Consider the following reaction: 2HCl + Ca (OH) 2 CaCl2 + 2H2O A scientist wants to neutralize 25 mL of 0.001 M Ca (OH) 2 with a volume of 0.005 M HCl. What is the minimum volume of HCl required?

The balanced chemical equation for the above reaction is as follows;

Ca (OH) 2 + 2HCl - > CaCl2 + 2H2O

Stoichiometry of Ca (OH) 2 to HCl is 1:2

Number of Ca (OH) 2 moles present - 0.001 mol/L / 1000 mL/L x 25 mL

Number of Ca (OH) 2 moles = 2.5 x 10^ (-5) mol

Number of HCl moles needed for neutralisation = 2.5 x 10^ (-5) mol x2 = 5 x 10 ^ (-5) mol

The molarity of HCl solution is 0.005 M

The solution contains 0.005 mol in 1000 mL

Therefore volume of 5x10^ (-5) mol in = 1000/0.005 x 5 x 10^ (-5) mol = 10 mL

10 mL of HCl is needed for neutralisation