Question

Calculate the solubility (in moles per liter) of Al(OH)3 (Ksp = 2 x 10–32) in each of the following. a. water Solubility = mol/L b. a solution buffered at pH = 6.0 Solubility = mol/L c. a solution buffered at pH = 11.0 Solubility = mol/L

Answer 1

Let the solubility of Alwe in wate, be a molle Mariongly soluble salt ALPH11) Aud?ug) 3ošt(ag) kos = XX1032 Ab tabaq um 30M [AUT] [or] 2 2410- [ALOHA (1)}" (actinity of pure tolid is Censidered to be unity) af eguil'snkin 2 x (3713 Si o 2x 70-34 1 - 27 x = 2x1532 re". 0,0741x/0+32 2.- 10.0741 X/0SM яго» г17 X-Ph لطاکار ) - /Pr 21: * T J Al(Olt), waster)

b) in a solution buffered ee het the solubility of Allot + ри» с. г. to, pH = 6 . Putla 10-ph = 10M arnociation Constant of waters so, con) - kú -14 og [ue] ASIOH) (0) 2 Alt + 304 (49), ksp = 2X10²32 Aft=t 10-8m Att tą - yn gootsyym Z 70-8 (10-93x yo 2x1632 Y = 2x10-32 To-zy 2x10d more c. het the solability of Allohg in a solution pholl be ' mole buffered a l кг 11 Эрон 14 – рн = ? a cor rolopona 103 AllOH) is = Alts carst 304 Curs lkeen. - At tat 2014 (cal, (ASPALCOH ۱۵n At tzter - 153432 7163M

(103) x 2 - 2810 32 zo 2x 10-22 = 2x10 23 mole 10-9