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A proton transfer favours the side with the weaker acid, and its equilibrium constant is ten to the power of the difference in pKa between the acid formed and the acid reacting; the hydrogencarbonate test uses exactly this.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will calculate the equilibrium constant of a proton transfer, say which side is favoured, match reactions to their direction, order the steps of a prediction, and explain the hydrogencarbonate test.
You can compare acids by pKa and explain the differences by resonance, induction and hybridization. This lesson uses the numbers to predict reactions.
In a proton transfer, an acid gives H+ to a base, making the acid's conjugate base and the base's conjugate acid. The favoured side is the side the equilibrium lies towards.
Write any proton transfer and there is an acid on each side:
$$\mathrm{CH_3COOH + OH^- \rightleftharpoons CH_3COO^- + H_2O}$$
Left acid: ethanoic acid, pKa 5. Right acid: water, pKa 16. The equilibrium favours the side with the weaker acid (higher pKa), here the right. The constant is
$$K = 10^{\,\text{pKa(acid formed)} - \text{pKa(acid reacting)}} = 10^{16-5} = 10^{11}$$
Run the same logic backwards: ethyne (25) with hydroxide gives water (16), a stronger acid, so $K = 10^{16-25} = 10^{-9}$ and almost nothing happens.
Another way: steps
To predict a proton transfer:
Sodium hydrogencarbonate is a mild base whose conjugate acid, carbonic acid, has pKa 6.4 and falls apart into water and carbon dioxide. Any acid stronger than 6.4 gives it a proton and the mixture fizzes; any weaker acid does not. So a carboxylic acid (pKa about 4–5) fizzes and a phenol (about 10) does not — a quick way to tell the two apart, and exactly the reason baking soda neutralises vinegar.
A proton transfer always goes toward the stronger acid. It goes toward the weaker one.
Compare the bases' pKa values. Compare the two acids.
K is the pKa difference. It is ten to that power.
Any base removes any proton. Only if its conjugate acid is weaker.
Left acid: phenol, pKa 10. Right acid: water, pKa 16.
One acid on each side.
Water is weaker, so the right side is favoured.
Weaker acid's side.
$K = 10^{16-10} = 10^6$: phenol dissolves in sodium hydroxide as phenoxide.
Ten to the difference.
Which acid is on each side?
Ethanol on the left, ammonia on the right.
Which side is favoured, and what is K?
The right side, with $K = 10^{22}$.
propanone (pKa about $20$) gives a proton to the conjugate base of ethyne, forming ethyne (pKa about $25$). What is the equilibrium constant for the transfer?
Answer:
In the equilibrium HC≡CH + (conjugate base of NH3) ⇌ (conjugate base of HC≡CH) + NH3, the pKa of HC≡CH is about $25$ and of NH3 about $38$. Which side is favoured?
Match each proton transfer to the side that is favoured. (pKa: ethanoic acid 5, water 16, ethanol 16, ethyne 25, ammonia 38.)
| products favoured | reactants favoured | about equal | |
|---|---|---|---|
| ethanoic acid + OH− ⇌ ethanoate + H2O | |||
| ethyne + OH− ⇌ acetylide + H2O | |||
| ethyne + NH2− ⇌ acetylide + NH3 | |||
| ethanol + OH− ⇌ ethoxide + H2O |
Put the steps for predicting the direction of a proton transfer in order.
Number the steps in order (write the number in the box):
Sodium hydrogencarbonate solution fizzes with benzoic acid (pKa 4.2) but not with phenol (pKa 10). Carbonic acid, formed when hydrogencarbonate takes a proton, has pKa 6.4 and then releases CO2. Why the difference?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An acid with pKa $12$ is mixed with the conjugate base of an acid with pKa $9$. What is the equilibrium constant for the proton transfer?
The equilibrium constant for the transfer is a.
You can predict which way a proton moves and by how much. Tell someone why phenol does not fizz with baking soda. Next unit: how electrons move in reactions — nucleophiles, electrophiles and curly arrows.
8. Your turn: ethanol (16) with the amide ion (ammonia 38), step 3