You met pH and pOH as a scale from 0 to 14. Here we compute them exactly with logarithms, starting with strong acids and bases, which ionize completely.
Key idea
$\text{pH} = -\log[\mathrm{H_3O^+}]$, $\text{pOH} = -\log[\mathrm{OH^-}]$, and at $25\ ^\circ\text{C}$, $[\mathrm{H_3O^+}][\mathrm{OH^-}] = K_w = 1.0 \times 10^{-14}$, so $\text{pH} + \text{pOH} = 14.00$. A strong acid gives $[\mathrm{H_3O^+}]$ equal to its concentration; a strong base gives $[\mathrm{OH^-}]$ equal to its concentration times the number of $\mathrm{OH^-}$ per formula unit.
Why it works
Water ionizes slightly: $\mathrm{2H_2O \rightleftharpoons H_3O^+ + OH^-}$, with $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]$. Taking $-\log$ of both sides turns the product into a sum: $\text{pH} + \text{pOH} = \text{p}K_w = 14.00$. The strong acids ($\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$, $\mathrm{HNO_3}$, $\mathrm{HClO_4}$ and the first proton of $\mathrm{H_2SO_4}$) give up their protons completely, so no equilibrium calculation is needed. The strong bases are the Group 1 hydroxides and $\mathrm{Ca(OH)_2}$, $\mathrm{Sr(OH)_2}$ and $\mathrm{Ba(OH)_2}$, which release two $\mathrm{OH^-}$ each.
Worked examples
Worked example: A strong acid
Find the pH of $0.0050\ \text{M}$ $\mathrm{HNO_3}$.
- Complete ionization: $[\mathrm{H_3O^+}] = 0.0050\ \text{M}$.
- $\text{pH} = -\log(0.0050) = 2.30$. Two significant figures in the concentration give two decimal places in the pH.
Answer: $\text{pH} = 2.30$.
Worked example: Two hydroxides per formula unit
Find the pH of $0.0020\ \text{M}$ $\mathrm{Ba(OH)_2}$.
- Each $\mathrm{Ba(OH)_2}$ gives two $\mathrm{OH^-}$: $[\mathrm{OH^-}] = 2(0.0020) = 0.0040\ \text{M}$.
- $\text{pOH} = -\log(0.0040) = 2.40$; $\text{pH} = 14.00 - 2.40 = 11.60$.
Answer: $\text{pH} = 11.60$.
Worked example: A very dilute acid
What is the pH of $1.0 \times 10^{-8}\ \text{M}$ HCl? (It is not 8.00.)
- An acid cannot make water basic. At this dilution water's own ionization matters: the charge balance is $$[\mathrm{H_3O^+}] = [\mathrm{Cl^-}] + [\mathrm{OH^-}] = 1.0 \times 10^{-8} + \dfrac{K_w}{[\mathrm{H_3O^+}]}$$
- With $h = [\mathrm{H_3O^+}]$: $h^2 - (1.0 \times 10^{-8})h - 1.0 \times 10^{-14} = 0$, so $h = 1.05 \times 10^{-7}\ \text{M}$ and $\text{pH} = 6.98$.
Answer: $\text{pH} = 6.98$, just on the acidic side of neutral.
$K_w$ depends on temperature, so "neutral" is pH 7.00 only at $25\ ^\circ\text{C}$. Neutral always means $[\mathrm{H_3O^+}] = [\mathrm{OH^-}]$:
| Temperature | $K_w$ | pH of pure water |
|---|---|---|
| 0 °C | $1.1 \times 10^{-15}$ | 7.48 |
| 25 °C | $1.0 \times 10^{-14}$ | 7.00 |
| 37 °C | $2.4 \times 10^{-14}$ | 6.81 |
| 50 °C | $5.5 \times 10^{-14}$ | 6.63 |
| 100 °C | $5.1 \times 10^{-13}$ | 6.15 |
Common mistake
Dropping the minus sign (pH would come out negative for any ordinary acid) or forgetting that $\mathrm{Ba(OH)_2}$ gives two hydroxide ions. Check: an acid must give pH below 7 (at 25 °C), a base above, and doubling a concentration changes the pH by only 0.30.
See the scale with real liquids in the Mixing Lab, where you can test kitchen liquids with four indicators.