Topic 13 · page 6 of 12

Kc from a titration

A classic experiment finds Kc by letting an ester and water reach equilibrium and then titrating the acid that formed. Subtract the blank for the acid catalyst, then use the amounts in the expression.

14 min read13.3

Read first: Kc from equilibrium amounts

An experiment that measures K 13.3

To get KcK_\text{c} from an experiment you need the amount of at least one substance at equilibrium. For a reaction that makes an acid, titration gives it. The standard example is the hydrolysis of ethyl ethanoate:

CHX3COOCX2HX5(l)+HX2O(l)⇌CHX3COOH(l)+CX2HX5OH(l) \ce{CH3COOC2H5(l) + H2O(l) <=> CH3COOH(l) + C2H5OH(l)}

The reaction is very slow on its own, so a little dilute hydrochloric acid is added as a catalyst. The catalyst only gets the mixture to equilibrium sooner. It does not change KcK_\text{c}.

PracticalHydrolysis of ethyl ethanoate

  1. Mix known amounts of ethyl ethanoate and water, and a measured volume of dilute hydrochloric acid, in a stoppered flask. Record the amounts.
  2. Leave the stoppered flask at constant temperature for about a week, so that equilibrium is reached.
  3. Cool a sample in ice to slow the reaction. Titrate it with standard sodium hydroxide, using phenolphthalein as the indicator.
  4. Titrate the same volume of the hydrochloric acid on its own (a blank).
  5. Subtract the blank from the first titration. The difference is the ethanoic acid formed.

Why each step is there 13.3

The calculation 13.3

Worked exampleEthyl ethanoate hydrolysis

(Illustrative data.) The mixture started with 0.0500 mol ethyl ethanoate and 0.200 mol water, which includes the water in the dilute acid. At equilibrium the whole mixture needs 0.0345 mol of NaOH. The hydrochloric acid catalyst alone needs 0.0050 mol of NaOH. Calculate KcK_\text{c}.

  1. Ethanoic acid at equilibrium = 0.0345 − 0.0050 = 0.0295 mol
  2. The ratio is 1 : 1 : 1 : 1, so ethanol formed = 0.0295 mol too.
  3. Ester left = 0.0500 − 0.0295 = 0.0205 mol. Water left = 0.200 − 0.0295 = 0.1705 mol.
  4. Kc=[CHX3COOH][CX2HX5OH][CHX3COOCX2HX5][HX2O]=0.0295×0.02950.0205×0.1705K_\text{c} = \dfrac{[\ce{CH3COOH}][\ce{C2H5OH}]}{[\ce{CH3COOC2H5}][\ce{H2O}]} = \dfrac{0.0295 \times 0.0295}{0.0205 \times 0.1705} = 0.249, no units

There are two terms on the top and two on the bottom, so the volume cancels and the moles can go straight into the expression. That is why this experiment does not need the volume of the mixture.

When only a sample is titrated 13.3

In many experiments only a fraction of the mixture is taken for the titration. Scale the result up to the whole mixture before using it.

Worked exampleEsterification with a scaled sample

(Illustrative data.) 0.200 mol of ethanoic acid and 0.200 mol of ethanol are mixed with a few drops of concentrated sulfuric acid as catalyst. The water in the catalyst is negligible. When equilibrium is reached, the whole mixture is washed into a volumetric flask with cold water and made up to 250.0 cm³. A 25.0 cm³ sample needs 14.3 cm³ of 0.500 mol dm⁻³ NaOH. A blank shows that the catalyst in the whole mixture needs 0.0050 mol of NaOH. Calculate KcK_\text{c} for CHX3COOH(l)+CX2HX5OH(l)⇌CHX3COOCX2HX5(l)+HX2O(l)\ce{CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l)}.

  1. NaOH in the sample = 0.500 × 0.0143 = 0.00715 mol
  2. Scale up: the sample is one tenth of the mixture, so the whole mixture needs 0.00715 × 10 = 0.0715 mol
  3. Ethanoic acid at equilibrium = 0.0715 − 0.0050 = 0.0665 mol
  4. Ethanoic acid used = 0.200 − 0.0665 = 0.1335 mol. The ratio is 1 : 1 : 1 : 1, so ester = water = 0.1335 mol, and ethanol = 0.200 − 0.1335 = 0.0665 mol.
  5. Kc=[CHX3COOCX2HX5][HX2O][CHX3COOH][CX2HX5OH]=0.1335×0.13350.0665×0.0665K_\text{c} = \dfrac{[\ce{CH3COOC2H5}][\ce{H2O}]}{[\ce{CH3COOH}][\ce{C2H5OH}]} = \dfrac{0.1335 \times 0.1335}{0.0665 \times 0.0665} = 4.03, no units

Check yourself

  1. Explain why a sample of the equilibrium mixture is cooled before the titration.
  2. Why is a blank titration with the hydrochloric acid needed?
  3. In an experiment on the hydrolysis, 0.0300 mol ethyl ethanoate and 0.250 mol water react. The total acid at equilibrium needs 0.0245 mol of NaOH, and the catalyst alone needs 0.0050 mol. Calculate KcK_\text{c}.
  4. In the experiment in question 3, which amounts would change if the blank was forgotten, and in which direction would KcK_\text{c} move?
  5. 0.100 mol of ethanoic acid and 0.100 mol of ethanol are left to reach equilibrium with a catalyst that, with the whole mixture, needs 0.0040 mol of NaOH. The mixture is made up to 100.0 cm³ and a 10.0 cm³ sample needs 8.00 cm³ of 0.500 mol dm⁻³ NaOH. Calculate the amount of ethanoic acid at equilibrium and KcK_\text{c}.
  6. A student says "the volume of the mixture does not matter in this calculation". Is she right? Explain.

Answers cover these until you have tried the questions

  1. Cooling slows the reaction, so the composition does not change while the sample is being titrated. The titration then measures the equilibrium composition.
  2. Sodium hydroxide also neutralises the hydrochloric acid catalyst. The blank measures that acid alone, and subtracting it leaves the NaOH used by the ethanoic acid.
  3. Ethanoic acid = 0.0245 − 0.0050 = 0.0195 mol = ethanol. Ester = 0.0300 − 0.0195 = 0.0105 mol. Water = 0.250 − 0.0195 = 0.2305 mol. KcK_\text{c} = (0.0195 × 0.0195) ÷ (0.0105 × 0.2305) = 3.8025 × 10⁻⁴ ÷ 2.4203 × 10⁻³ = 0.157, no units.
  4. The ethanoic acid would be 0.0245 mol instead of 0.0195 mol, so too large. The ester left and the water left would both be too small. The top of the expression grows and the bottom shrinks, so KcK_\text{c} would be too large: (0.0245² ÷ (0.0055 × 0.2255)) = 0.484.
  5. NaOH in the sample = 0.500 × 0.00800 = 0.00400 mol. Whole mixture = 0.00400 × 10 = 0.0400 mol. Ethanoic acid = 0.0400 − 0.0040 = 0.0360 mol. Used = 0.100 − 0.0360 = 0.0640 mol, so ester = water = 0.0640 mol and ethanol = 0.0360 mol. KcK_\text{c} = (0.0640 × 0.0640) ÷ (0.0360 × 0.0360) = 4.096 × 10⁻³ ÷ 1.296 × 10⁻³ = 3.16
  6. She is right for this equation, because the number of terms on the top (ester and water) equals the number on the bottom (acid and ethanol). The factors of 1/V cancel, so moles can be used. For a reaction such as PCl₅ ⇌ PCl₃ + Cl₂ the volume would matter.