6. What volume of 6M Ba (OH) 2 will be needed to neutralize 400.0 ml of a 2.0 M HCl solution?

133 mL

Explanation:

M = mol solute / L solution

mol solute = M * L solution

mol solute = (2.0 M * 0.4 L)

mol solute = 0.8 mol HCl = mol H

We need the same amount of OH moles to neutralize the solution

L solution = mol solute/M

L solution = 0.8 mol Ba (OH) 2 / 6M

L solution=0.133 L=133 mL