How many milliliters of 0.100 M Ba (OH) 2 are required to neutralize 20.0 mL of 0.250 M HCl?

Ba (OH) ₂ + 2HCl = BaCl₂ + 2H₂O

n (HCl) = c (HCl) v (HCl)

n{Ba (OH) ₂) }=c{Ba (OH) ₂}v{Ba (OH) ₂}=2n (HCl) = 2c (HCl) v (HCl)

v{Ba (OH) ₂}=2c (HCl) v (HCl) / c{Ba (OH) ₂}

v{Ba (OH) ₂}=2*0.250*0.020/0.100=0.100 L = 100 mL