If we add a small amount of acid (adding $H^{+}$ ions):

• The extra $H^{+}$ ions react with the conjugate base ($A^{-}$) in the reservoir to form more weak acid ($HA$).
• Equation: $H^{+}+A^{-}\to HA$
• Most of the added $H^{+}$ is removed, so the pH barely drops.

If we add a small amount of base (adding $O{H}^{-}$ ions):

• The added $O{H}^{-}$ ions react with the $H^{+}$ ions in the solution to form water.
• This temporarily lowers the $[H^{+}]$, but by Le Chatelier's principle, the weak acid ($HA$) in the reservoir dissociates to replace the lost $H^{+}$ ions.
• Alternatively, you can just think of the $O{H}^{-}$ reacting directly with the weak acid: $O{H}^{-}+HA\to H_{2}O+A^{-}$
• The $O{H}^{-}$ is removed, and the $H^{+}$ is restored, so the pH barely rises.

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Basic buffers

A basic buffer works on exactly the same principle, but it uses a weak base and the salt of that weak base.

A classic example is ammonia (${NH}_3$) mixed with ammonium chloride (${NH}_{4}Cl$).

• The weak base equilibrium: ${NH}_3+H_{2}O\rightleftharpoons {NH}_4^{+}+{OH}^{-}$
• The salt fully dissociates: ${NH}_{4}Cl\to {NH}_4^{+}+{Cl}^{-}$

Here, the reservoirs are the weak base (${NH}_3$) and its conjugate acid (${NH}_4^{+}$).

• If you add acid ($H^{+}$), it reacts with the weak base: $H^{+}+{NH}_3\to {NH}_4^{+}$.
• If you add base ($O{H}^{-}$), it reacts with the conjugate acid: $O{H}^{-}+{NH}_4^{+}\to {NH}_3+H_{2}O$.

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Applications of buffers

Buffer solutions are essential in many everyday products and natural systems:

• Biological systems: Your blood is buffered (mainly by the carbonic acid / hydrogencarbonate system) to stay at a pH of around 7.4. Even a small drop to 7.35 can cause severe health problems.
• Biological washing powders: These contain enzymes that only work effectively at a specific pH. Buffers are added to keep the pH exactly where the enzyme is most active.
• Shampoos: Hair is sensitive to high pH (which can cause the cuticles to open and hair to feel rough). Shampoos are often buffered to a slightly acidic pH (around 5.5) to keep hair smooth.

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Calculating the pH of an acidic buffer

Because a buffer is essentially a weak acid equilibrium that has been tampered with, we can calculate its pH using the same acid dissociation constant ($K_a$) we use for normal weak acids.

$$K_a=\frac{[H^{+}][A^{-}]}{[HA]}$$

If we rearrange this to make $[H^{+}]$ the subject, we get the master equation for acidic buffer calculations:

$$[H^{+}]=K_a\times \frac{[HA]}{[A^{-}]}$$

The two big mathematical approximations

When we plug numbers into this equation, we make two important approximations:

1. $[A^{-}]$ at equilibrium $\approx$ Initial $[A^{-}]$ from the salt. We assume all the $A^{-}$ comes from the fully dissociated salt, and the tiny amount produced by the weak acid dissociating is negligible.
2. $[HA]$ at equilibrium $\approx$ Initial $[HA]$. We assume the weak acid barely dissociates at all (especially since the added salt pushes the equilibrium even further left).

There are two main styles of buffer calculation you will face. Let's look at a worked example for each.

Calculation Type 1: Direct mixing

This is the simpler scenario. The question will tell you that a certain amount of weak acid is mixed with a certain amount of its salt. You just need to find the moles of each.

Calculation Type 2: Partial neutralisation

This scenario is harder. You are reacting an excess of weak acid with a strong base (like NaOH). The strong base reacts with some of the weak acid to make the salt. You have to figure out how much weak acid is left over, and how much salt was created.

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