Question

The pH of an aqueous solution of 0.462 M caffeine (a weak base with the formula C8H10N4O2) is

Answer 1

[B]=[C,H,N.02] Equation representing the ionization of B is, B + H2O + BH* + OH Baseionization constant K, of B =4.1x104 B + H2O = BH* + OH Initialconcentration (M): 0.462 0 0 Change(M) -X + x + x Equilibrium concentration (M): 0.462 - x [BH* [OH] Baseionization constant, K, = [B] 4.1*10-4 = (x)(x) 0.462-X 4.1x10-=- 0.462 - x We can assume that (0.462-x) M = 0.462M 1 4.1x10-4 = 0.462 x? =(4.1x10*)(0.462) x= (4.1x10-^)(0.462) = 1.38x10 PM From the equilibrium table, x = [OH ] = 1.38x10 PM pOH=- log [OH] =- log(1.38x10-2) =1.86 pH=14.00 -1.86 =12.13 Therefore, pH of the 0.462M aqueous base solution is 12.13 |