What Volume of a 0.53 M HCl solution is required to neutralize 16.5 mL of a .124 M Ca(OH)2 Solution?

Question

What Volume of a 0.53 M HCl solution is required to neutralize 16.5 mL of a .124 M Ca(OH)2 Solution?

Answer 1

2HCl + Ca(OH)2 ==> CaCl2 + 2H2O ... balanced equation

moles Ca(OH)2 present = 16.5 ml x 1 L/1000 ml x 0.124 mol/L = 0.002046 moles

moles HCl required = 0.002046 mol Ca(OH)2 x 2 mol HCl/mol Ca(OH)2 = 0.004092 moles HCl

Volume HCl: (x L)(0.53 mol/L) = 0.004092 moles and x = 0.007721 L = 7.72 ml of HCl