Answer :
(a) To completely neutralize 25.0 mL of 0.101 M Ba(OH)₂ solution, 38.5 mL of 0.155 M HCl are needed.
(b) To neutralize 25.0 g of NaOH using 3.50 M H₂SO₄, 35.8 mL of H₂SO₄ are required, Therefore, the correct answers are option A
(a) First, we determine the moles of Ba(OH)₂ in the solution:
Moles of Ba(OH)₂ = (0.101 mol/L) * (25.0 mL / 1000 mL)
= 0.00252 mol
Since the acid-base reaction is 1:2 for Ba(OH)₂ and HCl, we need twice the number of moles of HCl to neutralize it:
Moles of HCl = 2 * Moles of Ba(OH)₂
= 2 * 0.00252 mol
= 0.00504 mol
Now, we can find the volume of 0.155 M HCl needed using the formula:
Volume (mL) = Moles / Molarity
= 0.00504 mol / 0.155 mol/L
≈ 38.5 mL.
(b) For the neutralization of NaOH with H₂SO₄, we first find the moles of NaOH:
Moles of NaOH = Mass / Molar mass
= 25.0 g / 40.00 g/mol
= 0.625 mol
Since the acid-base reaction is 2:1 for H₂SO₄ and NaOH, we need half the number of moles of H₂SO₄:
Moles of H₂SO₄= 0.625 mol / 2
= 0.3125 mol
Now, we can find the volume of 3.50 M H₂SO₄ needed:
Volume (mL) = Moles / Molarity
= 0.3125 mol / 3.50 mol/L
≈ 35.8 mL.
Therefore, the correct answers are option A: (a) 38.5 mL and (b) 35.8 mL.
