Final answer:
Neutralize 10.00 mL of 0.100 M H2SO4, 20.00 mL of 0.100 M NaOH is required since the stoichiometry of the reaction requires two moles of NaOH for every mole of H2SO4.
Step-by-step explanation:
The question involves finding the volume and concentration of a base needed to neutralize a given volume and concentration of sulfuric acid (H2SO4). The balanced chemical equation for the neutralization of sulfuric acid with a base such as sodium hydroxide (NaOH) or potassium hydroxide (KOH) is:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
or
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)
To neutralize 10.00 mL of 0.100 M H2SO4, we need enough base to provide the same number of moles of OH- ions as there are moles of H+ in the acid. Since sulfuric acid is a diprotic acid (providing 2 H+ per molecule), we need twice as many moles of NaOH or KOH to neutralize it.
First, calculate the moles of H2SO4:
- 10.00 mL H2SO4 × 0.100 M = 0.00100 mol H2SO4
Now, calculating the moles of NaOH or KOH required:
- 0.00100 mol H2SO4 × 2 = 0.00200 mol NaOH/KOH
Finally, using the given concentrations to find the volume:
- (a) For 0.100 M NaOH: 0.00200 mol ÷ 0.100 M = 0.02000 L = 20.00 mL of NaOH (correct answer)
- (b) For 0.0200 M NaOH: 0.00200 mol ÷ 0.0200 M = 0.100 L = 100.0 mL of NaOH
- (c) For 0.100 M KOH: 0.00200 mol ÷ 0.100 M = 0.02000 L = 20.00 mL of KOH
- (d) For 0.100 M Ba(OH)2: Because barium hydroxide is dihydroxide, 0.00100 mol H2SO4 would require 0.00100 mol Ba(OH)2, and: 0.00100 mol ÷ 0.100 M = 0.01000 L = 10.00 mL of Ba(OH)2
The correct answer is (a) 20.00 mL of 0.100 M NaOH as it provides the 0.00200 mol of OH- ions needed to neutralize the sulfuric acid.