Metal ions in aqueous solutions are solvated, or hydrated by water molecules. Ty
ID: 1010567 • Letter: M
Question
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 H ion from its water ligands to the surrounding free water molecules. Shown below is the hydrolysis of Al(H2O)63 (pKa = 4.85) in water Calculate the pH of a 0.00107 M AlCl3 solution and determine what fraction of the aluminum is in the form Al(H2O)5OH2 .
Explanation / Answer
AlCl3 shall undergo hydrolysis in aqueous solution as
AlCl3 (aq) + 6 H2O (l) -------> [Al(H2O)6]3+ (aq) + 3 Cl- (aq) ……(1)
We note that AlCl3 and [Al(H2O)6]3+ have a 1:1 molar ratio, thus the starting concentration of [Al(H2O)6]3+ is 0.00107 M (same as the concentration of AlCl3). Now, [Al(H2O)6]3+ undergoes ionization as
[Al(H2O)6]3+ (aq) <======> [Al(H2O)5OH]2+ (aq) + H+ (aq)
initial 0.00107 0 0
change - x + x + x
equilibrium (0.00107 – x) x x
The equilibrium constant for the ionization reaction (i.e, the acid dissociation constant) is
Ka = [[Al(H2O)5OH]2+][H+]/[[Al(H2O)6]3+] where the first square braces […] indicate molar concentration.
Now, pKa = 4.85 (given)
===> Ka = 10-4.85 = 1.4125*10-5 (I will keep a few guard digits extra).
Therefore, 1.4125*10-5 = (x)(x)/(0.00107 – x)
Now, we use an approximation that x << 0.00107 M, hence we can neglect the x in the denominator. This gives
1.4125*10-5 = x2/0.00107
===> x2 = 1.5114*10-8
===> x = 1.2294*10-4
The molar concentration of proton at equilibrium is 1.2294*10-4 M and pH = -log10[H+] = - log10(1.2294*10-4) = 3.9103 3.91
The pH of the aqueous solution is 3.91 (ans)
The equilibrium concentration of [Al(H2O)5OH]2+ is 1.2294*10-4 M. The percent dissociation =
([[Al(H2O)5OH]2+]eqbm/[[Al(H2O)6]3+]initial)*100
= ((1.2294*10-4 M)/(0.00107 M))*100 = 11.4897 11.49
The dissociation of [Al(H2O)6]3+ in aqueous solution is 11.49% (ans)