Given a saturated solution of 25 drops of Ca(OH)2 in water that requires 34 drops of 0.0050 M HCl to titrate to the endpoint, calculate [OH-] and [Ca2+]. An ICE table will help you relate [OH-] to [Ca2+].
Frankly I get confused by drops. Let's say a drop is 0.1 mL (I know that's a huge drop but we'll be consistent).
Ca(OH)2 ==> Ca^2+ + 2OH^-
2HCl + Ca(OH)2 ==>2H2O + CaCl2
millimols HCl = 34 x 0.1 x 0.005 = 0.017 mmols HCl.
mmols Ca(OH)2 must be 1/2 that = 0.0085
M Ca(OH)2 = mmols/mL = 0.0085/2.5 = 0.0034 M Ca(OH)2.
Therefore, (Ca^2+) = 0.0034M
(OH^-) = 2x that = ?