Question

The pH of an aqueous solution of 0.572 M caffeine (a weak base with the formula C₈H₁₀N₄O₂) is:

Answer 1

Let α be the dissociation of the weak base caffeine
                            BOH    B ⁺ + OH^-

initial conc.            c               0         0

change               -cα            +cα      +cα

Equb. conc.         c(1-α)        cα      cα

Dissociation constant, K_b = (cα x cα) / ( c(1-α)               

                                          = c α² / (1-α)

In the case of weak bases α is very small so 1-α is taken as 1

So K_b = cα²

==> α = √ ( K_b / c )

Given K_b = 4.1x10^-4

          c = concentration = 0.572 M

Plug the values we get α = 0.0268
So the concentration of [OH^-] = cα

                                               = 0.572 x0.0268
                                               = 0.0153 M

pOH = - log [OH^-]

        = - log 0.0153

        = 1.81

So pH = 14 - 1.81 = 12.19