200. mL of Ca(OH)2 are required to neutralize 100. mL of 0.10 M Acetic acid. What is the molarity of the Ca(OH)2

Question

200. mL of Ca(OH)₂ are required to neutralize 100. mL of 0.10 M Acetic acid. What is the molarity of the Ca(OH)₂

Answer 1

Ca(OH)2 + 2CH3COOH ==> Ca(CH3COO-)2 + 2H2O

moles CH3COOH used = 100 ml x 1 L / 1000 ml x 0.10 mol/L = 0.0100 moles CH3COOH

moles Ca(OH)2 present = 0.0100 mol CH3COOH x 1 mol Ca(OH)2 / 2 mol CH3COOH = 0.005 mol Ca(OH)2

Molarity of Ca(OH)2 = 0.005 mol / 0.200 L = 0.0250 M