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

Apr 5, 2017

As the $\left[{H}^{+}\right]$ ion decreases the $\left[O {H}^{-}\right]$ ion increase.

#### Explanation:

The Keq for water is ${10}^{-} 14$ This means that the Hydrogen ion concentration times the Hydroxide concentration must always be the same ${10}^{-} 14$

$\left[{H}^{+}\right] \times \left[O {H}^{-}\right] = {10}^{-} 14$ This means that the two ions are inversely proportional. If one goes up the other goes down.

At the neutral point of water, both the Hydrogen ion and Hydroxide ion are equal at ${10}^{-} 7$

$10 - 7 \times {10}^{-} 7 = {10}^{-} 14$

If the Hydrogen ion decreases from ${10}^{-} 7$ to${10}^{-} 8$ The Hydroxide ion must increase from ${10}^{-} 7$ to ${10}^{-} 6$

${10}^{-} 8 \times {10}^{-} 6 = {10}^{-} 14$

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

Apr 5, 2017

If the $\left[{H}^{+}\right]$ decreases, the $\left[O {H}^{-}\right]$ increases.

#### Explanation:

Keep in mind that chemist writes $\left[{H}^{+}\right]$ because it is simple, but in the solution, this will always form $\left[{H}_{\text{3"O^"+}}\right]$.

$\to$Why does the $\left[O {H}^{-}\right]$ increases when the $\left[{H}_{\text{3"O^"+}}\right]$ decreases?
This follows from the rule:
$p H + p O H = 14$
This is the same as:
K_"w"=[H_"3"O^"+"]*[OH^"-"]=1*10^(-14)
Therefore when $\left[{H}_{\text{3"O^"+}}\right]$, the $\left[O {H}^{-}\right]$ must increase to get the same ${K}_{\text{w}}$ value.

$\to$How is $p H + p O H = 14$ established?
In water, the following (ionization) reaction occurs:
$2 {H}_{\text{2"O -> H_"3"O^"+" + OH^"-}}$

Therefore the equilibrium can be written like
K_"c"=([H_"3"O^"+"]*[OH^"-"])/[H_"2"O]
Since water is the solvent here, we do not take the concentration of water into consideration (is very big). We obtain the following expression:
K_"c"=[H_"3"O^"+"]*[OH^"-"]

The ${K}_{\text{c}}$ in this equation represents a special number because we talk about the ionisation of water. Therefore we denote ${K}_{\text{c}}$ as ${K}_{\text{w}}$. The value of the ${K}_{\text{w}}$ is measured at 25°C.
K_"w" (25°C) = 1*10^(-14)
This means we can say:
K_"c"=K_"w"=[H_"3"O^"+"]*[OH^"-"]=1*10^(-14)

To get from the $\left[{H}_{\text{3"O^"+}}\right]$ (concentration ${H}_{\text{3"O^"+}}$) to the pH, we use the following formula:
$p H = - \log \left[{H}_{\text{3"O^"+}}\right]$
The same is true for the $\left[O {H}^{\text{-}}\right]$, since we define pOH as
$p O H = - \log \left[O {H}^{\text{-}}\right]$

Now if we take the Log from both sides of the ${K}_{\text{w}}$ equation, we get:
$\log \left(1 \cdot {10}^{- 14}\right) = \log \left(\left[{H}_{\text{3"O}}\right] \cdot \left[O {H}^{-}\right]\right)$
A mathematics rule tells us that multiplying inside the logarithm function is the same as adding these logarithms. Therefore we get
$\log \left({10}^{- 14}\right) = \log \left[{H}_{\text{3"O}}\right] + \log \left[O {H}^{-}\right]$

And now we can use the definitions of pOH and OH! We get:
$\log \left({10}^{- 14}\right) = - p H - p O H$
with $\log \left({10}^{- 14}\right) = - 14$ we get the function
$- p H - p O H = - 14$
Which is the same as
$p H + p O H = 14$