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 = KaKw=10−510−14= 10–9
\ pKb = –log Kb = –log 10–9 = 9