The Henderson-Hasselbalch Equation · 亨德森-哈塞尔巴尔赫方程
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Henderson-Hasselbalch equation/ˈhendəsn ˈhæsɪlbæltʃ ɪˈkweɪʒn/ | 亨德森-哈塞尔巴赫方程 | hēng dé sēn - hā sāi ěr bā hè fāng chéng |
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.
求缓冲 pH 的一个公式
- 给定缓冲溶液的成分,它的 pH 是多少?
- 一个整洁的方程立刻给出答案。
- 它把酸的强度与混合比例结合起来。
- 代入两个数,读出 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.
缓冲方程
- 亨德森-哈塞尔巴赫方程是:
- 它由碱与酸的比例给出缓冲溶液的 pH。
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.
比例的作用
- 碱和酸相等给出 $\log 1 = 0$,所以 $\text{pH} = \text{p}K_a$。
- 碱多于酸会升高 pH。
- 酸多于碱会降低 pH。
If a buffer has more conjugate base than acid, its pH is... · 如果缓冲溶液中共轭碱多于酸，其pH是……
$\log(\text{ratio} > 1)$ is positive, raising the pH. · $\log(\text{ratio} > 1)$为正，提高pH。
When acid and conjugate base are equal, the log term is zero. · 当酸和共轭碱相等时，对数项为零。
$\log(1) = 0$, so $\text{pH} = \text{p}K_a$. · $\log(1) = 0$，所以 $\text{pH} = \text{p}K_a$。
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.
解缓冲问题
- 代入 $\text{p}K_a$ 和两个浓度。
- 它们比值的对数把 pH 上下移动。
- 你也可以反过来求达到目标 pH 所需的比例。
In the equation, the ratio inside the log is... · 在方程中，对数内的比值是……
It is · 它是 $[\text{base}]/[\text{acid}]$, conjugate base on top. · 它是$[\text{base}]/[\text{acid}]$，共轭碱在分子上。
You can rearrange the equation to solve for the base-to-acid . · 你可以重排方程以求出碱与酸的。
Solving for the ratio lets you design a buffer at a target pH. · 求解比值可让你设计目标pH的缓冲溶液。
A buffer has $\text{p}K_a = 4.7$ with equal acid and base. What is its pH?
- The ratio is 1, so $\log 1 = 0$.
- $\text{pH} = 4.7 + 0 = 4.7$.
一个缓冲溶液 $\text{p}K_a = 4.7$,酸和碱相等。它的 pH 是多少?
- 比值是 1,所以 $\log 1 = 0$。
- $\text{pH} = 4.7 + 0 = 4.7$。
Ratio sets the pH · 比值设定pH
Using pH = pKa + log([base]/[acid]), sort each buffer by its pH relative to pKa. · 使用pH = pKa + log([base]/[acid])，根据各缓冲溶液的pH相对于pKa的位置对其进行排序。
A buffer has $\text{p}K_a = 5.0$ with equal acid and base. Its pH? · 具有等量酸和碱的缓冲溶液，其$\text{p}K_a = 5.0$是多少？其pH呢？
$\log 1 = 0$, so $\text{pH} = \text{p}K_a = 5.0$. · $\log 1 = 0$，所以 $\text{pH} = \text{p}K_a = 5.0$。
A buffer has $\text{p}K_a = 4$ and $\log\frac{[\text{base}]}{[\text{acid}]} = 1$. Its pH? · 具有$\text{p}K_a = 4$和$\log\frac{[\text{base}]}{[\text{acid}]} = 1$的缓冲溶液，其pH是多少？
$\text{pH} = 4 + 1 = 5$.
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.
比值是 $[\text{base}]/[\text{acid}]$,共轭碱在上面——把它反过来会把修正项反过来。当两者相等时,对数为零,给出 $\text{pH} = \text{p}K_a$。而且该方程假设是真正的缓冲溶液,两种物种都以可观的量存在。
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.
亨德森-哈塞尔巴赫方程 $\text{pH} = \text{p}K_a + \log\frac{[\text{base}]}{[\text{acid}]}$ 由 $\text{p}K_a$ 和碱酸比给出缓冲溶液的 pH。等量给出 $\text{pH} = \text{p}K_a$;碱多则升高,酸多则降低。