calculate the L of 5.00 M HCI required to neutralize 500.0 mL of 1.00 M Ca
(OH)2
Ca(OH)2 + 2HCl = CaCl2 + 2H2O
So, each mole of Ca(OH)2 neutralizes 2 moles of HCl
500mL of 1.00M Ca(OH)2 = 0.50 moles
so, you need 1.00 moles of HCl
That would be 200mL of 5.0M acid