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# Help:Equations/Conjugate Acid-Base Pairs and pH

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$K_{a}\text{(HA)}=\frac{[\text{ H}_{\text{3}}\text{O}^{\text{+}}][\text{ A}^{-}\text{ }]}{[\text{ HA }]}$
$K_{a}\text{(HA)}=\frac{[\text{ H} _{\text{3}}\text{O}^{\text{+}}][\text{ A}^{-}\ text{ }]}{[\text{ HA }]}$
$K_{b}\text{(A}^{-}\text{)}=\frac{[\text{ HA }][\text{ OH}^{-}\text{ }]}{[\text{ A}^{-}\text{ }]\text{ }}$
$K_{b}\text{(A}^{-}\text{)} =\frac{[\text{ HA }][\text{ OH}^{-}\text{ }]} {[\text{ A}^{-}\text{ }]\text{ }}$
\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}
\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}
$K_{a}\text{(HA)}\times K_{b}\text{(A}^{-}\text{)}=K_{w}$
$K_{a}\text{(HA)}\times K_{b} \text{(A}^{-}\text{)}=K_{w}$
\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}
\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}
$\text{p}K_{a}\left(\text{HA}\right) + \text{ }\text{p}K_{b}\text{(A}^{-}\text{)}=\text{p}K_{w}$
$\text{p}K_{a}\left(\text {HA}\right) + \text{ }\text{p}K_{b}\text {(A}^{-}\text{)}=\text{p}K_{w}$