The Henderson-Hasselbalch Equation
Introduced| English | Meaning |
|---|---|
| Henderson-Hasselbalch equation/ˈhendəsn ˈhæsɪlbæltʃ ɪˈkweɪʒn/ | Henderson-Hasselbalch equation |
One formula for buffer pH
- Given a buffer's ingredients, what is its pH?
- A single tidy equation answers it instantly.
- It combines the acid's strength with the mixing ratio.
- Plug in two numbers and read out the pH.
The buffer equation
- The Henderson-Hasselbalch equation 亨德森-哈塞尔巴赫方程 is:
- It gives a buffer's pH from the ratio of base to acid.
At half-equivalence (HA) = (A⁻), so the pH equals the pKa
What the ratio does
- Equal base and acid give $\log 1 = 0$, so $\text{pH} = \text{p}K_a$.
- More base than acid raises the pH.
- More acid than base lowers the pH.
If a buffer has more conjugate base than acid, its pH is...
Difficulty 1/5 · 100 coins + a small bonus on first completion today
$\log(\text{ratio} > 1)$ is positive, raising the pH.
When the acid and conjugate base concentrations are equal, $\text{pH} = 7$.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
False. With equal concentrations the log term is zero, so $\text{pH} = \text{p}K_a$, which is 7 only by coincidence.
Solving buffer problems
- Plug in $\text{p}K_a$ and the two concentrations.
- The log of their ratio shifts the pH up or down.
- You can also solve for the ratio needed to hit a target pH.
In the equation, the ratio inside the log is...
Difficulty 1/5 · 100 coins + a small bonus on first completion today
It is $[\text{base}]/[\text{acid}]$, conjugate base on top.
You can rearrange the equation to solve for the base-to-acid ____.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Solving for the ratio lets you design a buffer at a target pH.
A buffer has $\text{p}K_a = 4.7$ with equal acid and base. What is its pH?
- The ratio is 1, so $\log 1 = 0$.
- $\text{pH} = 4.7 + 0 = 4.7$.
Ratio sets the pH
Using pH = pKa + log([base]/[acid]), sort each buffer by its pH relative to pKa.
A buffer has $\text{p}K_a = 5.0$ with equal acid and base. Its pH?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
$\log 1 = 0$, so $\text{pH} = \text{p}K_a = 5.0$.
A buffer has $\text{p}K_a = 4$ and $\log\frac{[\text{base}]}{[\text{acid}]} = 1$. Its pH?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
$\text{pH} = 4 + 1 = 5$.
The ratio is $[\text{base}]/[\text{acid}]$, conjugate base on top — flipping it flips the correction. When the two are equal, the log is zero, giving $\text{pH} = \text{p}K_a$. And the equation assumes a real buffer, with both species present in decent amounts.
The Henderson-Hasselbalch equation $\text{pH} = \text{p}K_a + \log\frac{[\text{base}]}{[\text{acid}]}$ gives a buffer's pH from its $\text{p}K_a$ and the base-to-acid ratio. Equal amounts give $\text{pH} = \text{p}K_a$; more base raises it, more acid lowers it.