Answer
$[OH^-] \approx 0.169M$
$pH \approx 13.228$
Work Step by Step
1. Find the nº of moles of $OH^-$ given by $KOH:$
$n(moles) = concentration * (L)$
$n(moles) = 0.105 * 0.005$
$n(moles) = 5.25 \times 10^{-4} moles (KOH)$
Since KOH is a strong base:
$n(KOH) = n(OH^-) = 5.25 \times 10^{-4}$
2. Find the nº of moles of $OH^-$ given by $Ca(OH)_2$:
$n(moles) = 9.5 \times 10^{-2} * 0.015$
$n(moles) = 1.425 \times 10^{-3}(Ca(OH)_2)$
Since $Ca(OH)_2$ is a strong base:
$n(OH^-) = 2*n(Ca(OH)_2) = 2*1.425 \times 10^{-3} = 2.85 \times 10^{-3}$
3. Find the total nº of moles of $OH^-$:
$n(OH^-)_{total} = 5.25 \times 10^{-4} + 2.85 \times 10^{-3}$
$n(OH^-)_{total} = 3.375 \times 10^{-3} moles$
4. Find the concentration of $OH^-$
$Total Volume: 5ml + 15ml = 20ml$
$concentration = \frac{n(moles)}{V(L)}$
$C = \frac{3.375 \times 10^{-3}}{0.020}$
$C \approx 0.169 M = [OH^-]$
5. Find the pH:
$pOH = -log(0.169) = 0.772$
$pH = 14 - pOH = 14 - 0.772 = 13.228$