6D.15
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6D.15
Postby vanessanguyen3I » Sat Jan 23, 2021 8:54 pm
The textbook problem is: Calculate the pH of (b) 0.055 M AlCl3 (aq). Can someone explain how the answer key came up with the original equation for this problem? Where does Al(H2O)6 3+ come from?
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Re: 6D.15
Postby Silvi_Lybbert_3A » Sat Jan 23, 2021 9:09 pm
If you think back to coordination compounds you know when AlCl3 dissolves in water it becomes Cl- ions and Al(H2O)6 3+ ions because the Al can accept the lone pairs on the oxygens from six water molecules. It forms an octahedral complex. Once Al+3 dissolves in water by being surrounded by six water molecules, one of these water molecules might give a hydrogen to a nearby water molecule. This can occur because the water molecules attached to the Al can hydrogen bond with other near by water molecules, and one of those hydrogen bonds might become a covalent bond as the nearby water molecule becomes H3O+ and the water molecule attached to the Al loses a hydrogen and becomes a negatively charged OH- group. So the aluminum coordination compound acts as an acid: [Al(H2O)6]3+ + H20 <---> H3O+ + [Al(H2O)5OH]2+
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