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If 50 mL of 0.1 M sulfuric acid is neutralized by sodium hydroxide, how many mol

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Question: If 50 mL of 0.1 M sulfuric acid is neutralized by sodium hydroxide, how many moles of NaOH are required?

Options:

  1. 0.005
  2. 0.01
  3. 0.02
  4. 0.03

Correct Answer: 0.01

Solution: 

H2SO4 dissociates to give 2 H+ ions. Moles of H2SO4 = 0.1 M * 0.050 L = 0.005 moles. Therefore, moles of NaOH required = 0.005 moles * 2 = 0.01 moles.

If 50 mL of 0.1 M sulfuric acid is neutralized by sodium hydroxide, how many mol

Practice Questions

Q1
If 50 mL of 0.1 M sulfuric acid is neutralized by sodium hydroxide, how many moles of NaOH are required?
  1. 0.005
  2. 0.01
  3. 0.02
  4. 0.03

Questions & Step-by-Step Solutions

If 50 mL of 0.1 M sulfuric acid is neutralized by sodium hydroxide, how many moles of NaOH are required?
  • Step 1: Understand that sulfuric acid (H2SO4) dissociates into 2 hydrogen ions (H+) when it reacts with sodium hydroxide (NaOH).
  • Step 2: Calculate the volume of sulfuric acid in liters. Since we have 50 mL, we convert it to liters: 50 mL = 0.050 L.
  • Step 3: Use the molarity (M) of sulfuric acid to find the number of moles. Moles of H2SO4 = Molarity * Volume in liters = 0.1 M * 0.050 L.
  • Step 4: Calculate the moles of H2SO4: 0.1 * 0.050 = 0.005 moles.
  • Step 5: Since each mole of H2SO4 produces 2 moles of H+, we need to multiply the moles of H2SO4 by 2 to find the moles of NaOH required: 0.005 moles * 2 = 0.01 moles.
  • Step 6: Conclude that 0.01 moles of NaOH are required to neutralize the sulfuric acid.
  • Stoichiometry – Understanding the mole ratio between reactants in a neutralization reaction.
  • Acid-Base Neutralization – Recognizing the reaction between an acid and a base to form water and salt.
  • Concentration and Volume Calculations – Calculating moles from molarity and volume.
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