Question

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

Answer 1

133 mL

Step-by-step 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