# Chapter 16 - Acid-Base Equilibria - Exercises - Page 718: 16.45d

$[OH^-] \approx 0.169M$ $pH \approx 13.228$

#### 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: $pOH = -log(0.169) = 0.772$ $pH = 14 - pOH = 14 - 0.772 = 13.228$

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