5.1 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

A strong acid is defined as a substance which completely dissociates to give all the maximum possible ${H^ + }$,

For example: $$HCl{\text{ }} \to {\text{ }}{{\text{H}}^ + }{\text{ + C}}{{\text{l}}^ - }$$

a) When $\left[ {HCl} \right]{\text{ }} \geqslant {\text{ 1}}{{\text{0}}^{ - 6}}{\text{ M}}$, then contribution from the autoprotolysis of water is neglected in comparison to the ${H^ + }$ coming from the acid.

Therefore, $$\left[ {HCl} \right]{\text{ = }}\left[ {{{\text{H}}^ + }} \right]{\text{ }}$$

Example 1. Find the $pH$ of ${10^{ - 3}}{\text{ M}}{\text{ HCl}}$.

Solution: Here, $$\left[ {HCl} \right]{\text{ = }}\left[ {{{\text{H}}^ + }} \right]{\text{ }} = {\text{ 1}}{{\text{0}}^{ - 3}}$$

$$\therefore {\text{ pH = - log 1}}{{\text{0}}^{ - 3}}{\text{ = 3}}$$

b) When $\left[ {HCl} \right]{\text{ < }}{10^{ - 6}}{\text{ M}}$ then contribution of ${H^ + }$ obtained from the autoprotolysis of water can't be neglected and needs to be included in the calculations.

But the contribution from water is not ${\text{ }}{10^{ - 7}}{\text{ M}}$ because it would give less than this amount in the presence of $HCl$.

This is because of common ion effect, which states that when a common ion is added to an equilibrium mixture, the equilibrium shifts in that direction where the common ion gets consumed.

Example 2. Find the $\left[ {H_{total}^ + } \right]{\text{ }}$ in ${10^{ - 7}}{\text{ M}}{\text{ HCl}}$.

Solution: Given that $$\left[ {HCl} \right]{\text{ = 1}}{{\text{0}}^{ - 7}}{\text{ M}}$$

Also water dissociates to give ${H^ + }$ and $O{H^ - }$. Let us assume, the contribution from water be $\left[ {H_{water}^ + } \right]$ i.e., $x$.

Then, $$\left[ {{\text{H}}_{total}^ + } \right]{\text{ = x + }}{10^{ - 7}}{\text{ M}}$$

$${H_2}O{\text{ }} \rightleftharpoons {\text{ }}{{\text{H}}^ + }{\text{+ O}}{{\text{H}}^ - }$$

We know that,

$${K_w}{\text{ = }}\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]{\text{ = 1}}{{\text{0}}^{ - 14}}$$$${K_w}{\text{ = (x + 1}}{{\text{0}}^{ - 7}})x$$$$x{\text{ = 0}}{\text{.7 }} \times {\text{ 1}}{{\text{0}}^{ - 7}}$$

Therefore, $$\left[ {H_{total}^ + } \right]{\text{ = 1}}{{\text{0}}^{ - 7}}{\text{ + (0}}{\text{.7 + 1}}{{\text{0}}^{ - 7}}){\text{ = 1}}{\text{.7 }} \times {\text{ 1}}{{\text{0}}^{ - 7}}{\text{ M}}$$