6.2 Salt of weak base + strong acid

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

$${\text{N}}{{\text{H}}_4}{\text{OH + HCl }} \to {\text{ N}}{{\text{H}}_4}Cl{\text{ + }}{{\text{H}}_2}O$$

The salt would dissolve as $${\text{N}}{{\text{H}}_4}Cl{\text{ }} \to {\text{ }}N{H_4}^ + {\text{ + C}}{{\text{l}}^ - }{\text{ }}$$

and the hydrolysis of $N{H_4}^ + {\text{ }}$ is shown as

$$N{H_4}^ + {\text{ + }}{H_2}O \rightleftharpoons{\text{N}}{{\text{H}}_4}{\text{OH}}{\text{ + }}{{H^ + }}$$

${C(1 - \alpha )}$ ${C\alpha }$ ${C\alpha }$

The hydrolysis equilibrium is represented by the hydrolysis constant

$${K_h} = \frac{{\left[ {{\text{N}}{{\text{H}}_4}{\text{OH}}} \right]\left[ {{H^ + }} \right]}}{{\left[ {N{H_4}^ + } \right]}}$$

Now, $$Kw = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$$ and $${K_b} = \frac{{\left[ {N{H_4}^ + } \right]\left[ {O{H^ - }} \right]}}{{\left[ {{\text{N}}{{\text{H}}_4}{\text{OH}}} \right]}}$$

Therefore, $$\frac{{Kw}}{{{K_b}}} = \frac{{\left[ {{\text{N}}{{\text{H}}_4}{\text{OH}}} \right]\left[ {{H^ + }} \right]}}{{\left[ {N{H_4}^ + } \right]}} = {K_h}$$

$$\frac{{K_w}}{{{K_b}}} = \frac{{C\alpha \times C\alpha }}{{C(1 - \alpha )}} = \frac{{C{\alpha ^2}}}{{1 - \alpha }}$$

Assuming that $\alpha \ll 1$, we can neglect $\alpha $ in the denominator,

$$C{\alpha ^2} = \frac{{K_w}}{{{K_b}}}$$

$$\alpha = \sqrt {\frac{{K_w}}{{{K_b}C}}} $$

Finally, $$\left[ {{H^ + }} \right] = C\alpha = C \times \sqrt {\frac{{K_w}}{{{K_b}C}}} = \sqrt {\frac{{K_wC}}{{{K_b}}}} {\text{ }}$$

Note: Whenever the salt of a weak base and strong acid is dissolved in water, the solution is acidic.