5.1.3(i)-(m) - Buffer solutions and blood pH

What a buffer solution does

A buffer minimises pH changes when small amounts of acid or base are added. It does not keep pH exactly constant, and its capacity is finite: it needs both members of its conjugate pair to remain available.

The acidic buffers in this OCR section contain a weak acid and its conjugate base. A convenient general pair is:

Here, $\ce{HA}$ is the weak acid and $\ce{A-}$ is its conjugate base. For the common ethanoic acid buffer:

The important point is that the mixture contains significant amounts of both forms of the conjugate acid-base pair. The weak acid alone does not provide much $\ce{A-}$, because it only partially dissociates. A buffer needs a reservoir of $\ce{HA}$ and a reservoir of $\ce{A-}$.

Making an acidic buffer

OCR specifies two ways to form an acidic buffer.

The first method is to mix a weak acid with a soluble salt of the weak acid. For example:

The salt dissociates fully in water:

The sodium ion is a spectator ion. The buffer action comes from $\ce{CH3COOH}$ and $\ce{CH3COO-}$, not from $\ce{Na+}$.

The second method is to react an excess of weak acid with a strong alkali. The strong alkali makes some conjugate base in the same beaker:

If $\ce{HA}$ is in excess, some $\ce{HA}$ remains after neutralisation and some $\ce{A-}$ has been formed. That gives the two species needed for a buffer. If the strong alkali is not limiting and neutralises all the weak acid, there is no reservoir of $\ce{HA}$ left, so the solution is not the acid buffer described here.

Worked example:

25.0 $\,\mathrm{cm}^{3}$ of 0.100 $\,\mathrm{mol} \,\mathrm{dm}^{-3}$ $\ce{CH3COOH}$ is mixed with 10.0 $\,\mathrm{cm}^{3}$ of 0.100 $\,\mathrm{mol} \,\mathrm{dm}^{-3}$ $\ce{NaOH}$.

The $\ce{OH-}$ reacts with the same amount of $\ce{CH3COOH}$, forming the same amount of $\ce{CH3COO-}$.

So after neutralisation:

• $\ce{CH3COOH}$ left = $2.50\times10^{-3}$ - $1.00\times10^{-3}$ = $1.50\times10^{-3}$ $\,\mathrm{mol}$
• $\ce{CH3COO-}$ formed = $1.00\times10^{-3}$ $\,\mathrm{mol}$

This is a buffer because both $\ce{CH3COOH}$ and $\ce{CH3COO-}$ are present.

How the conjugate pair controls pH

An acidic buffer controls pH because each member of the conjugate acid-base pair has a job.

If a small amount of acid is added, the added $\ce{H+}$ is mostly removed by the conjugate base:

For the ethanoic acid/ethanoate buffer:

This reduces the increase in $[\ce{H+}]$, so the pH falls only slightly.

If a small amount of base is added, the added $\ce{OH-}$ is neutralised by the weak acid:

For the ethanoic acid/ethanoate buffer:

This reduces the decrease in $[\ce{H+}]$, so the pH rises only slightly.

A common mistake is to say that the equilibrium "stops pH changing". The more accurate OCR idea is that the conjugate pair removes most of the added $\ce{H+}$ or $\ce{OH-}$, so the pH change is minimised.

The same action can be expressed through $\ce{HA <=> H+ + A-}$. Added acid drives this equilibrium left; removal of $\ce{H+}$ by $\ce{OH-}$ allows more HA to dissociate, shifting it right. Name the equilibrium before describing its shift. The remaining reservoirs are large relative to the small addition, so their ratio changes only a little.

[DIAGRAM: asset_name: Acid buffer response; asset_slug: 05_01_03d_buffer_solutions_and_blood_ph__diagram_01; recommended_method: drawn_chem; description: Ethanoic acid-ethanoate buffer removes added acid or alkali through balanced reactions while pH changes slightly.]

Calculating pH from $K_a$ and the conjugate pair

For a weak acid buffer:

Rearrange this expression to make $[\ce{H+}]$ the subject:

Then convert $[\ce{H+}]$ to pH:

In OCR buffer calculations, $[\ce{HA}]$ and $[\ce{A-}]$ are the equilibrium concentrations of the conjugate acid-base pair. In practice, the tiny extra dissociation of $\ce{HA}$ is usually negligible compared with the much larger amounts of $\ce{HA}$ and $\ce{A-}$ already present, so the concentrations after mixing are used.

Worked example:

A buffer contains 0.100 $\,\mathrm{mol} \,\mathrm{dm}^{-3}$ $\ce{CH3COOH}$ and 0.150 $\,\mathrm{mol} \,\mathrm{dm}^{-3}$ $\ce{CH3COO-}$. The $K_a$ of $\ce{CH3COOH}$ is $1.74\times10^{-5}$ $\,\mathrm{mol} \,\mathrm{dm}^{-3}$ at 25 $^{\circ}\mathrm{C}$. Calculate the pH.

Choose the equation:

Substitute:

Now calculate pH: