What is stoichiometry with acid and base dissociation?

1 Answer
Aug 12, 2015


I have to guess what you are asking so I apologize if I am wrong. Stoichiometry is simply a word that means "in the right measure".


So, an example of stoichiometry. If you went out to the shops with a £20 note, and bought 4L of milk costing £1-20, when you received the change you would immediately know whether or not you had been short-changed. Either the grocer had made a mistake, or was trying to short change you. Nevertheless, you would have cause for complaint if you didn't get £18-80 precisely. This is a stoichiometric transaction. In all chemical reactions, mass is conserved in the same way as change: if you start with 10 g of total reactant, at most you are going to get 10 g of product in total, because mass is conserved in every chemical reaction. In practice, you are not even going to get that.

So, because atoms have finite and definite mass, it follows that they will combine to preserve the same mass, If you have an unknown quantity of sodium hydroxide solution, and you react this with a known quantity of hydrochloric acid, you can calculate both the amount and mass of the sodium hydroxide, by reference to the known amount of hydrochloric acid you use. We can represent this reaction in words,

Sodium hydroxide + hydrogen chloride#rarr# sodium chloride + water.

Or, more conveniently, we can represent it with symbols:

#(i) NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)#

In fact, we can be even more concise than this, because we are not really interested in the #NaCl(aq)#; we are interested in what is called the neutralization of the acid by the base:

#(ii) OH^-# #+# #H^+ rarr H_2O#

or (where protium ion #H^+# is represented as the hydronium ion, #H_3O^+, # and I assure you they are the same).

#(iii) OH^-# #+# #H_3O^+ rarr 2H_2O#

In one way or another, equation (ii) and (iii) are all that is occurring in an acid/base reaction, and represents acid/base stoichiometry in water. Hydroxide anion, #OH^-#, is neutralized by the protium ion, #H^+#, to give #H_2O#.

(ii) and (iii) are the same reactions (you'll see both in texts, though your teacher may have a personal preference). Equations (ii) and (iii) are essentially all you need to know about acid/base stoichiometry. You still have to know which acids give 2 equivalents of #H^+#, sulfuric acid, #H_2SO_4# does so, and which bases give rise to 2 equiv #OH^-#, e.g. #Mg(OH)_2, Ca(OH)_2#. Practice is the key.