## General Chemistry: Principles and Modern Applications (10th Edition)

Solubility = $83.0 mg/100ml$
1. Calculate $[OH^-]$ pH + pOH = 14 12.35 + pOH = 14 pOH = 1.65 $[OH^-] = 10^{- 1.65}$ $[OH^-] = 2.23 \times 10^{- 2}$ 2. Since each $Ca(OH)_2$ has 2 hydroxides, its concentration will be half of the concentration of $[OH^-]$ $[Ca(OH)_2] = [OH^-]/2 = (2.23 \times 10^{- 2})/2 = 1.12 \times 10^{-2}M$ ** Notice that, since the solution is saturated, this concentration represents the solubility of the compound. 3. Convert the number of moles in grams: ** Molar mass: $40.1 (Ca) + 2*16(O) + 2*1(H) = 74.1g/mol$ mm (g/mol) = $\frac{mass(g)}{n(moles)}$ 74.1 = $\frac{mass(g)}{ 0.0112}$ mass(g) = 74.1$\times$ 0.0112 mass(g) = 0.82992 Therefore: $0.0112 mol/L = 0.82992 g/L (Ca(OH)_2)$ 4. Now that we have the concentration in g/L, we need to convert into g/100ml: $\frac{ 0.82992}{1000ml} \div \frac{10}{10} = \frac{ 0.082992}{100ml}$ 5. Convert grams into miligrams: $\frac{ 0.082992g}{100ml} \times \frac{1000mg}{1g} \approx \frac{83.0mg}{100ml}$ Therefore: $Ca(OH)_2 = 83.0 \frac{mg}{100ml}$