Solveeit Logo

Question

Question: If K<sub>a</sub> = 10<sup>–5</sup> for a weak acid, then pK<sub>b</sub> for its conjugate base would...

If Ka = 10–5 for a weak acid, then pKb for its conjugate base would be -

A

10–10

B

9

C

10–9

D

5

Answer

9

Explanation

Solution

Kb = KwKa=1014105\frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{10^{- 5}}= 10–9

\ pKb = –log Kb = –log 10–9 = 9