Answer
$C_9H_7NH^+(aq) + H_2O(l) C_9H_7N(aq) + H_3O^+(aq)$
$pKb (Quinoline) = 4.8$
Work Step by Step
1. Write the reaction where the protonated quinoline reacts with water, and gives a proton to it.
$C_9H_7NH^+(aq) + H_2O(l) C_9H_7N(aq) + H_3O^+(aq)$
2. Calculate the pKa of protonated quinoline:
$pKa = -log(Ka) = -log(6.3 \times 10^{-10}) = 9.2$
3. For a conjugate acid-base pair:
$pKa+ pKb = pKw = 14$
$pKb = 14 - pKa = 14 -9.2 = 4.8$
