Solveeit Logo
Login

Question

Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.

Answer

c = 0.005
pH = 9.95
pOH = 4.05
pH = log (4.105)
4.05 = - log [OH-] = [OH-] = 8.91 × 10-5 = ca = 8.91 × 10-5 = a = 8.91×10−55×10−3\frac{8.91 \times 10^{-5}}{5\times10^{-3}} = 1.782 × 10-2
Thus, Kb = ca2 = 0.005 × (1.782)2 × 10-4 = 0.005 × 3.1755 × 10-4 = 0.0158 × 10-4
Kb = 1.58 × 10-6
Pkb = -log kb = -log(1.58 × 10-6) = 5.80