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How are pH and pOH defined and calculated for strong acids and bases, and how do they relate through the water equilibrium?

Topic 8.2 pH and pOH of Strong Acids and Bases: calculate pH and pOH from concentration for strong acids and bases, using the autoionisation of water and the relationship pH plus pOH equals 14 at 25 degrees Celsius.

A focused answer to AP Chemistry Topic 8.2, covering the definitions of pH and pOH, the autoionisation of water and Kw, the relationship pH plus pOH equals 14 at 25 degrees Celsius, and calculating pH for strong acids and bases, with full worked examples.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. pH, pOH and the autoionisation of water
  3. The pH plus pOH relationship
  4. Strong acids and bases
  5. Try this

What this topic is asking

The College Board (Topic 8.2) wants you to calculate pH and pOH from concentration for strong acids and bases, using the autoionisation of water (KwK_w) and the relationship pH plus pOH equals 14 at 25 degrees Celsius. Because strong acids and bases ionize completely, these calculations are direct.

pH, pOH and the autoionisation of water

The pH scale compresses the wide range of hydronium concentrations into a convenient logarithmic scale. A lower pH means a higher hydronium concentration (more acidic). Water's slight autoionisation links the hydronium and hydroxide concentrations through KwK_w.

The pH plus pOH relationship

This relationship lets you move freely between the acid and base scales. Note it holds at 25 ∘C25\ ^\circ\text{C}; at other temperatures KwK_w differs, so the sum differs from 14, though the AP exam usually works at 25 ∘C25\ ^\circ\text{C}.

Strong acids and bases

Because a strong acid ionizes completely, [H3O+][\text{H}_3\text{O}^+] equals the initial acid concentration (for a monoprotic acid like HCl), so pH =−log⁡(concentration)= -\log(\text{concentration}). For a strong base, [OH−][\text{OH}^-] equals the base concentration times the number of hydroxide ions per formula unit (for example, Ca(OH)2\text{Ca(OH)}_2 gives two), so you find pOH first and then pH =14.00−pOH= 14.00 - \text{pOH}. The complete ionization of strong acids and bases is what makes these calculations a one-step substitution.

The common strong acids worth recognizing are hydrochloric, hydrobromic, hydroiodic, nitric, perchloric and sulfuric acids, and the common strong bases are the hydroxides of the group 1 metals and the heavier group 2 metals. For these, the assumption of complete ionization is excellent, so the hydronium or hydroxide concentration follows directly from the formula and the concentration. There is one further subtlety: in extremely dilute solutions (around 10−710^{-7} M or lower), the small amount of hydronium produced by the autoionisation of water becomes significant and you can no longer take [H3O+][\text{H}_3\text{O}^+] to be exactly the acid concentration. On the AP exam the concentrations are almost always high enough that this correction is unnecessary, but it is a reminder that the simple formula relies on the acid contribution dominating the water's own ionization.

Try this

Q1. Calculate the pH of a 0.00100.0010 M HNO3_3 solution (a strong acid). [2 points]

  • Cue. [H3O+]=0.0010[\text{H}_3\text{O}^+] = 0.0010 M; pH =−log⁡(0.0010)=3.00= -\log(0.0010) = 3.00.

Q2. A solution has pOH =4.0= 4.0 at 25 ∘C25\ ^\circ\text{C}. Calculate its pH. [1 point]

  • Cue. pH =14.0−4.0=10.0= 14.0 - 4.0 = 10.0.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). A solution is made 0.01000.0100 M in HCl, a strong acid. (a) Calculate the [H3O+][\text{H}_3\text{O}^+] and the pH. (b) Calculate the [OH−][\text{OH}^-] using Kw=1.0×10−14K_w = 1.0 \times 10^{-14}. (c) Calculate the pOH and verify that pH plus pOH equals 14. (d) Justify why [H3O+][\text{H}_3\text{O}^+] equals the HCl concentration for a strong acid.
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A 4-point quantitative FRQ on pH of a strong acid.

(a) pH (1 point): HCl is strong, so [H3O+]=0.0100[\text{H}_3\text{O}^+] = 0.0100 M; pH =−log⁡(0.0100)=2.00= -\log(0.0100) = 2.00.
(b) Hydroxide (1 point): [OH−]=Kw[H3O+]=1.0×10−140.0100=1.0×10−12[\text{OH}^-] = \dfrac{K_w}{[\text{H}_3\text{O}^+]} = \dfrac{1.0 \times 10^{-14}}{0.0100} = 1.0 \times 10^{-12} M.
(c) pOH (1 point): pOH =−log⁡(1.0×10−12)=12.00= -\log(1.0 \times 10^{-12}) = 12.00; pH + pOH =2.00+12.00=14.00= 2.00 + 12.00 = 14.00, as required.
(d) Justify (1 point): a strong acid ionizes completely, so each mole of HCl gives one mole of H3O+\text{H}_3\text{O}^+; therefore [H3O+][\text{H}_3\text{O}^+] equals the initial HCl concentration.

Markers reward the pH, the hydroxide concentration from KwK_w, the pOH with the sum check, and the complete-ionization reasoning.

AP 2021 (style)1 marksSection I (multiple choice). A solution has pH =11.0= 11.0 at 25 ∘C25\ ^\circ\text{C}. Its [H3O+][\text{H}_3\text{O}^+] is (A) 1.0×10−31.0 \times 10^{-3} M (B) 1.0×10−111.0 \times 10^{-11} M (C) 11.011.0 M (D) 1.0×1031.0 \times 10^{3} M. Justify your choice.
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A 1-point conceptual MCQ. The answer is (B).

pH =−log⁡[H3O+]= -\log[\text{H}_3\text{O}^+], so [H3O+]=10−pH=10−11.0=1.0×10−11[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-11.0} = 1.0 \times 10^{-11} M. The trap is (A), which is the hydroxide concentration (since pOH =3.0= 3.0), not the hydronium concentration.

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