How many milliliters of 0.100 m ba(oh)2 are required to neutralize 20.0 ml of 0.250 m hcl?

25.0 mL of 0.100 M Ba(OH)₂ are required.

Explanation:

The reaction that takes place is:

Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O

The formula for concentration is:

C = n / V

0.020 L * 0.250 M = 0.005 mol HCl

Then we convert them to moles of Ba(OH)₂:

0.005 mol HCl * = 0.0025 mol  Ba(OH)₂

0.0025 mol Ba(OH)₂ / 0.100 M = 0.025 L = 25 mL