Worked example 1
Find the pH and percent ionization of 0.150 M acetic acid (Ka = 1.8 × 10⁻⁵).
Try it first: Check C/Ka before deciding how to solve.
0 of 4 steps revealed.
What you'll be able to do: Calculate the pH and percent ionization of a weak acid solution using Ka and an ICE table.
Most acids you meet in biology, food and the lab are weak. Their pH is not the concentration written on the bottle, it is the answer to an equilibrium problem.
These are recommended, not required. You can start this lesson at any time.
Acetic acid in water reaches an equilibrium in which most molecules remain intact. The equilibrium constant for that proton transfer is called the acid ionization constant, Ka.
Acetic acid has Ka = 1.8 × 10⁻⁵, so at 0.10 M about 1.3% of the acid molecules are ionized at equilibrium (roughly 1 in 75).
Ka = [H₃O⁺][A-] / [HA]
pKa = -log Ka turns awkward exponents into a readable scale. A smaller pKa means a stronger acid: HF (pKa 3.17) is stronger than acetic acid (4.74), which is far stronger than HCN (9.21).
Let x be the amount of acid that ionizes. Initially [HA] = C and both products are zero; at equilibrium [HA] = C - x and [H₃O⁺] = [A-] = x. Substituting gives Ka = x²/(C - x), an exact quadratic.
For HA + H₂O ⇌ H₃O⁺ + A⁻, Ka ≈ [H₃O⁺][A⁻]/[HA]. With only HA initially, and water autoionization negligible, the table is C, 0, 0 initially; −x, +x, +x in change; C − x, x, x at equilibrium. Rearranging gives x² + Kax − KaC = 0, so x = (−Ka + √(Ka² + 4KaC))/2. The negative root is unphysical. If strong acid or A⁻ was initially added, the product entries no longer both equal x and this particular quadratic must be rebuilt.
Ka = x² / (C - x)
For the simple monoprotic-acid setup HA + H₂O ⇌ H₃O⁺ + A−, C/Ka > 400 is a conservative screen for trying the small-x approximation x = sqrt(Ka × C).
The shortcut depends on the equilibrium expression and initial concentrations, so do not apply that ratio as a universal rule to every ICE table.
Solve with the approximation, then check whether the change is less than 5% of each nonzero initial concentration it replaces; otherwise solve exactly.
Percent ionization = 100 x x/C. Diluting a weak acid raises the percent ionization while lowering [H₃O⁺], which surprises students. Le Chatelier explains it: dilution favours the side with more dissolved particles, which is the ionized side.
Within the small-x regime, x ≈ √(KaC), while the ionized fraction is x/C ≈ √(Ka/C). Lowering C reduces the absolute hydronium concentration but increases the fraction ionized. For example, a hundredfold dilution would lower x by roughly tenfold and raise percent ionization roughly tenfold if both solutions still meet the approximation. When that prediction approaches a large fraction, solve the full equation instead; percent ionization can never exceed 100%. At extreme dilution, include water autoionization as well.
percent ionization = 100 × [H₃O⁺] / C(initial)
H₂CO₃ loses protons in two steps with Ka1 = 4.3 × 10⁻⁷ and Ka2 = 4.7 × 10⁻¹¹. Because Ka2 is roughly ten thousand times smaller, the second step contributes almost nothing to the pH and can normally be ignored.
Successive ionizations have separate equations and constants. If Ka₁ greatly exceeds Ka₂ in an acid-only solution, H₃O⁺ from the first step suppresses the second, often allowing the first-step pH calculation. Check that assumption rather than summing separate pH values. It can fail when constants are not well separated, when the first step is strong as in sulfuric acid, or when a salt starts at an intermediate protonation state. Also note that tabulated “carbonic acid” constants commonly refer to the combined dissolved CO₂/H₂CO₃ system; keep the stated convention.
The estimate x≈sqrt(KaC) assumes x is small compared with starting concentration C. Check x/C, often against a 5% classroom criterion; if it is too large, solve the quadratic x²/(C-x)=Ka. The estimate may give a roughly plausible pH but an inaccurate percent ionization, especially in dilute weak-acid solutions.
The 5% criterion is a classroom accuracy target for the omitted concentration term, not a fundamental discontinuity in nature. At C = 1.00 × 10⁻⁴ M and Ka = 1.00 × 10⁻⁵, the shortcut x = 3.16 × 10⁻⁵ M is 31.6% of C, so retain C − x. The quadratic gives x = 2.70 × 10⁻⁵ M and percent ionization 27.0%. Back-substitute to verify Ka = x²/(C − x). A roughly plausible pH does not prove that the approximate ionization fraction was accurate.
The equilibrium constant for proton transfer from an acid to water.
pKa = -log Ka; smaller pKa means a stronger acid.
For the simple monoprotic setup C/Ka > 400 is a conservative trial screen; validate the computed x/C and the neglect of water and other initial ions.
Rises on dilution even though pH rises too.
Successive steps have separate Ka values. A well-separated first step often dominates an acid-only solution, but that assumption must be checked.
Ka = [H₃O⁺][A-] / [HA]
[H₃O⁺] ≈ sqrt(Ka x C)
percent = 100 × [H₃O⁺] / C
Find the pH and percent ionization of 0.150 M acetic acid (Ka = 1.8 × 10⁻⁵).
Try it first: Check C/Ka before deciding how to solve.
0 of 4 steps revealed.
A 0.0100 M solution of a weak acid has pH 3.40. Find Ka.
Try it first: Convert the pH into x, then read the ICE table backwards.
0 of 3 steps revealed.
Compare the percent ionization of 0.100 M and 0.00100 M HF (Ka = 6.8 × 10⁻⁴).
Try it first: Predict the direction of the change with Le Chatelier before calculating.
0 of 3 steps revealed.
For HA with C=1.00×10⁻⁴ M and Ka=1.00×10⁻⁵, find the percent ionization.
Try it first: First test sqrt(KaC) against C.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Setting [H₃O⁺] equal to the acid concentration. | That is true only for strong acids. | Use Ka and an ICE table for anything weak. |
| Using the small-x shortcut without checking. | For acids with large Ka or low C, it can be off by tens of percent. | Apply the 5 percent rule and solve the quadratic if it fails. |
| Confusing percent ionization with acid strength. | Percent ionization depends on concentration; Ka does not. | Compare Ka or pKa when ranking acids. |
| Assuming a larger pKa means a stronger acid. | The log is negative, so the ordering reverses. | Smaller pKa is stronger. |
| Adding the pH of the second ionization step for a polyprotic acid. | pH values do not add, and Ka2 is usually negligible anyway. | Solve the first step and check whether the second contributes. |
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After the lesson · Question bank
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A weak acid only partially transfers its proton, so HA + H₂O ⇌ H₃O⁺ + A− has Ka = [H₃O⁺][A−]/[HA]. A larger Ka, or smaller pKa, means a stronger weak acid. Use an ICE table to find pH. For a simple monoprotic acid HA starting at concentration C with negligible initial products, C/Ka > 400 is a conservative screen for trying x ≈ sqrt(Ka × C); verify the change is under 5% of C, or solve the quadratic. Other equilibrium stoichiometries and starting mixtures need their own check. Percent ionization is 100 × [H₃O⁺]/C(initial), and dilution can increase percent ionization even as pH rises.
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