Chapter 16 - Acids and Bases - Exercises - Strong Acids, Strong Bases, and pH - Page 739: 14

$pH = 11.7$

1. Find the concentration of NaOH after the dilution: $C_1 * V_1 = C_2 * V_2$ $ 0.606* 0.125= C_2 * 15$ $ 0.07575 = C_2 * 15$ $C_2 = 0.00505M$ 2. Since NaOH is a strong base, $[NaOH] = [OH^-] = 0.00505M$ 3. Calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 5.05 \times 10^{- 3})$ $pOH = 2.296$ $pH + pOH = 14$ $pH + 2.296 = 14$ $pH = 11.7$