How pH relates to pKa
Rearranging the Ka expression gives:
So comparing the pH of the solution with the pKa of the weak acid tells you which form dominates:
- pH < pKa: [HA] > [A⁻], so the protonated form酸式 dominates.
- pH = pKa: [HA] = [A⁻], half each (the half-equivalence point of a titration, U8.5).
- pH > pKa: [A⁻] > [HA], so the deprotonated form碱式 dominates.
Each unit that the pH moves away from pKa multiplies the ratio by 10: at pH = pKa + 1, A⁻ is about 91%, and at pH = pKa + 2, about 99%.
Here the pH is set by other substances in the solution (a buffer, for example), and the weak acid is just one component. Drag the pH and watch the fraction of each form change; for a polyprotic acid, each time the pH passes a pKa, the main form has one fewer H⁺:
Indicators
An acid–base indicator指示剂 is itself a weak acid, HIn, whose HIn and In⁻ forms have different colors. When the pH is more than 1 below the indicator’s pKa, HIn dominates and you see the acid color; when it is more than 1 above, In⁻ dominates and you see the base color. So an indicator changes color over a range of about pKa ± 1.
Example 1: HF at pH 4.00
A solution contains a small amount of HF (Ka = 6.4 × 10⁻⁴), and its pH is held at 4.00. Which form of HF dominates? What fraction is HF molecules?
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Calculate the pKa.
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Compare the pH with the pKa to find the main form. pH 4.00 > pKa 3.19: there is less H₃O⁺ than at the “half-ionized” point, so the equilibrium lies on the deprotonated side, and F⁻ dominates.
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Get the ratio from the Ka expression. Rearrange the Ka expression and take logs:
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Substitute the values.
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Convert to percentages. Count [HF] as 1 part, so [F⁻] is 6.5 parts, 7.5 parts in all:
Example 2: phosphate at pH 7.40
For H₃PO₄, pKa₁ = 2.12, pKa₂ = 7.21 and pKa₃ = 12.38 (calculated from the OpenStax Ka values). In a solution at pH 7.40, what is the main phosphorus-containing form, and what is the ratio?
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Line up the three ionizations with their pKa values. Each time the pH rises past a pKa, the main form loses one more proton:
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Find the interval. 7.21 < 7.40 < 12.38: the pH is between pKa₂ and pKa₃, but very close to pKa₂ (only 0.19 away) and far from pKa₃.
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Find the relevant pair. The nearest pKa is pKa₂, which belongs to H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻, so both of these forms are present in large amounts.
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Find the ratio.
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Convert to percentages. Counting only these two forms, there are 1 + 1.55 = 2.55 parts:
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Conclusion. Both HPO₄²⁻ and H₂PO₄⁻ are present in large amounts, with slightly more HPO₄²⁻; H₃PO₄ (the pH is more than 5 units above pKa₁) and PO₄³⁻ (about 5 units below pKa₃) are negligible.
- CH₃COOH (pKa 4.74), pH 3.0
pH < pKa.
- CH₃COOH (pKa 4.74), pH 7.0
pH > pKa, so CH₃COO⁻ dominates.
- NH₄⁺ (pKa 9.26), pH 7.0
pH < pKa, so NH₄⁺ dominates.
- NH₄⁺ (pKa 9.26), pH 11.0
pH > pKa, so NH₃ dominates.
- HClO (pKa 7.54), pH 7.54
pH = pKa.
- HF (pKa 3.19), pH 2.0
pH < pKa.
- HCN (pKa 9.31), pH 8.0
pH < pKa, so HCN dominates.
- HCOOH (pKa 3.74), pH 5.0
pH > pKa, so HCOO⁻ dominates.
Common mistakes
Mistake 1: “If the solution is acidic (pH < 7), the weak acid is mostly HA.” Compare with the pKa of that acid, not with 7. At pH 5.0, HCOOH (pKa 3.74) is mostly HCOO⁻.
Mistake 2: “When the pH is 1 above pKa, there is twice as much A⁻ as HA.” The ratio is 10pH − pKa: 1 unit above means 10 times as much A⁻ as HA.
Mistake 3: “At pH = pKa, [H₃O⁺] is half the acid concentration.” pH = pKa means [HA] = [A⁻]; then [H₃O⁺] = Ka, whatever the total concentration of the acid.
Quick check
A solution at pH 5.74 contains a little CH₃COOH (pKa 4.74). What is [CH₃COO⁻] ÷ [CH₃COOH]?
The ratio = 105.74 − 4.74 = 10.
An indicator HIn has pKa = 9.0; HIn is colorless and In⁻ is red. When it is added to a solution at pH 7.0, the solution is:
The pH is 2 below pKa, so [In⁻] ÷ [HIn] = 0.01 and nearly all of it is colorless HIn.
For H₂CO₃, pKa₁ = 6.37 and pKa₂ = 10.33. At pH 8.3, the main carbon-containing form is:
The pH is between the two pKa values and more than 1 from each, so HCO₃⁻ dominates.
For a monoprotic weak acid at pH = pKa, which statement is correct?
The ratio 10pH − pKa = 1, so [HA] = [A⁻].
A solution at pH 4.00 contains a weak acid with pKa = 6.00. About what percentage of it is present as A⁻?
[A⁻] ÷ [HA] = 10−2 = 0.01, so A⁻ is about 0.01 ÷ 1.01 ≈ 1%.
For H₃PO₄, pKa₃ = 12.38. At pH 12.38, which two forms have equal concentrations?
pKa₃ belongs to HPO₄²⁻ ⇌ H⁺ + PO₄³⁻, so the two are equal when pH = pKa₃.
AP-style free response
Swimming pools are often disinfected with chlorine compounds. The main disinfecting species is hypochlorous acid, HClO (Ka = 2.9 × 10⁻⁸, OpenStax), which is a much more effective disinfectant than ClO⁻.
(a) Write the equation for the ionization of HClO, and calculate its pKa.
(b) When the pool water has pH 7.00, calculate [ClO⁻] ÷ [HClO].
(c) At pH 7.00, what percentage of the chlorine species (HClO + ClO⁻) is HClO?
(d) If the pH of the pool rises to 8.00, what percentage is HClO?
(e) Using your answers to (c) and (d), explain why the pH of the pool water shouldn’t be allowed to get too high.
Model answer and scoring
(a) HClO(aq) + H₂O(l) ⇌ H₃O⁺(aq) + ClO⁻(aq); pKa = −log(2.9 × 10⁻⁸) = 7.54.
(b) [ClO⁻] ÷ [HClO] = 107.00 − 7.54 = 10−0.54 = 0.29.
(c) 1 ÷ (1 + 0.29) = 78%.
(d) The ratio = 108.00 − 7.54 = 2.9, so HClO is 1 ÷ (1 + 2.9) = 26%.
(e) The higher the pH (the further above pKa), the more the equilibrium favors ClO⁻, so the more effective disinfectant, HClO, makes up a smaller share. When the pH rises from 7.00 to 8.00, the share of HClO falls from 78% to 26%, and disinfection becomes much less effective.
Scoring: (a) equation and pKa; (b) ratio; (c) percentage; (d) new ratio and percentage; (e) explanation using the relationship between pH and pKa.
Summary
- [A⁻] ÷ [HA] = 10pH − pKa: the protonated form dominates when pH < pKa, the deprotonated form when pH > pKa, and they are equal when pH = pKa.
- Each unit that pH moves from pKa multiplies the ratio by 10; at 2 units away, about 99% is one form.
- For a polyprotic acid, use pKa₁, pKa₂, … to find the main form in each range; an indicator changes color over about pKa ± 1.
AP tip: When asked which form dominates, write the comparison: “pH (…) > pKa (…), so A⁻ dominates.” For a quantitative answer, find the ratio with 10pH − pKa, then convert it to a percentage.