume is 25.0 mL. The other volumes would not supply the exact amount of acid needed to reach the equivalence point under these conditions.

At the equivalence point for a strong base–strong acid titration, moles of base equal moles of acid. The NaOH solution provides 25.0 mL × 0.100 M = 2.50 × 10^-3 moles of OH-. To neutralize that amount, you must add the same number of moles of H+, and with 0.100 M HCl, the required volume is (2.50 × 10^-3 mol) / (0.100 mol/L) = 0.0250 L = 25.0 mL. So the necessary volume is 25.0 mL. The other volumes would not supply the exact amount of acid needed to reach the equivalence point under these conditions.

More Multiple Choice Questions
Subscribe

Get the latest from Passetra

You can unsubscribe at any time. Read our privacy policy