Question 16.9: What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.00...

Analyze We are asked to calculate the pH of two solutions of strong bases.
Plan We can calculate each pH by either of two equivalent methods. First, we could use Equation 16.16 to calculate $[H^{+}]$ and then use Equation 16.17 to calculate the pH. Alternatively, we could use $[OH^{-} ]$ to calculate pOH and then use Equation 16.20 to calculate the pH.

$K_{w} = [H_{3}O^{+}][OH^{-}] = [H^{+}][OH^{-}] = 1.0 × 10^{-14}      (at   25  °C)$         [16.16]

pH = – log $[H^{+}]$                           [16.17]

pH + pOH = 14.00      (at  25  °C)                           [16.20]
Solve
(a) NaOH dissociates in water to give one $OH^{-}$ ion per formula unit. Therefore, the $OH^{-}$ concentration for the solution in (a) equals the stated concentration of NaOH, namely 0.028 M.
Method 1:

pH = -log(3.57 × $10^{-13}$) = 12.45

Method 2:

pOH = – log(0.028) = 1.55
pH = 14.00 – pOH = 12.45

(b) $Ca(OH)_{2}$ is a strong base that dissociates in water to give two $OH^{-}$ ions per formula unit. Thus, the concentration of $OH^{-}$(aq) for the solution in part (b) is 2 × (0.0011 M) =0.0022 M.
Method 1:

pH = -log(4.55 × $10^{-12}$) = 11.34

Method 2:
pOH = -log(0.0022) = 2.66
pH = 14.00 – pOH = 11.34