# If pK_a=7.4, what is pK_b?

Oct 28, 2016

${K}_{a} \times {K}_{b}$ $=$ ${K}_{w}$, where ${K}_{w} = {10}^{-} 14$ under standard conditions.

#### Explanation:

Thus ${K}_{b} = {K}_{w} / {K}_{a}$ $=$ ${10}^{-} \frac{14}{4 \times {10}^{-} 8}$.

We can make it a bit more simply if we take logarithms to the base 10 of each side:

${\log}_{10} {K}_{b} = {\log}_{10} {K}_{w} - {\log}_{10} {K}_{a}$

OR

$- {\log}_{10} {K}_{w} = - {\log}_{10} {K}_{a} - {\log}_{10} {K}_{a}$

But $- {\log}_{10} {K}_{w}$ $=$ $- {\log}_{10} {10}^{-} 14$ $=$ $- \left(- 14\right)$, and $- {\log}_{10} {K}_{a} = p {K}_{a}$, and $- {\log}_{10} {K}_{p} = p {K}_{b}$

And thus, finally, $p {K}_{w} = 14 = p {K}_{a} + p {K}_{b}$

Since $p {K}_{a} = - {\log}_{10} \left(4.0 \times {10}^{-} 8\right) = - \left(- 7.40\right) = 7.4$.

$p {K}_{b} = 14 - 7.40 = 6.60$.

And (finally!) ${K}_{b} = {10}^{- 6.60}$ $=$ ??