# --- Chemistry ---

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What is the most common chemical method for removing H_30^+ ions in aqueous solution? Write a net ionic equation which describes this method.

What happens to the H concentration when the OH concentration goes up?

We havn't started Acids/Bases yet in class, in fact, this question comes from an equilibrium lab.

I know that NaOH is a base, so if I added it, it would remove the H_30 ions? Would my net equation look something like this?:

H_30(aq) + NaOH(aq) --> Na_2O + H_40_2

Well, if you havent had acids and bases, forget it.

The point you will learn is that the product of the OH concentration and H3O ions is constant. So if you add OH ions, the H3O goes down.

In this case, you would get..
H_30(aq) + NaOH(aq) --> Na_2O + 2H_20 .

As a practical matter, this wont work, but you have not learned it yet.

In any water solution, the product of the H3O+ and the OH- concentration is constant. If the hydronium ion concentration is to go down, OH has to go up. I doubt if you have learned that yet.

One final question(s) about what you've just taught me and some stuff I've just looked up on the internet: If a solution contains hydronium then that is the definition of an acid? Therefore the only way to remove an acid is to add a base or to neutralise the two by having the ammount of H30=OH? If then OH > H30 does that mean that the H30 ions have been removed? Should I say you neutralise the solution to remove the ions or that you have more base than acid?

You are too far ahead of yourself. If the product of OH and H+ is always constant, then increasing OH reduces H+. I would avoid the term neutralization. If the product is constant, removing on H+ by a reaction, leaving OH consant, then the water will produce another H+ in its place. The product is constant. Pure Water is neutral, having H+ exactly equal to the OH. At 25C, the product of the ions is 1*10^-14 moles/liter, that is

conc H+ * Conc OH- = 1*10^-14 moles/liter,

This is constant of nature.

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