Question

How many milliliters of 0.120 M HCl are needed to completely neutralize 60.0 mL of 0.105 M Ba(OH)2 solution?

Answer 1

Molarity = No.of moles / Volume in litre

Hence, for Ba(OH)₂, no. of moles = Molarity X Volume in litre

no. of moles = 0.105 X (60/1000)

= 6.3x10^-3 moles

Now, Using Law of Equivalent M₁V₁ = M₂V₂

Here, let M₁V₁ is for Ba(OH)₂ and M₂V₂ for HCl, then this equation becomes: M₁V₁ = 2 M₂V₂ as for each molecule of Ba(OH)₂, two molecule of HCl are consumed to neutralize it.

Hence, volume of HCl required = M₁V₁ / 2 M₂

  = 6.3x10^-3 / ( 2x0.120M)

= 0.02625 L

= 26.25 mL