Question

If the Ka of the conjugate acid is 4.83 × 10^-8 , what is the pKb for the base?

Answer 1

Explanation:

We are given: Ka = 4.83*10^-8

We use Ka to find pKa:

$\begin{gathered} pKa\text{ = -log\lbrack Ka\rbrack} \\ \\ \text{ = -log\lbrack4.83}\times10^{-8}] \\ \\ \text{ = 7.32} \end{gathered}$

We know: pKa + pKb = 14

$\begin{gathered} pKa+pKb\text{ = 14} \\ \\ \therefore pKb\text{ = 14-pKa} \\ \\ \text{ = 14-7.32} \\ \\ \text{ = 6.68} \end{gathered}$

Answer:

pKb = 6.68