What is the reaction of an ion with #H_2O# to produce #H^+(aq)# and #OH^(-)(aq)#?

1 Answer
Apr 12, 2017

Answer:

We write.........#H_2O(l) rightleftharpoonsH^(+) + HO^-#

Explanation:

OR, MORE COMMONLY..........

#2H_2O rightleftharpoonsH_3O^+ +HO^-#

This is the so-called #"autoprotolysis reaction of water"#, which has been meticulously studied, and underlies our conception of acid/base chemistry.

The #"hydronium ion"#, #H_3O^+#, and #"hydroxide ion"#, #HO^-#, are conceived to be the characteristic cation and characteristic anion of the #"WATER SOLVENT"#, i.e. #"the acidium ion"#, and #"the hydroxide ion"#. I write #"conceived"# because the acidium ion in water solution is probably #H_5O_2^+# or #H_7O_3^+#, a cluster of several water molecules WITH AN EXTRA #H^+# associated with the cluster. We write #H^+#, or #H_3O^+# to represent this species, i.e. as a shorthand.

In water at #298*K#,

#2H_2O rightleftharpoonsH_3O^+ +HO^-# #K_w=[H_3O^+][HO^-]=10^-14#.

Given that this is a bond-breaking reaction, how do you think this equilibrium would evolve AT HIGHER TEMPERATURES? (If you are not an undergrad, don't bother with this question).