Buffer solutions
| English | Chinese | Pinyin |
|---|---|---|
| buffer | 缓冲溶液 | huǎn chōng róng yè |
| weak acid | 弱酸 | ruò suān |
| conjugate base | 共轭碱 | gòng è jiǎn |
Solutions that resist change
- A buffer 缓冲溶液 resists a change in pH when a little acid or alkali is added.
- It is made from a weak acid 弱酸 and its conjugate base 共轭碱.
- It is vital in living things — like your blood.
A buffer solution is made from:
A weak acid and its conjugate base (e.g. ethanoic acid + sodium ethanoate) make a buffer.
A buffer keeps the pH almost constant when a small amount of acid or alkali is added.
That is exactly what a buffer does — it resists changes in pH.
A buffer resists changes in pH when small amounts of acid or base are added.
It contains a weak acid and its conjugate base.
How a buffer works
It holds a store of both partners:
- added $\text{H}^+$ is removed by the conjugate base: $\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$.
- added $\text{OH}^-$ is removed by the weak acid: $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$.
So the pH barely changes.
A buffer mops up added acid or alkali
The buffer region
On the curve, a buffer is the flat stretch where added acid or base barely moves the pH — then it jumps at the equivalence point.
When a little acid (H⁺) is added to a buffer, it is removed by:
The conjugate base mops up added H⁺; the weak acid mops up added OH⁻, so the pH barely changes.
Match each buffer idea.
Each item links the term to its correct meaning.
Finding the pH and uses
- Put the concentrations of the acid and its salt into the $K_a$ expression: $[\text{H}^+] = K_a \times \dfrac{[\text{acid}]}{[\text{salt}]}$.
- Buffers matter in life: $\text{HCO}_3^-$ keeps the pH of blood close to 7.4.
In blood, the pH is kept close to 7.4 by the buffer ion:
The hydrogencarbonate ion buffers blood, keeping its pH around 7.4.
Working out the pH
A buffer is 0.10 mol/dm³ ethanoic acid ($K_a = 1.8 \times 10^{-5}$) + 0.10 mol/dm³ sodium ethanoate.
- $[\text{H}^+] = K_a \times \dfrac{[\text{acid}]}{[\text{salt}]} = 1.8 \times 10^{-5} \times \dfrac{0.10}{0.10} = 1.8 \times 10^{-5}$
- $\text{pH} = -\log(1.8 \times 10^{-5}) = \mathbf{4.74}$
When acid and salt concentrations are equal, $[\text{H}^+] = K_a$, so the pH equals $\text{p}K_a$.
You've got it
- a buffer = weak acid + its conjugate base; it resists pH change
- added $\text{H}^+$ is mopped up by A⁻; added $\text{OH}^-$ by HA
- find its pH from $[\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{salt}]}$ (equal concentrations ⇒ pH = p$K_a$)
- blood is buffered by $\text{HCO}_3^-$ near pH 7.4