Question #b01a3

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Question #b01a3

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Answer 1 · 2015-10-05T17:19:25

The 1st case involves hydrolysis. The 2nd case is just dilution so is not really a reaction.

Explanation:

Aluminium chloride is a white solid which reacts vigorously with water to give hydrated $A l^{3 +}$ $Al^(3+)$ ions and $C l^{-}$ $Cl^-$ ions:

$A l_{2} C l_{6} + 12 H_{2} O \to 2 {[A l {(H_{2} O)}_{6}]}_{6}^{3 +} + 6 C l^{-}$ $Al_2Cl_6+12H_2Orarr2[Al(H_2O)_(6)]^(3+)+6Cl^(-)$

The small, highly charged $A l^{3 +}$ $Al^(3+)$ weakens the $O - H$ $O-H$ bonds resulting in a hydrolysis reaction:

${[A l {(H_{2} O)}_{6}]}_{6}^{3 +}$ $[Al(H_2O)_(6)]^(3+)$ $+ H_{2} O \to {[A l {(H_{2} O)}_{5} O H]}_{5}^{2 +} + H_{3} O^{+}$ $+H_2Orarr[Al(H_2O)_(5)OH]^(2+)+H_3O^(+)$

The resulting solution is, therefore, quite acidic.

Hydrated aluminium chloride cannot exist as a solid for this reason so adding water to its solution will simply dilute it.

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Question #b01a3

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