## Chemistry and Chemical Reactivity (9th Edition)

$K_{sp} (Ca(OH)_2)= (5.52 \times 10^{-5})$
1. Calculate the molar mass $(Ca(OH)_2)$: 40.08* 1 + 2 * ( 16* 1 + 1.008* 1 ) = 74.096g/mol 2. Calculate the number of moles $(Ca(OH)_2)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 1.78}{ 74.096}$ $n(moles) = 0.0240$ 3. Find the concentration in mol/L $(Ca(OH)_2)$: $0.024$ mol in 1L: $0.024 M (Ca(OH)_2)$ 4. Write the $K_{sp}$ expression: $Ca(OH)_2(s) \lt -- \gt 1Ca^{2+}(aq) + 2OH^{-}(aq)$ $K_{sp} = [Ca^{2+}]^ 1[OH^{-}]^ 2$ 5. Determine the ions concentrations: $[Ca^{2+}] = [Ca(OH)_2] * 1 = [0.024] * 1 = 0.024$ $[OH^{-}] = [Ca(OH)_2] * 2 = 0.048$ 6. Calculate the $K_{sp}$: $K_{sp} = (0.024)^ 1 \times (0.048)^ 2$ $K_{sp} = (0.024) \times (2.3 \times 10^{-3})$ $K_{sp} = (5.52 \times 10^{-5})$