Answer
This $Ca(OH)_2$ solution has $pH = 13.00$
Work Step by Step
1. Calculate the concentration $(mol/L)$
$C(M) = \frac{n(mol)}{V(L)} = \frac{0.0050 mol}{0.100L} = 0.050M$ of $Ca(OH)_2$.
2. Since $Ca(OH)_2$ is a strong base and it has 2 $OH^-$ in each molecules: $[OH^-] = 2 * [Ca(OH)_2] = 2 * 0.050 = 0.10M$
3. Calculate the pOH and then the pH value:
$pOH = -log[OH^-]$
$pOH = -log( 0.10)$
$pOH = 1.00$
$pH + pOH = 14$
$pH + 1 = 14$
$pH = 13.00$
