12.18-12.24 - Buffers And Neutralisation

What A Buffer Solution Does

Pure water has no useful reserve of acid or base. Add a little strong acid and $[\ce{H+}]$ rises sharply; add a little strong alkali and $[\ce{H+}]$ falls sharply. A buffer solution is different.

The most common A level model is a weak acid and its conjugate base:

The solution must contain appreciable amounts of both:

• $\ce{HA}$, the weak acid, which can supply $\ce{H+}$
• $\ce{A-}$, the conjugate base, which can remove added $\ce{H+}$

For example, ethanoic acid and sodium ethanoate make a buffer:

The sodium ions from sodium ethanoate are spectator ions. The important particles are $\ce{CH3COOH}$ molecules and $\ce{CH3COO-}$ ions.

How added acid is resisted

If a small amount of strong acid is added, it adds $\ce{H+}$ ions. The conjugate base removes most of those added $\ce{H+}$ ions:

For an ethanoic acid/ethanoate buffer:

The $\mathrm{pH}$ still changes a little because the ratio of $\ce{A-}$ to $\ce{HA}$ has changed. It does not change dramatically because most of the added $\ce{H+}$ has been converted into weak acid molecules.

How added alkali is resisted

If a small amount of strong alkali is added, it adds $\ce{OH-}$ ions. The weak acid neutralises most of the added hydroxide:

For an ethanoic acid/ethanoate buffer:

Again, the $\mathrm{pH}$ changes a little because the ratio of conjugate base to weak acid changes. A buffer resists $\mathrm{pH}$ change; it does not make $\mathrm{pH}$ impossible to change.

Common-error contrast

Do not explain buffer action by writing only "equilibrium shifts left" or "equilibrium shifts right". Good answers name the species that react with the added acid or alkali. Added $\ce{H+}$ is removed by $\ce{A-}$; added $\ce{OH-}$ is removed by $\ce{HA}$.

Calculating The pH Of A Buffer

For a weak acid buffer:

Rearrange before substituting:

The useful chemistry is in the ratio. More $\ce{A-}$ means more capacity to remove $\ce{H+}$, so $[\ce{H+}]$ is lower and $\mathrm{pH}$ is higher. More $\ce{HA}$ means more weak acid present, so $[\ce{H+}]$ is higher and $\mathrm{pH}$ is lower.

The second form is the same equation written in logarithmic form. Questions often give enough data to use either route, but the $K_{\mathrm a}$ expression makes the chemical assumption visible.

The main assumptions

For standard buffer calculations:

• the weak acid is only slightly dissociated, so the amount of $\ce{HA}$ at equilibrium is close to the amount placed in the mixture
• the salt is fully dissociated, so it supplies the conjugate base ions
• if $\ce{HA}$ and $\ce{A-}$ are in the same final solution, their concentration ratio equals their mole ratio
• any strong acid or strong base added reacts completely with $\ce{A-}$ or $\ce{HA}$ before the buffer $\mathrm{pH}$ calculation is used

Worked example: pH from given data

$50.0 \mathrm{cm^{3}}$ of $0.100 \mathrm{mol\,dm^{-3}}$ ethanoic acid is mixed with $25.0 \mathrm{cm^{3}}$ of $0.100 \mathrm{mol\,dm^{-3}}$ sodium ethanoate. For ethanoic acid, $K_{\mathrm a} = 1.74\times 10^{-5} \mathrm{mol\,dm^{-3}}$. Calculate the $\mathrm{pH}$ of the buffer.

Find the amounts mixed:

They are in the same final volume, so the concentration ratio is the mole ratio:

Now calculate $[\ce{H+}]$:

Convert to $\mathrm{pH}$:

$\mathrm{pH} = 4.46$

Chemical interpretation: the buffer contains twice as much weak acid as conjugate base, so the $\mathrm{pH}$ is below the acid's $\mathrm pK_{\mathrm a}$ value of about 4.76.

Worked example: after adding strong acid

The same buffer is treated with $5.00 \mathrm{cm^{3}}$ of $0.100 \mathrm{mol\,dm^{-3}} \ce{HCl}$. Estimate the new $\mathrm{pH}$.

The added strong acid provides:

This reacts with ethanoate ions:

Update the moles before using the buffer equation:

Now:

$\mathrm{pH} = 4.32$

The $\mathrm{pH}$ has fallen, but only from 4.46 to 4.32. The stoichiometry explains why: most added $\ce{H+}$ has been converted into weak acid rather than remaining free in solution.

Designing A Buffer Of A Given pH

To prepare a buffer of a chosen $\mathrm{pH}$, choose a weak acid whose $\mathrm pK_{\mathrm a}$ is reasonably close to the desired $\mathrm{pH}$, then calculate the required ratio of conjugate base to weak acid.

The logarithmic form is the most efficient route:

Rearrange to find the ratio:

That ratio is not a decoration. It tells you how much conjugate base reserve is needed compared with weak acid reserve.

Worked example: finding the required concentration

Design an ethanoic acid/ethanoate buffer with $\mathrm{pH} = 5.00$. Use $\mathrm pK_{\mathrm a} = 4.76$. If the final ethanoic acid concentration is to be $0.100 \mathrm{mol\,dm^{-3}}$, calculate the final ethanoate ion concentration required.

Start with the ratio:

Now use the chosen acid concentration:

$[\ce{CH3COO-}] = 1.74\times 0.100$

Chemical interpretation: because the target $\mathrm{pH}$ is above $\mathrm pK_{\mathrm a}$, the mixture needs more conjugate base than weak acid. A buffer with equal concentrations of acid and conjugate base would have $\mathrm{pH} = \mathrm pK_{\mathrm a}$, not 5.00.

A buffer can also be prepared by partial neutralisation. Starting with 0.0100 mol HA and adding 0.00400 mol OH⁻ leaves 0.00600 mol HA and forms 0.00400 mol A⁻. Both remain, so the mixture is a buffer. At exactly 0.0100 mol OH⁻, all the formal HA has been consumed and the usual acid-buffer calculation no longer applies. Beyond that point, calculate excess strong alkali instead.

Diluting both components equally approximately preserves pH because their ratio is unchanged. It lowers the buffer capacity per unit volume: an equal-volume sample of the diluted buffer contains fewer moles of each component, so the same added amount of strong acid or alkali changes their ratio more. Simply adding water to the whole original buffer does not remove any of its acid or conjugate-base reserve. Very large dilution eventually invalidates the buffer approximations.

From concentrations to a practical mixture

If the final volume is $250.0 \mathrm{cm^{3}}$, the required amounts in the final solution are:

If $1.00 \mathrm{mol\,dm^{-3}}$ stock solutions of ethanoic acid and sodium ethanoate were available, those amounts would require:

The mixture would then be made up to $250.0 \mathrm{cm^{3}}$ with deionised water in a volumetric flask. Dilution to the mark changes both concentrations, but it does not change the ratio if both components are diluted into the same final volume.