## Chemistry (7th Edition)

$pH = 11.774$
1. Calculate the molar mass: 40.08* 1 + 16* 1 = 56.08g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.25}{ 56.08}$ $n(moles) = 4.458\times 10^{- 3}$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 4.458\times 10^{- 3}}{ 1.5}$ $C(mol/L) = 2.972\times 10^{- 3}$ 4. Analyzing the CaO's reaction with water, we got: $1 CaO(aq) + H_2O(l) \ -- \gt Ca^{2+}(aq) + 2OH^-(aq)$ So, since CaO is a strong base that gives $2OH^-$ in its ionization reaction: $[OH^-] = 2 * [CaO]= 2 * 2.972 \times 10^{-3} = 5.944 \times 10^{-3}M$ 5. Calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 5.944 \times 10^{- 3})$ $pOH = 2.226$ $pH + pOH = 14$ $pH + 2.226 = 14$ $pH = 11.774$