How much sr(oh)2 • 8 h2o (m = 265.76) is needed to prepare 250.0 ml of solution in which [oh–] = 0.100 m?

The required is the amount of Sr(OH)2·8H2O needed to prepare a solution having a concentration of 0.100 M OH-. We start with the concentration of the hydroxide ion to determine the number of moles hydroxide ion for a given volume of 250.0 mL solution.

0.100 mol/L (0.250 L) = 0.4 mol hydroxide ion needed.

We know that, for every mole of Sr(OH)2·8H2O in solution, two hydroxide ions is formed. Thus,

0.4 mol OH- ( 1 mol Sr(OH)2·8H2O / 2 mol OH-) ( 265.76 g / mol) = 53.152 g 

Thus, 53.152 grams of Sr(OH)2·8H2O is needed to make a solution with 0.100 M OH-.