Ecuaciones clave

Kw = [H3O+][OH−] = 1,0 $\times$ 10−14 (a 25 °C)

$\text{pH}=\text{−log}[{\text{H}}_3{\text{O}}^{\text{+}}]$

pOH = −log[OH−]

[H3O+] = 10−pH

[OH−] = 10−pOH

pH + pOH = pKw = 14,00 a 25 °C

$K_{\text{a}}=\frac{[{\text{H}}_3{\text{O}}^{\text{+}}]\left[{\text{A}}^{\text{–}}\right]}{\left[\text{HA}\right]}$

$K_{\text{b}}=\frac{[{\text{HB}}^{\text{+}}]\left[{\text{OH}}^{\text{–}}\right]}{\left[\text{B}\right]}$

Ka $\times$ Kb = 1,0 $\times$ 10−14 = Kw

$\text{Porcentaje de ionización}=\frac{{[{\text{H}}_3{\text{O}}^{\text{+}}]}_{\text{eq}}}{{[\text{HA]}}_0}\times 100$

pKa = −log Ka

pKb = −log Kb

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\frac{\left[{\text{A}}^{\text{–}}\right]}{\left[\text{HA}\right]}$