Questions: Weak Acid Ionization

Question 1

A 1.0 M acetic acid solution (Ka = 1.8 × 10⁻⁵) has a percent ionization of about 1.3%. If this solution is diluted to 0.001 M, what happens to the percent ionization?

• A. It decreases to about 0.13% — dilution spreads the same ions over more volume
• B. It stays at 1.3% — the Ka is fixed so the fraction ionized cannot change
• C. It increases to about 13% — dilution shifts equilibrium toward the products side
• D. It increases to 100% — at sufficiently low concentration, weak acids fully ionize

Answer: (C)

This is the key counterintuitive result of weak acid equilibria. Dilution decreases the concentration of all species, and Le Chatelier's principle predicts the system shifts toward the side with more particles (ionized products: H⁺ + A⁻ vs. undissociated HA) to partially restore the lost concentration. The result: a *larger fraction* of acid molecules ionize, even though the absolute [H⁺] decreases. At 1.0 M, only 1.3% ionizes; at 0.001 M, about 13% ionizes. The Ka is constant (not the percent ionization), so as C₀ decreases, [H⁺]eq/C₀ — the percent ionization — must increase.

Question 2

You are setting up an ICE table for 0.10 M acetic acid (Ka = 1.8 × 10⁻⁵). Using the small-x approximation, you find x = 1.34 × 10⁻³ M. Should you accept this approximation?

• A. No — always use the quadratic formula for accuracy
• B. Yes — x/C₀ = 1.34 × 10⁻³ / 0.10 = 1.34%, which is well below the 5% threshold
• C. No — the approximation is only valid when Ka > 10⁻³
• D. Yes — the approximation is always valid for acetic acid regardless of concentration

Answer: (B)

The 5% rule checks whether the approximation introduces unacceptable error: if x/C₀ < 0.05 (5%), the approximation is valid. Here, 1.34 × 10⁻³ / 0.10 = 0.0134 = 1.34%, comfortably below 5%. The approximation (0.10 − x ≈ 0.10) is justified. Option D is wrong because the approximation fails for dilute solutions of the same acid — at 0.001 M acetic acid, x/C₀ ≈ 13%, requiring the quadratic. The validity of the approximation depends on the ratio Ka/C₀, not on the acid identity alone.

Question 3

Diluting a weak acid decreases both the absolute [H⁺] concentration and the percent ionization.

• T. True
• F. False

Answer: False

Dilution decreases the absolute [H⁺] (fewer moles of H⁺ per liter) but increases the percent ionization (a larger fraction of acid molecules have donated a proton). These two effects move in opposite directions. The absolute [H⁺] decreases because even though more molecules ionize proportionally, the total acid concentration has dropped so much that the absolute count falls. The percent ionization increases because Le Chatelier's principle drives the equilibrium toward the products side (more particles) when the solution is diluted. For strong acids, dilution does decrease both — because strong acids are always 100% ionized regardless of concentration.

Question 4

A 'weak acid' is simply a dilute solution of an acid — calling it 'weak' means there isn't much acid present.

• T. True
• F. False

Answer: False

'Weak' refers to the degree of ionization, not the concentration. A weak acid is one that only partially ionizes in water — it establishes an equilibrium between HA and H⁺ + A⁻ with Ka << 1. A strong acid (like HCl) ionizes essentially completely regardless of concentration. You can have a concentrated weak acid (e.g., 5 M acetic acid) or a dilute strong acid (e.g., 0.001 M HCl) — the weakness/strength describes the acid's intrinsic tendency to donate protons, quantified by Ka. This distinction matters because pH calculations for weak vs. strong acids require completely different methods.

Question 5

Explain why percent ionization increases as a weak acid solution is diluted, even though the absolute [H⁺] decreases.

Model answer: Percent ionization is [H⁺]eq / C₀ × 100. When the solution is diluted, C₀ decreases faster than [H⁺]eq does, because the equilibrium shifts right (toward more ionization) in response to dilution. Le Chatelier's principle predicts this shift: dilution reduces the concentration of all species, but the products side (H⁺ + A⁻) has more particles, so shifting toward products partially counteracts the dilution. The equilibrium constant Ka is fixed, so as C₀ falls, a larger fraction of the remaining acid must ionize to satisfy Ka. The absolute [H⁺] does fall (numerator decreases), but the denominator C₀ decreases proportionally more, raising the ratio.

A mathematical way to see this: Ka = [H⁺][A⁻]/[HA] ≈ x²/(C₀ − x) ≈ x²/C₀. Solving: x ≈ √(Ka·C₀). Percent ionization = x/C₀ ≈ √(Ka·C₀)/C₀ = √(Ka/C₀). As C₀ decreases, √(Ka/C₀) increases — percent ionization scales with 1/√C₀. So halving the concentration increases percent ionization by a factor of √2 ≈ 1.41. This is why weak and strong acids behave differently on dilution: for a strong acid (x = C₀ always), percent ionization is always 100% regardless of concentration.