7.12.1 Designing Buffer Solution

Knowledge of pKa, pKb  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 pH we use weak acid and its salt formed with strong base.

We develop the equation relating the pH, the equilibrium constant, Ka of weak acid and ratio of concentration of weak acid and its conjugate base. For the general case where the weak acid HA ionises in water,

HA + H2O ⇔  H3O+ + A

For which we can write the expression

Ka = [H3O+][A]/[HA]

Rearranging the expression we have,

[H3O+] = Ka {[HA]/[A]}

 

Taking logarithm on both the sides and rearranging the terms we get

pKa = pH – log {[A]/[HA]}

or pH = pKa + log {[A]/[HA]}

pH = pKa + log {[conjugate base, A]/[acid, HA]}

 

The above expression is known as Henderson–Hasselbalch equation. The quantity {[A]/[HA]} is the ratio of 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 [HA] is negligibly different from concentration of acid taken to form buffer.

Also, most of the conjugate base, [A], 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, the Henderson–Hasselbalch  equation takes the form:

pH = pKa + log {[salt]/[Acid]}

 

In the equation pH = pKa + log {[A]/[HA]} , if the concentration of [A] is equal to the concentration of [HA], then pH = pKa because value of log 1 is zero.

Thus if we take molar concentration of acid and salt (conjugate base) same, the pH of the buffer solution will be equal to the pKa of the acid.

So for preparing the buffer solution of the required pH we select that acid whose pKa is close to the required pH.

For acetic acid pKa value is 4.76, therefore 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,

pOH = pKb + log {[conjugate acid, BH+]/[base, B]}

pH of the buffer solution can be calculated by using the equation pH + pOH =14.

We know that pH + pOH = pKw and pKa + pKb = pKw.

 

On putting these values in equation pOH = pKb + log {[conjugate acid, BH+]/[base, B]}  it takes the form as follows:

pKw - pH = pKw - pKa + log {[conjugate acid, BH+]/[base, B]}

pH = pKa + log {[conjugate acid, BH+]/[base, B]}

If molar concentration of base and its conjugate acid (cation) is same then pH of the buffer solution will be same as pKa for the base.

pKa value for ammonia is 9.25; therefore a buffer of 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, the above equation becomes:

pH = 9.25 + log {[conjugate acid, BH+]/[base, B]}

pH of the buffer solution is not affected by dilution because ratio under the logarithmic term remains unchanged.

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