Is the relationship between Ka and Kb the same for all acid base pairs? [closed]








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What is the relationship between $K\_\mathrm{a}$ and $K\_\mathrm{b}$ for acid/base reactions. I know for water $K\_\mathrm{w}=K\_\mathrm{a}\cdot K\_\mathrm{b}$, where $K\_\mathrm{w}$ is $10^{-14}$. Is this true for other acid/base pairs?




First of all, let me put a statement to clarify the proper meanings of $K\_\mathrm{a}$ and $K\_\mathrm{b}$ for this case.  

**The $K\_\mathrm{a}$ is the acid dissociation constant of a weak acid, and the $K\_\mathrm{b}$ is the dissociation constant of its conjugate base. Alternatively, we can also say, $K\_\mathrm{b}$ is the base dissociation constant of a weak base, and $K\_\mathrm{a}$ is the dissociation constant of its conjugate acid.**


We can deduce the same relationship, i.e. $K\_\mathrm{a}\cdot K\_\mathrm{b} = K\_\mathrm{w}$, for these cases also. Without loss of generality, let the weak acid be $\ce{HA}$ and its conjugate base is $\ce{A^-}$ (you can also do the same [not identical] derivation considering base $\ce{B}$ and its conjugate acid $\ce{BH^+}$). 


Considering acid dissociation, 
\begin{align}
\ce{HA &<=> H^+ + A^-}& 
K\_\mathrm{a} &= \ce{\frac{[H^+][A^-]}{[HA]}},
\end{align} 
and for the basic action of its conjugate base, consider,
\begin{align}
\ce{A^- + H2O &<=> HA + OH^-}&
K\_\mathrm{b} &= \ce{\frac{[HA][OH^-]}{[A^-]}}.
\end{align} 
Multiplying both the equations, we will have, 
$$
K\_\mathrm{a}\cdot K\_\mathrm{b} 
= \ce{\frac{[A^-][H^+]}{[HA]}}\cdot \ce{\frac{[HA][OH^-]}{[A^-]}} 
= \ce{[H^+][OH^-]} = K\_\mathrm{w}$$ 
So, it is proved that for any acid/conjugate base pair (or, base/conjugate acid pair) the identical relation, $K\_\mathrm{a}\cdot K\_\mathrm{b} = K\_\mathrm{w}$ holds.


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