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

$[H^+] = 4.05 \times 10^{-14}M$ $pH = 13.39$
1. Find the molar mass of $Ba(OH)_2 {\circ} 8 H_2O$ Ba: 137.3 * 1 = 137.3 O: 16.0 * 10 = 160.0 H: 1.0 * 18 = 18.0 137.3 + 160 + 18 = 315.3 2. Find the number of moles. $mm(g/mol) = \frac{mass(g)}{n(moles)}$ $315.3= \frac{ 3.9}{n(moles)}$ $n(moles) = \frac{ 3.9}{ 315.3}$ $n(moles) = 0.01236$ 3. Find the concentration. $Concentration(M) = \frac{n(moles)}{Volume(L)}$ $Concentration(M) = \frac{ 0.01236}{ 0.1}$ $Concentration(M) = 0.1236M$ 4. Calculate $[OH^-]$ ** This base has 2 "OH"s. $[OH^-] = 2* [Base]$ $[OH^-] = 0.247$ 5. Calculate $[H^+]$ $[H^+] * [OH^-] = 10^{-14}$ $[H^+] = \frac{10^{-14}}{[OH^-]}$ $[H^+] = \frac{10^{-14}}{ 0.247}$ $[H^+] = 4.05 \times 10^{-14}$ 6. Find the pH. $pH = -log[H^+] = -log( 4.05 \times 10^{-14})$ $pH = 13.39$