Back to the on-screen lesson ·
A weak acid and its conjugate base together: how each absorbs one kind of intruder, the Henderson–Hasselbalch equation, and a buffer's limited capacity.
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.
By the end of this lesson you will be able to explain how a buffer absorbs added acid and added alkali, calculate a buffer's pH from its $pK_a$ and the ratio of its partners before and after an addition, recognise which mixtures are buffers, give the pH range a buffer covers, and say when a buffer's capacity has been exceeded.
Last lesson the weak-acid titration curve had a long, almost flat stretch before equivalence, where the flask held both the acid and its conjugate base, and halfway along it the pH equalled the $pK_a$. From the strong and weak acids lesson you have $K_a$, and from Chemistry 1 the idea that amounts in the same volume can be compared directly. A buffer is that flat stretch, bottled.
A buffer is a solution that keeps its pH nearly constant when small amounts of acid or alkali are added. An acidic buffer is a weak acid with its conjugate base, usually supplied as a sodium salt; a basic buffer is a weak base with its conjugate acid, such as ammonia with ammonium chloride. The Henderson–Hasselbalch equation gives a buffer's pH from its $pK_a$ and the ratio of its two partners. Buffer capacity is how much acid or alkali a buffer can absorb before its pH changes sharply.
Take ethanoic acid and sodium ethanoate, both at $0.10$ mol/L. The solution contains plenty of weak acid, $\mathrm{CH_3COOH}$, and plenty of its conjugate base, $\mathrm{CH_3COO^-}$.
Add a little hydrochloric acid. The added hydronium ions do not stay free; the conjugate base takes them:
$$\mathrm{CH_3COO^- + H_3O^+ \rightarrow CH_3COOH + H_2O}$$
Add a little sodium hydroxide instead. The acid takes the hydroxide:
$$\mathrm{CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O}$$
Either way the intruder is turned into more of one partner and less of the other. So what does the pH depend on? Rearranging $K_a$ and taking logarithms gives the Henderson–Hasselbalch equation:
$$\mathrm{pH} = pK_a + \log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$$
The pH follows the ratio of conjugate base to acid, and a small addition moves that ratio only a little. With $0.010$ mol of each in a litre, adding $0.001$ mol of hydrochloric acid changes the ratio from $10:10$ to $9:11$; the pH falls from $4.76$ to $4.67$. The same acid added to a litre of pure water would take the pH from 7 to 3.
Three ratios are worth knowing without a calculator:
| $[\mathrm{A^-}] : [\mathrm{HA}]$ | $\log_{10}$ of the ratio | pH |
|---|---|---|
| $1 : 1$ | $0$ | $pK_a$ |
| $10 : 1$ | $+1$ | $pK_a + 1$ |
| $1 : 10$ | $-1$ | $pK_a - 1$ |
Because both partners are in the same solution, the ratio of their concentrations equals the ratio of their amounts, so buffer arithmetic can be done in moles or millimoles without dividing by the volume.
Another way: picture
Think of two goalkeepers guarding one goal, one for each kind of shot. Hydronium ions are stopped by the conjugate base, hydroxide ions by the acid. Every save costs the keeper some energy — the partner is used up a little — and a keeper who has made enough saves collapses, after which every shot goes in.
Another way: steps
To work out a buffer's pH after an addition:
A buffer has limits. It can absorb added acid only while it has conjugate base left, and added alkali only while it has acid left. A buffer with $5$ mmol of ethanoate can take up $5$ mmol of hydronium; the sixth millimole stays free and the pH plunges. More concentrated buffers have more capacity, and a buffer is strongest when the two partners are equal, because then it is equally far from running out of either.
A buffer is useful while the ratio stays between about $1:10$ and $10:1$ — beyond that, one partner is too scarce to absorb much. That puts its working range at $pK_a \pm 1$. So a chemist who needs a buffer at pH 4.5 chooses an acid with a $pK_a$ near 4.5 — ethanoic acid at $4.76$ is close — and sets the ratio to fine-tune it. A buffer at pH 9 would use ammonia and ammonium chloride instead, since ammonium's $pK_a$ is $9.25$.
Blood must stay between pH 7.35 and 7.45. Its main buffer is carbonic acid, from dissolved carbon dioxide, with hydrogencarbonate ions:
$$\mathrm{H_2CO_3 + H_2O \rightleftharpoons H_3O^+ + HCO_3^-}$$
At body temperature the effective $pK_a$ is about $6.1$, so at pH 7.40 the ratio of hydrogencarbonate to carbonic acid is about $20:1$ — typically $24$ mmol/L of hydrogencarbonate to $1.2$ mmol/L of carbonic acid. That is outside the ideal range for a closed buffer, and it works anyway because the system is open: the lungs breathe out extra carbon dioxide, removing carbonic acid, and the kidneys adjust the hydrogencarbonate. Holding your breath lets carbon dioxide build up and pushes the pH down; breathing very fast pushes it up.
A buffer prevents every pH change. It makes a change small, not zero, and only while it has capacity. Past that, the pH moves as sharply as in water.
A buffer is a solution at pH 7. A buffer can be made at almost any pH; its pH is set by the $pK_a$ and the ratio.
Any acid with its salt is a buffer. Hydrochloric acid with sodium chloride is not: chloride cannot take a proton back. The acid must be weak.
The ratio goes acid over base. In the Henderson–Hasselbalch equation the conjugate base is on top. More base means a higher pH.
Diluting a buffer changes its pH a lot. Dilution changes both concentrations by the same factor, so the ratio, and the pH, barely change; what falls is the capacity.
A buffer has $12$ mmol of ethanoic acid and $12$ mmol of sodium ethanoate, pH $4.76$. $4$ mmol of hydrochloric acid is added.
Amounts first.
The ethanoate takes the hydronium: ethanoate $12 - 4 = 8$ mmol, ethanoic acid $12 + 4 = 16$ mmol.
The added acid reacts completely with the conjugate base.
Ratio $8 : 16 = 0.5$, and $\log_{10} 0.5 = -0.30$, so the pH is $4.76 - 0.30 = 4.46$.
A drop of 0.30 where water would have dropped by several units.
A buffer is needed at pH 5.20, and the candidates are methanoic acid ($pK_a$ 3.75) and ethanoic acid ($pK_a$ 4.76).
The target must be within one unit of the $pK_a$.
Methanoic acid covers 2.75 to 4.75, which misses 5.20; ethanoic acid covers 3.76 to 5.76, which includes it.
Choose the acid whose range contains the target.
$5.20 - 4.76 = 0.44$, so the ratio of ethanoate to ethanoic acid is $10^{0.44} \approx 2.75$.
More base than acid, because the target is above the $pK_a$.
Ratio of base to acid: $3 : 30$, a tenth, whose logarithm is $-1$.
Conjugate base on top.
So the pH is $4.20 - 1 = \ldots$
…$3.20$, at the very edge of its range; with only $3$ mmol of benzoate it can absorb very little added acid.
A buffer is made from $3$ mmol of ethanoic acid ($pK_a = 4.76$) and $30$ mmol of sodium ethanoate in the same solution. What is its pH?
Answer:
A buffer holds $22$ mmol of methanoic acid and $18$ mmol of its conjugate base, from sodium methanoate. A technician wants the pH to equal the $pK_a$, $3.75$, which needs equal amounts of the two. Use the additions to get there.
This task has no paper form; do it on a device.
A buffer contains $2$ mmol of ethanoic acid and $2$ mmol of sodium ethanoate, pH $4.76$. Then $4$ mmol of hydrochloric acid is added. What happens to the pH?
A buffer is made from ethanoic acid and sodium ethanoate in equal amounts. A small amount of acid or alkali is added. For each quantity, say how it changes.
| change | |
|---|---|
| amount of weak acid after a little acid is added | |
| amount of conjugate base after a little acid is added | |
| pH after a little acid is added | |
| amount of weak acid after a little alkali is added | |
| pH after a little alkali is added |
Sort each mixture: is it a buffer or not?
| a buffer | not a buffer | |
|---|---|---|
| ethanoic acid with sodium ethanoate | ||
| hydrochloric acid with sodium chloride | ||
| ammonia with ammonium chloride | ||
| ethanoic acid on its own | ||
| carbonic acid with sodium hydrogencarbonate |
Buffers made from benzoic acid ($pK_a = 4.20$) and sodium benzoate are useful while the ratio of the two lies between one to ten and ten to one. Give the pH range such buffers cover.
This task has no paper form; do it on a device.
Blood is buffered mainly by carbonic acid and hydrogencarbonate ions. At pH $7.40$ and body temperature the ratio of hydrogencarbonate to carbonic acid is $20$ to $1$. A patient's blood has $22$ mmol/L of hydrogencarbonate and a pH of $7.40$. What is the concentration of carbonic acid, in mmol/L?
Answer: mmol/L
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A buffer contains $6$ mmol of methanoic acid ($pK_a = 3.75$) and $27$ mmol of sodium methanoate. Then $3$ mmol of sodium hydroxide is added. What is the new pH?
Answer:
You can explain and calculate a buffer's pH, and you know its limits. Say out loud why adding twice as much acid as a buffer's conjugate base makes the pH plunge. Next: choosing an indicator whose colour change lands inside the steep part of a titration curve.
10. Your turn: a buffer has $30$ mmol of benzoic acid ($pK_a$ 4.20) and $3$ mmol of sodium benzoate. What is its pH, and is it a good buffer against added acid?, step 3