Help:Equations/Conjugate Acid-Base Pairs and pH

<math>K_{a}\text{(HA)}=\frac{[\text{ H}
_{\text{3}}\text{O}^{\text{+}}][\text{ A}^{-}\
text{ }]}{[\text{ HA }]}</math>

<math>K_{b}\text{(A}^{-}\text{)}
=\frac{[\text{ HA }][\text{ OH}^{-}\text{ }]}
{[\text{ A}^{-}\text{ }]\text{ }}</math>

<math>\begin{align}
K_{a}\text{(HA)}\times K_{b}\text{(A}^{-}
\text{)}&=\frac{[\text{H}_{\text{3}}\text{O}
^{\text{+}}][\text{A}^{-}]}{[\text{HA }]}\
times \frac{[\text{HA}][\text{OH}^{-}]}
{[\text{A}^{-}]}\\
\text{ }\\
\text{ }&=[\text{H}_{\text{3}}\text{O}^
{\text{+}}][\text{ OH}^{-}]\end{align}</math>

<math>K_{a}\text{(HA)}\times K_{b}
\text{(A}^{-}\text{)}=K_{w}</math>

<math>\begin{align}
\text{p}K_{a}&=-\text{log}\frac{K_{a}
\text{(HA)}}{\text{mol dm}^{-\text{3}}}-\
text{log}\frac{K_{b}\text{(A}^{-}\text{)}}
{\text{mol dm}^{-\text{3}}}\\
\text{ }\\
\text{ }&=-\text{log}\frac{\text{10}^{-\text
{14}}\text{ mol}^{\text{2}}\text{ dm}^{-\text
{6}}}{\text{mol}^{\text{2}}\text{ dm}^{-6}}\
end{align}</math>

<math>\text{p}K_{a}\left(\text
{HA}\right) + \text{ }\text{p}K_{b}\text
{(A}^{-}\text{)}=\text{p}K_{w}</math>