6.12 Buffer Solutions

Many body fluids e.g., blood or urine have definite \(\mathrm{pH}\) and any deviation in their \(\mathrm{pH}\) indicates malfunctioning of the body. The control of \(\mathrm{pH}\) is also very important in many chemical and biochemical processes. Many medical and cosmetic formulations require that these be kept and administered at a particular \(\mathrm{pH}\). The solutions which resist change in \(\mathrm{pH}\) on dilution or with the addition of small amounts of acid or alkali are called Buffer Solutions. Buffer solutions of known \(\mathrm{pH}\) can be prepared from the knowledge of \(\mathrm{p} K_{\mathrm{a}}\) of the acid or \(\mathrm{p} K_{\mathrm{b}}\) of base and by controlling the ratio of the salt and acid or salt and base. A mixture of acetic acid and sodium acetate acts as buffer solution around \(\mathrm{pH} 4.75\) and a mixture of ammonium chloride and ammonium hydroxide acts as a buffer around \(\mathrm{pH} 9.25\). 

Designing Buffer Solution

Knowledge of \(\mathrm{p} K_a, \mathrm{p} K_b\) and equilibrium constant help us to prepare the buffer solution of known pH. Let us see how we can do this.

Preparation of Acidic Buffer

To prepare a buffer of acidic \(\mathrm{pH}\) we use weak acid and its salt formed with strong base. We develop the equation relating the \(\mathrm{pH}\), the equilibrium constant, \(K_a\) of weak acid and ratio of concentration of weak acid and its conjugate base. For the general case where the weak acid HA ionizes in water,
\(
\mathrm{HA}+\mathrm{H}_2 \mathrm{O} \leftrightharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{A}^{-}
\)
For which we can write the expression
\(
K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}
\)
Rearranging the expression we have,
\(
\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=K_a \frac{[\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}
\)
Taking logarithm on both sides and rearranging the terms we get –
\(
\mathrm{p} K_a=\mathrm{pH}-\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}
\)
or
\(
\mathrm{pH}=\mathrm{p} K_a+\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \dots(7.39)
\)
\(
\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\text { Conjugate base, } \mathrm{A}^{-}\right]}{[\text {Acid, } \mathrm{HA}]} \dots(7.40)
\)
The expression (7.40) is known as Henderson-Hasselbalch equation. The quantity \(\frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\) is the ratio of the concentration of conjugate base (anion) of the acid and the acid present in the mixture. Since acid is a weak acid, it ionises to a very little extent and concentration of \([\mathrm{HA}]\) is negligibly different from the concentration of acid taken to form buffer. Also, most of the conjugate base, \(\left[\mathrm{A}^{-}\right]\), comes from the ionisation of salt of the acid. Therefore, the concentration of conjugate base will be negligibly different from the concentration of salt. Thus, equation (7.40) takes the form:
\(
p \mathrm{H}=p K_a+\log \frac{[\text { Salt }]}{[\text { Acid }]}
\)
In the equation (7.39), if the concentration of \(\left[\mathrm{A}^{-}\right]\)is equal to the concentration of \([\mathrm{HA}]\), then \(\mathrm{pH}=\mathrm{p} K_a\) because value of \(\log 1\) is zero. Thus if we take molar concentration of acid and salt (conjugate base) same, the \(\mathrm{pH}\) of the buffer solution will be equal to the \(\mathrm{p} K_a\) of the acid. So for preparing the buffer solution of the required \(\mathrm{pH}\) we select that acid whose \(\mathrm{p} K_a\) is close to the required \(\mathrm{pH}\). For acetic acid \(\mathrm{pK}_{\mathrm{a}}\) value is 4.76, therefore \(\mathrm{pH}\) of the buffer solution formed by acetic acid and sodium acetate taken in equal molar concentration will be around 4.76.

A similar analysis of a buffer made with a weak base and its conjugate acid leads to the result,
\(
\mathrm{pOH}=\mathrm{p} K_{\mathrm{b}}+\log \frac{\left[\text { Conjugate acid, } \mathrm{BH}^{+}\right]}{[\text {Base }, \mathrm{B}]} \dots(7.41)
\)
\(\mathrm{pH}\) of the buffer solution can be calculated by using the equation \(\mathrm{pH}+\mathrm{pOH}=14\).

We know that \(\mathrm{pH}+\mathrm{pOH}=\mathrm{pK}_{\mathrm{w}}\) and \(\mathrm{p} K_{\mathrm{a}}+\mathrm{p} K_{\mathrm{b}}=\mathrm{p} K_{\mathrm{w}}\). On putting these values in equation (7.41) it takes the form as follows:
\(
\mathrm{p} K_{\mathrm{w}}-\mathrm{pH}=\mathrm{p} K_{\mathrm{w}}-\mathrm{p} K_a+\log \frac{\left[\text { Conjugate acid, } \mathrm{BH}^{+}\right]}{[\text {Base, } \mathrm{B}]}
\)
or
\(
\mathrm{pH}=\mathrm{p} K_a+\log \frac{\left[\text { Conjugate acid, } \mathrm{BH}^{+}\right]}{[\text {Base, } \mathrm{B}]} \dots(7.42)
\)
If molar concentration of base and its conjugate acid (cation) is same then \(\mathrm{pH}\) of the buffer solution will be same as \(\mathrm{pK}_a\) for the base. \(\mathrm{p} K_a\) value for ammonia is 9.25; therefore a buffer of \(\mathrm{pH}\) close to 9.25 can be obtained by taking ammonia solution and ammonium chloride solution of same molar concentration. For a buffer solution formed by ammonium chloride and ammonium hydroxide, equation (7.42) becomes:
\(
\mathrm{pH}=9.25+\log \frac{\left[\text { Conjugate acid, } \mathrm{BH}^{+}\right]}{[\text {Base, } \mathrm{B}]}
\)
\(\mathrm{pH}\) of the buffer solution is not affected by dilution because ratio under the logarithmic term remains unchanged.