Answer
$[OH^-] = 4.2 \times 10^{-4}$
$pH = 10.623$
Work Step by Step
1. Calculate the concentration of $Ca(OH)_2$ after the dilution:
$C_i * V_i = C_f * V_f$
$0.0105 * 0.010 = C_f * 0.500$
$C_f = \frac{0.000105}{0.500}$
$C_f = 2.1 \times 10^{-4}M$
2. Since $Ca(OH)_2$ is a strong base, and each mol of $Ca(OH)_2$ has 2 $OH^-$:
$[OH^-] = 2*[Ca(OH)_2]$
$[OH^-] = 2*2.1\times10^{-4} = 4.2 \times 10^{-4}M$
3. Find the pH:
$pOH = -log[OH^-]$
$pOH = -log(4.2\times10^{-4}) = 3.377$
$pH = 14 - pOH$
$pH = 14 - 3.377 = 10.623$
