# As the #H^+# in a solution decreases, what happens to the #OH^-#?

##### 2 Answers

As the

#### Explanation:

The Keq for water is

At the neutral point of water, both the Hydrogen ion and Hydroxide ion are equal at

If the Hydrogen ion decreases from

So as the Hydrogen (acid) ion decreases there is a corresponding increase in the Hydroxide (basic) ion.

If the

#### Explanation:

Keep in mind that chemist writes

**Why does the #[OH^-]# increases when the #[H_"3"O^"+"]# decreases?**

This follows from the rule:

This is the same as:

Therefore when

**How is #pH+pOH =14# established? **

In water, the following (ionization) reaction occurs:

Therefore the equilibrium can be written like

Since water is the solvent here, we do not take the concentration of water into consideration (is very big). We obtain the following expression:

The

This means we can say:

To get from the

The same is true for the

Now if we take the Log from both sides of the

A mathematics rule tells us that multiplying inside the logarithm function is the same as adding these logarithms. Therefore we get

And now we can use the definitions of pOH and OH! We get:

with

Which is the same as