# Why does a buffer work best at the pH closest to its pKa?

I think I'm having trouble understanding exactly how the $$\mathrm{p}K_\mathrm{a}$$ relates here on a conceptual level. For example I know that carbonic acid works somewhat well as a buffer at $$\mathrm{pH}$$ near 7, since its $$\mathrm{p}K_\mathrm{a}$$ is relatively close, but why is that exactly?

• A good way to think about pKa is that it is the pH at which the molecule is 50% ionized. Aug 28, 2020 at 16:36

A buffer works best when there is the same amount of weak acid/base and its conjugate. If you look at the Henderson Hasselbalch equation, and set the concentration of the weak acid/base equal to each other, $$\mathrm{pH} = \mathrm{p}K_\mathrm{a}$$.

I propose this simple graphic answer as a complement to the more thorough replies above (comments welcome to confirm validity of my answer). Trace pH Vs % [HA] and [A-]. 0% means you only have the acid [HA] and 100% you only have [A-]. Addition of acid or base, to the buffer solution, affects the ratio [A-]/[AH] and consequently, pH. This has been explained quantitatively in previous answers. On the plot, you see two drops in HA % due to addition of base. Compare the corresponding changes in pH and notice that operating around pH =pKa lowers the impact on the pH..

Why does a buffer work best at the $$\mathrm{pH}$$ closest to its $$\mathrm{p}K_\mathrm{a}$$?

For the novices like you who just begin to learn chemistry, this question can be easily explain by using Henderson Hasselbalch equation. Suppose you have an aqueous weak acid solution ($$\ce{HA}$$). It will be in ionic equilibrium as follows:

$$\ce{HA (aq) <=> H+ (aq) + A- (aq)} \tag1$$

Thus by definition:

$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\tag2$$

Take $$-\log$$ on both side of the equation $$(2)$$:

$$-\log K_\mathrm{a} = -\log \left(\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \right) \ \Rightarrow \ -\log K_\mathrm{a} = -\log [\ce{H+}] - \log\left(\frac{[\ce{A-}]}{[\ce{HA}]} \right)$$

$$\therefore \ \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\left(\frac{[\ce{A-}]}{[\ce{HA}]} \right)\tag3$$

The equation $$(3)$$ is called Henderson Hasselbalch equation, which is valid for calculation of buffer solutions (e.g., a solution of a weak acid and its congugate base or a weak base and its congugate acid). By this equation, if you know the $$\mathrm{p}K_\mathrm{a}$$ of any weak acid $$\ce{HA}$$ and molar ratio $$\frac{[\ce{A-}]}{[\ce{HA}]}$$ of the solution where $$[\ce{A-}]$$ is the concentration of the weak acid's conjugate base, you can calculate the $$\mathrm{pH}$$ of the solution.

For example, suppose you have a buffer solution of acetic acid/sodium acetate mixture in $$1:1$$ mole ratio (say $$\pu{0.1 M}$$). The $$\mathrm{p}K_\mathrm{a}$$ is given as $$4.75$$. What is the $$\mathrm{pH}$$ of the solution?

You can apply the values to the equation $$(3)$$: $$\mathrm{pH} = 4.75 + \log \left(\frac{\pu{0.1 M}}{\pu{0.1 M}}\right) = 4.75$$.

If the molar ratio is $$1:2$$ (say $$\pu{0.1 M}$$ in acetate and say $$\pu{0.2 M}$$ in acetic acid), then $$\mathrm{pH} = 4.75 + \log \left(\frac{\pu{0.1 M}}{\pu{0.2 M}}\right) = 4.75 - 0.301 = 4.45$$. Thus, as long as if you keep $$[\ce{A-}]: [\ce{HA}] \approx 1:1$$, $$\mathrm{pH} \approx \mathrm{p}K_\mathrm{a}$$, because $$\log 1 \rightarrow 0$$ in the equation $$(3)$$.

In your example, $$\mathrm{p}K_\mathrm{a1}$$ of carbonic acid (which is a polyprotic acid) is $$6.35$$, thus $$\mathrm{pH}$$ of $$1:1$$ carbonic acid : bicarbonate solution is $$6.35$$. If you want it at exactly at $$7.00$$, the mole ratio should be: $$7.00 = 6.35 + \log \left(\frac{[\ce{HCO3-}]}{[\ce{H2CO3}]}\right) \ \Rightarrow \ \left(\frac{[\ce{HCO3-}]}{[\ce{H2CO3}]}\right) = 10^{(7.00 - 6.35)} = 4.47$$

When the pH is equal to the pKa, you have high concentrations of both conjugate acid and conjugate base. You need the weak base to react with added acid, and the weak acid to react with added base to stabilize the pH.

When the pH is very different from the pKa, one of those two species will be at very very low concentration (there is an exponential relationship).

For a more quantitative explanation, see the other answers, which introduce the Henderson Hasselbalch relationship, and look up buffer capacity. The pKa of carbonic acid is 6.35, so it buffers better at pH 6 than at pH 7.

Buffers are made with weak acids or weak bases, that dissociate slower in water. In your example, carbonic acid is a weak acid written as $$\ce{HA}$$ and will dissociate into $$\ce{A-}$$ and $$\ce{H+}$$ when a strong base is poured into its solution. During this titration, the $$\mathrm{pH}$$ of the solution will increase steadily as $$\ce{HA}$$ dissociates, because:

1. Dissociation is forming $$\ce{A-}$$, which is a conjugate base and $$\ce{HA}$$ concentration is reducing; and

2. You are adding a base ($$\ce{OH-}$$) to the solution, and there is also $$\ce{H+}$$ being formed through the weak acid's dissociation. But it will react with the $$\ce{OH-}$$ to neutralize each other and the solution will be resisting a huge increase to $$\mathrm{pH}$$ but increasing steadily with the increase of $$\ce{A-}$$.

This will happen throughout the buffer region and in the middle of that region is the point where the amount of $$\ce{HA}$$ (original weak acid) becomes equal to the amount of $$\ce{A-}$$ (conjugate base). If you can get a weak acid or weak base in an environment with $$\mathrm{pH}$$ close to its $$\mathrm{p}K_\mathrm{a}$$, it is guaranteed you'll be within this buffer region since $$\mathrm{p}K_\mathrm{a}$$ is in the midpoint of that region.

• > that dissociate slower in water It is not about speed, but equilibrium. They dissociate just partially, regardless of speed. Aug 27, 2020 at 16:10