Question

2nd part only the pH is correct.

Metal ions in aqueous solutions are solvated, or hydrated, by water molecules. Typically this primary hydration sphere is composed of six water molecules. The hydrated metal ion acts as a weak acid, undergoing a stepwise hydrolysis in which it donates an Htion from its water ligands to the surrounding free water molecules. Shown is the hydrolysis of Al(H,0) (pKA = 4.85) in water. Al(H,0)(aq) + H20(1) - 1,0+ (aq) + Al( HO), OH (aq) Calculate the pH of a 0.00267 M AICI, solution and determine what fraction of the aluminum is in the form Al(H,O), OH . pH = 3.70 "AH,0),(OH) = .075

Answer 1

pka = 4.85 pka-log Ka = 4.85 Ka= 1.4125x10-5 AICH2O) = 3 + + 40 130t + Altos 01424 4 to c c cd egin: C-cd cd cd Ka = cd² d= 1 kg = 1 1. 41256 2=0.072 (:27 0.05) + approximateen formula cannot be worked out. Ka= (62) (cd) = e de 7(1-) 1- ka - Kad = ca ² ca tkad - Ka=0 & – Ka & J Kaj? - 4(c) (-ka) 2 (0) = - 1.4125x105 + √(1-4125x16 5)2 - 4 (0.00267) (-7.4125 X155) 210.60267) 2=0.07014 (neglecting we value) [A1 (140), OH)²+ = C& = 0.00267(0.07014) = (1.873 8104 M