If 42.2 mL of 0.206 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution?

Question

If 42.2 mL of 0.206  M  HCl solution is needed to neutralize a solution of Ca(OH)2 , how many grams of Ca(OH)2 must be in the solution?
Express the mass in grams to three significant digits.

Answer 1

42.2 mL, or 0.0422 L, of 0.206 M HCl contains 0.0086932 moles of HCl. 2 moles of HCl are needed to neutralize 1 mole of Ca(OH)2, so the solution contains 0.0043466 moles of Ca(OH)2. The molecular mass of Ca(OH)2 is 74.054 grams ( 40.04 + 2 * (15.9994 + 1.0078) ). 0.004366 moles times 74.054 grams per mole  =  0.32332 grams. Rounded to three significant figures, the answer is 0.323 grams of Ca(OH)2.