Question 18.p.3: NaOH is a strong base that dissociates completely in water. ......
NaOH is a strong base that dissociates completely in water. Subtract pH from 14.00 to find the pOH, and calculate inverse logs of pH and pOH to find [\text{H}_3\text{O}^+]\text{ and [OH}^-], respectively.
pH + pOH = 14.00
pOH = 14.00 – 9.52 = 4.48
\text{pH} = -\log [\text{H}_3\text{O}^+] \\ [\text{H}_3\text{O}^+] = 10^\text{-pH} = 10^{-9.52} = 3.01995 \times 10^{-10} = 3.0 \times 10^{-10} \ M \\ \text{pOH} = -\log [\text{OH}^-] \\ [\text{OH}^-] = 10^{- \text{pOH}} = 10^{-4.48} = 3.3113 \times 10^{-5} = 3.3 \times 10^{-5} \ M
Check: At 25°C, [\text{H}_3\text{O}^+] [\text{OH}^-]\text{ should equal 1.0 }\times 10^{-14}. (3.0 \times 10^{-10}) (3.3 \times 10^{-5}) = 9.9 \times 10^{-15}≈ 1 .0 \times 10^{-14}. The significant figures in the concentrations reflect the number of significant figures after the decimal point in the pH and pOH values.