# Question 9271e

Apr 23, 2015

You can't have a 0.250-M calcium hydroxide solution in 100.0 mL of water.

Calcium hydroxide has a solubility of 0.173 g/100mL at ${20}^{\circ} \text{C}$, which means that you cannot dissolve more than that amount in your volume without creating a saturated solution.

If you calculate the molarity of a 100.0-mL solution prepared by dissolving 0.173 g of calcium hydroxide, you'd get

0.173cancel("g") * ("1 mole "Ca(OH)_2)/(74.093cancel("g")) = "0.00233 moles "# $C a {\left(O H\right)}_{2}$

$C = \frac{n}{V} = \text{0.00233 moles"/(100 * 10^(-3)"L") = "0.0233 M}$

As you can see, this value is more than 10 times smaller than your target molarity of 0.250 M.