Acids and Bases - Neutralizing a Base with an Acid

What volume of 0.075 M HCl is required to neutralize 100 ml of 0.01 M Ca(OH)₂ solution?

Solution

HCl is a strong acid and will dissociate completely in water to H⁺ and Cl^-. For every mole of HCl, there will be one mole of H⁺. Since the concentration of HCl is 0.075 M, the concentration of H⁺ will be 0.075 M.

Ca(OH)₂ is a strong base and will dissociate completely in water to Ca²⁺ and OH^-. For every mole of Ca(OH)₂ there will be two moles of OH^-. The concentration of Ca(OH)₂ is 0.01 M so [OH^-] will be 0.02 M.

To the solution will be neutralized when the number of moles of H⁺ equals the number of moles of OH^-.

Step 1: Calculate the number of moles of OH^-.

Molarity = moles/volume

moles = Molarity x Volume

moles OH^- = 0.02 M/100 ml
moles OH^- = 0.02 M/0.1 L
moles OH^- = 0.002 moles

Step 2: Calculate the Volume of HCl needed

Molarity = moles/volume

Volume = moles/Molarity

Volume = moles H⁺/0.075 M

moles H⁺ = moles OH^-

Volume = 0.002 moles/0.075 M
Volume = 0.0267 L
Volume = 26.7 ml of HCl

Answer

26.7 ml of 0.075 M HCl is needed to neutralize 100 ml of 0.01 M Ca(OH)₂ solution.