7.1 Acidic buffer

3.1 Degree of Ionization (Dissociation)

5.1 Strong acid
5.2 Weak acid
5.3 Strong Acid + Weak Acid
5.4 Two weak acids
5.5 Dibasic and polyprotic weak acids

6.1 Salts of weak acid + strong base
6.2 Salt of weak base + strong acid
6.3 Salt of weak acid and weak base

7.1 Acidic buffer
7.2 Basic buffer
7.3 Salt buffer

8.1 Calculating Solubility of a salt

Aqueous solution of mixture of weak acid and salt of the same weak acid with any type of strong base is called acidic buffer solution. For example, $$C{H_3}COOH + C{H_3}COONa$$

In the solution we have the following dissociations

$$C{H_3}COOH \rightleftharpoons {C{H_3}CO{O^ - }} {\text{ + }} {H^ + }$$

and $$C{H_3}COO{\text{Na }} \to {\text{ }}C{H_3}CO{O^ - }{\text{ + }}O{H^ - }{\text{ }}$$

Because of the acetate coming from the salt, it creates a common ion effect due to which the dissociation of acetic acid decreases and we can assume that the entire acid remains as $C{H_3}COOH$.

So, there are two important constituents, $C{H_3}COOH$ and ${C{H_3}CO{O^ - }}$ in the solution.

$$C{H_3}CO{O^ - }{\text{ + }}{H^ + } \rightleftharpoons {\text{ }}C{H_3}COOH$$

This reaction proceeds practically to completion.

Now, let us assume that that we have $20$ moles of $C{H_3}CO{O^ - }$ in the buffer and $10$ moles of ${H^ + }$ added from outside. $10$ moles of $C{H_3}CO{O^ -}$ will be consumed to produce $10$ moles of $C{H_3}COOH$. ${C{H_3}CO{O^ - }}$ is left in the solution that would create a common ion effect and would further suppress the dissociation of $C{H_3}COOH$.

Because of this acetic acid would give only say $2$ moles of ${H^ + }$. Therefore, the buffer system has resisted the $pH$ change and the $pH$ change corresponds to only $2$ moles of ${H^ + }$ and not $10$ moles which was added.

$$C{H_3}COOH + O{H^ - } \rightleftharpoons {\text{ }}C{H_3}CO{O^ - }{\text{ + }}{H_2}O$$

Because of high $K$ value , the forward reaction proceeds practically to completition. Again, the buffer resists the $pH$ change in the similar way as explained above.

$$C{H_3}COOH \rightleftharpoons {C{H_3}CO{O^ - }} {\text{ + }} {H^ + }$$

We have $${K_a}{\text{ = }}\frac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}$$

In a buffer $$\left[ {C{H_3}CO{O^ - }} \right] = \left[ {Salt} \right];{\text{ }}\left[ {C{H_3}COOH} \right] = \left[ {Acid} \right]$$

$$\therefore {\text{ }}{K_a} = \frac{{\left[ {Salt} \right]\left[ {{H^ + }} \right]}}{{\left[ {Acid} \right]}}$$ or, $$\left[ {{H^ + }} \right] = {K_a}\frac{{\left[ {Acid} \right]}}{{\left[ {Salt} \right]}}$$

Taking log on both sides and multiplying by $ - 1$

$$pH = p{K_a} + \log \frac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}}$$