7.3 Salt 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

A salt buffer is a solution of a salt which itself can act as a buffer. Such a salt is the salt of a weak acid and a weak base. For example

$$C{H_3}COO{\text{N}}{{\text{H}}_4}{\text{ }} \rightleftharpoons {\text{ }}N{H_4}^ + {\text{ + }}C{H_3}CO{O^ - }{\text{ }}$$

When an acid is added, it reatcs with $C{H_3}CO{O^ - }$ to produce ${C{H_3}COOH}$ and when base is added, it reacts with $N{H_4}^ + $ to produce $N{H_4}OH$.

$$pH = 7 + \frac{1}{2}p{K_a} - \frac{1}{2}p{K_b}$$

Example 6.What is the $pH$ of $50{\text{ ml }}0.05N{\text{ }}NaOH\;and\,0.1N{\text{ }}C{H_3}COOH$ mixture?

Solution. $$C{H_3}COOH {\text{ + }}NaOH \to C{H_3}COONa{\text{ + }}{H_2}O$$

millimoles before reaction $2.5$ $5$ $0$

millimoles after reaction $2.5-2.5$ $5-2.5$ $2.5$

$$\therefore \left[ {C{H_3}COONa} \right] = \frac{{2.5}}{{100}}M$$$$\left[ {C{H_3}COOH} \right] = \frac{{2.5}}{{100}}M$$

$$pH = p{K_a} + \log \frac{{\left[ {C{H_3}COONa} \right]}}{{\left[ {C{H_3}COOH} \right]}} = 4.74 + \log 1 = 4.74$$

Example 7. How many moles of $HCl$ will be required to prepare one litre of buffer solution containing ($NaCN$ and $HCN$) of $pH = 8.5$ using $0.01g$ formula mass of $NaCN$.

(${K_a}(HCN) = 4.1 \times {10^{ - 10}}$)

Solution. $$NaCN{\text{ + }}HCl\to NaCl{\text{ + }}HCN$$

$a$ $a$ $a$

$$\left[ {HCN} \right] = a$$ $$\left[ {NaCN} \right] = (0.01 - a)$$ $$pH = p{K_a} + \log \frac{{\left[ {NaCN} \right]}}{{\left[ {HCN} \right]}}$$$$8.5 = \log \frac{{0.01 - a}}{a} - \log 4.1 \times {10^{ - 10}}$$$$\therefore \log \frac{{0.01 - a}}{a} = 8.5 + 0.617 - 10$$$$\therefore a = 0.0089\ moles$$