Topic 14 · page 10 of 18

Four monoprotic curves

Strong or weak acid, strong or weak base, gives four different curves. The start pH, the equivalence pH and the size of the vertical section all depend on which partner is weak.

15 min read14.15

Read first: Reading a titration curve

Four combinations 14.15

A monoprotic acid gives one H⁺ per molecule, and a monobasic base takes one. Each can be strong or weak, so there are four combinations. All four use 25.0 cm³ of 0.100 mol dm⁻³ solution in the flask and 0.100 mol dm⁻³ titrant from the burette.

Combination Example Start pH Equivalence volume Equivalence pH Vertical section
strong acid + strong base HCl + NaOH 1.00 25.0 cm³ 7.00 about 3 to 11
weak acid + strong base CH₃COOH + NaOH 2.88 25.0 cm³ 8.73 about 6.5 to 11
strong acid + weak base HCl added to NH₃ 11.12 25.0 cm³ 5.28 about 7.6 to 3
weak acid + weak base CH₃COOH + NH₃ 2.88 25.0 cm³ 7.00 none

"Vertical section" here is the pH range within ±0.5 cm³ of the equivalence volume. If you reverse the roles (for example, acid added to alkali), the curve is the mirror image and the same ideas apply.

Weak acid with strong base 14.15

A titration curve for 25.0 cm³ of 0.100 mol per cubic decimetre ethanoic acid with 0.100 mol per cubic decimetre sodium hydroxide. It starts at pH 2.88, rises slowly through a flat region passing pH 4.76 at 12.5 cm³, jumps from about pH 6.5 to 11 around 25.0 cm³, and levels off near pH 12.5. The equivalence point is at 25.0 cm³ and pH 8.73.
Weak acid with strong base. The pH at half the equivalence volume equals the pKa.

Compare it with the strong acid curve on the last page.

At equivalence the flask contains only sodium ethanoate. The ethanoate ion is the conjugate base of a weak acid, so it takes protons from water and makes OH⁻:

CHX3COOX−(aq)+HX2O(l)⇌CHX3COOH(aq)+OHX−(aq) \ce{CH3COO-(aq) + H2O(l) <=> CH3COOH(aq) + OH-(aq)}

That is why a solution of the salt is alkaline and the equivalence point is above pH 7.

Strong acid with weak base 14.15

A falling titration curve for 25.0 cm³ of 0.100 mol per cubic decimetre ammonia with 0.100 mol per cubic decimetre hydrochloric acid. It starts at pH 11.12, passes pH 9.25 at 12.5 cm³, drops from about pH 7.6 to 3.0 around 25.0 cm³, and levels off near pH 1.5. The equivalence point is at 25.0 cm³ and pH 5.28.
Strong acid added to a weak base. The equivalence point is below 7.

This is the mirror image of the last case. The start is lower than for a strong base (11.12 instead of 13.00), the vertical section is short and low (about pH 7.6 to 3), and the equivalence point is below 7, at 5.28. At equivalence the flask contains only ammonium chloride, and NH₄⁺ is a weak acid:

NHX4X+(aq)+HX2O(l)⇌NHX3(aq)+HX3OX+(aq) \ce{NH4+(aq) + H2O(l) <=> NH3(aq) + H3O+(aq)}

The flat region before equivalence is a buffer of NH₃ and NH₄⁺. At half the equivalence volume the pH is 9.25, the pKa\text{p}K_\text{a} of NH₄⁺.

Weak acid with weak base 14.15

A titration curve for 25.0 cm³ of 0.100 mol per cubic decimetre ethanoic acid with 0.100 mol per cubic decimetre ammonia. It starts at pH 2.88 and rises in an S shape that never becomes vertical. At the equivalence point, 25.0 cm³, the pH is 7.00, and the curve levels off near pH 9.25.
Weak acid with weak base. There is no vertical section, so no indicator will give a sharp end point.

Both partners are weak, so nothing in the flask resists the change at equivalence, and the pH moves smoothly through it. There is no vertical section. The equivalence pH of 7.00 is a coincidence of this pair: the ethanoate ion and the ammonium ion are equally weak as a base and an acid, so their effects cancel.

Check yourself

  1. Describe the main features of the curve for 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid titrated with 0.100 mol dm⁻³ NaOH.
  2. Explain why the equivalence point of a weak acid with a strong base is above pH 7.
  3. Explain why the equivalence point of a strong acid with a weak base is below pH 7.
  4. Which combination of acid and base gives no vertical section, and what does that mean for the choice of indicator?
  5. List three differences and one similarity between the curves for 0.100 mol dm⁻³ HCl and 0.100 mol dm⁻³ ethanoic acid, each titrated with 0.100 mol dm⁻³ NaOH.
  6. Calculate the volume of 0.0500 mol dm⁻³ NaOH needed to reach equivalence with 20.0 cm³ of 0.100 mol dm⁻³ ethanoic acid.

Answers cover these until you have tried the questions

  1. It starts at pH 2.88, rises quickly at first, then slowly through a flat (buffer) region. There is a vertical section from about pH 6.5 to 11 around 25.0 cm³, with the equivalence point at pH 8.73. Above 25 cm³ it levels off near pH 12.5.
  2. At equivalence the flask contains the salt of a weak acid and a strong base, here sodium ethanoate. The ethanoate ion is a weak base and accepts H⁺ from water, making OH⁻. So [OH⁻] > [H⁺] and the pH is above 7.
  3. At equivalence the flask contains the salt of a strong acid and a weak base, here ammonium chloride. The ammonium ion is a weak acid and donates H⁺ to water, making H₃O⁺. So [H⁺] > [OH⁻] and the pH is below 7.
  4. Weak acid with weak base. The pH changes smoothly through equivalence, so no indicator gives a sharp end point. A pH meter is used instead.
  5. Differences: the weak acid starts at a higher pH (2.88 against 1.00); its vertical section is shorter and higher (about 6.5 to 11 against 3 to 11); its equivalence pH is above 7 (8.73 against 7.00). It also has a buffer region before equivalence. Similarity: both need 25.0 cm³ of NaOH to reach equivalence, and after equivalence both curves are identical, set by the excess NaOH.
  6. Moles of CH₃COOH = 0.100 × 0.0200 = 2.00 × 10⁻³ mol. The ratio is 1 : 1, so moles of NaOH = 2.00 × 10⁻³ mol. Volume = 2.00 × 10⁻³ ÷ 0.0500 = 0.0400 dm³ = 40.0 cm³.