Question

Calculate the solubility (in moles per liter) of Al(OH)3 (Ksp = 2 x 10-2) in each of the following. a. water Solubility = mol/L b. a solution buffered at pH = 5.0 Solubility = mol/L c. a solution buffered at pH = 10.0 Solubility = moll Submit Answer Try Another Version 6 item attempts remaining

Answer 1

Ansacy Given that, Ksp = 2x10-32 the dissociabon of Alcoh), is given as follows: Al(OH); (5) A3+(29) +30+569) @ water For water PH=7 is, in. PoH = 7 the concentration of ott pOH =-109[OH-] 7--109 [ow ] ToH] = 107 The expression for Kop is, *sp = [1134] [OH-]3 2X1032 = [1134] [1073 2X1632 = (A13+3x10-21 [A13+) = 2x1032 16-2] = 2x10m .: The solubility of AI COH)2 in water is [2x101 moll

© a solution buffered at pH: poll = 14-59 the centration of oil is calculaled af, poll ... 109 Coll") 92-109 Com] [of] = 109 The expe Ski oh fox K 57 ;s, Ksp = {01373c0H-73 21032= [A138] [16973 2x10 32 = ( 13+] x 1027 [A134) = 2X1072 10-27 = 2x10-5M. is the solubility of Alloh )z ot PH = 5 Taxio-5 mol/ is ©a.Soliton busse ved at pH = 10 pot = 14-10=4, the conceritation of Oh" is , POH --109[OH-] 4 = -log[OH-]

[04] = 10km Thue expression for Ksp is, Kgp = [A13+3 [om-33 2x1053 = [Ayat] [10473 2X10:32 =[A13+] x 1012 [A13+) = 2X1032 . 10-12 . [A13+] =-2X1020m These fose, the solubility of Al(OH)3 at pH = lo is / 2x1077omov /2/