Chemistry > Ionic Equilibrium > 5.0 Determination of $pH$ of acids and bases
Ionic Equilibrium
1.0 Reversible Reaction
2.0 $pH$ Scale
3.0 Arrehenius Theory of Electrolyte Ionization (Dissociation)
4.0 Ionization of Water
5.0 Determination of $pH$ of acids and bases
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
6.0 Salt Hydrolysis
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
7.0 Buffer Solution
8.0 Solubility and Solubility Product
5.3 Strong Acid + Weak Acid
5.2 Weak acid
5.3 Strong Acid + Weak Acid
5.4 Two weak acids
5.5 Dibasic and polyprotic weak acids
6.2 Salt of weak base + strong acid
6.3 Salt of weak acid and weak base
Suppose we have a strong acid $HA$ and a weak acid $HB$
Strong acid of concentration ${C_1}$ will dissociate completely as
$$HA{\text{ }} \to {\text{ }}{{\text{H}}^ + }{\text{ + }}{{\text{A}}^ - }$$
and for the weak acid of concentration ${C_2}$ the following equilibrium exists:
$$HB \rightleftharpoons {{\text{H}}^ + }{\text{ + }}{{\text{B}}^ - }$$
${C_2}(1 - \alpha )$ ${C_2}\alpha + {C_1}$ ${C_2}\alpha $
The dissociation constant for the weak acid is thus given by
$${K_a}{\text{ = }}\frac{{\left[ {{H^ + }} \right]\left[ {{B^ - }} \right]}}{{\left[ {HB} \right]}}{\text{ = }}\frac{{({C_2}\alpha + {C_1}){C_2}\alpha }}{{{C_2}(1 - \alpha )}}{\text{ }}$$
From here, we can calculate $\alpha $ and then find the pH after determining the $\left[ {{H^ + }} \right]$
$$\left[ {{H^ + }} \right] = {C_2}\alpha + {C_1}$$$$pH = - \log \left[ {{H^ + }} \right]$$