Question

1) Drive the formula for the ionization constant $\left(K_b\right)$ for weak bases.
2) Drive the relation $pH+pOH=pKw$.

Answer 1

Finding $K_b$ for $K_b$ reacting as a base:
1. Since $A^{-}$ is a base we can also write the reversible reaction for $A^{-}$ acting as a base by accepting a proton from water:
$A_{\left(aq\right)}^{-}+H_2{O}_{\left(l\right)}\leftrightharpoons H{A}_{\left(aq\right)}+{OH}_{\left(aq\right)}^{-}$
The products of this reaction are $HA$ and ${OH}^{-}$. We can write out the equilibrium constant $K_b$for the reaction when $A^{-}$ acts as base
$K_b$=$\frac{\left[HA\right]\left[{OH}^{-}\right]}{{[A}^{-}]}$
Even though this almost looks like the reverse of $HA$ acting as an acid, they are actually very different reactions. When the conjugate base,$A^{-}$ acta as a base, one of the product is ${OH}^{-}$.
2. $pH$ and $Kw$:
$pH$ is defined for aqueous solutions only, and is temperature dependent
$pH=-Log\left[H_3{O}^{+}\right]$
${10}^{-pH}=\left[H_3{O}^{+}\right]$
It derives from the auto ionization of water.
$Kw=$ $\left[H_3{O}^{+}\right]$$\left[{OH}^{-}\right]$
$pKw=pH+p{OH}^{-}$.
$pH>7$ is basic , $pH=7$ is neutral , $pH<7$ is acidic.