Question

Calculate concentration of species in a solution containing a complex ion. In the presence of excess OH^ -, the Al^ 3+ (aq) ion forms a hydroxide complex ion, Al(OH)^ -_4. Calculate the concentration of free Al^ 3+ ion when 1.64 times 10^-2 mol Al(CH_3COO)_3(s) is added to 1.00 L of solution in which [OH^ -] is held constant (buffered at pH 12.10). For Al(OH)_4^ -, K_f = 1.1 times 10^33.

Answer 1

pH = 12.10

pOH = 1.90

[OH-] = 0.01259 M

Al+3 + 4 OH-          ----------------> [Al(OH)4]-

1               4                                            1

1.64x10^-2      0.01259                              1.64x10^-2

Kf = [[Al(OH)4]-] / [Al+3][OH-]^4

1.1 x 10^33 =   1.64x10^-2 / [Al+3] (0.01259)^4

[Al+3] = 5.93 x 10^-28 M