## Chemistry and Chemical Reactivity (9th Edition)

The pH of this solution is $11.477$.
Since $Ba(OH)_2$ is a strong base, and it has 2 $OH^-$ in each molecule: $[OH^-] = 2 * [Ba(OH)_2] = 2 * 0.0015 = 0.0030M$ 1. Calculate the pOH. $pOH = -log[OH^-]$ $pOH = -log( 3 \times 10^{- 3})$ $pOH = 2.523$ 2. Now, use the pOH value to find the pH. $pH + pOH = 14$ $pH + 2.523 = 14$ $pH = 11.477$