Question

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?

Answer 1

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