# How do you calculate something on a pH scale?

Mar 18, 2014

Sorry, but this might get lengthy, so bear with me please. :)

#### Explanation:

$\text{pH}$, or potential of hydrogen, is measured on a scale from 0 to 14, with 0 being the most ACIDIC and 14 the most BASIC.

To find $\text{pH}$ from the concentration of ${\text{H"_3"O}}^{+}$ (or just simply ${\text{H}}^{+}$) you need to use the formula:

"pH"= -log["H"_3"O"^+]

The [${\text{H"_3"O}}^{+}$] is just the concentration (in molarity) found through calculations (I'll cover that soon).

If you have the concentration of ${\text{OH}}^{-}$, however, simply find the $\text{pOH}$ from the expression:

"pOH" = -log["OH"^-]

After you get this value, you use the formula:

$\text{pH= 14-pOH}$

OK, so let's start with the basics of determining your ${\text{H"_3"O}}^{+}$ or ${\text{OH}}^{-}$ concentrations.

Molarity is the standard unit for concentration in chemistry, and is simply moles of substance over liters of solution.

$M = \text{moles"/"liters}$

So whenever I say concentration, I mean molarity.

You find the concentration of ${\text{H"_3"O}}^{+}$ by first writing out your acid dissociation equation:

${\text{HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A}}^{-}$

… where $\text{HA}$ is simply the acid you're dissolving in water.

If you have a STRONG acid, then it dissociates completely in water.

The concentration of ${\text{H"_3"O}}^{+}$ is the same as the concentration of the initial acid.

Now, you were probably given the ${K}_{a}$ of the acid, telling you that it is a WEAK acid.

That means that it does NOT dissociate completely in water.

The ${K}_{a}$ at this point is just a number to plug into your equation.

To find the concentration of ${\text{H"_3"O}}^{+}$ from the ${K}_{a}$ and your equation, simply plug the numbers that you have into this expression:

${K}_{a} = \left(\left[\text{A"^-]["H"_3"O"^+])/(["HA}\right]\right)$

You can do the exact same thing if it's a BASIC solution.

Just replace [${\text{H"_3"O}}^{+}$] with [${\text{OH}}^{-}$] and don't forget to change the $\text{pH}$ to $\text{pOH}$.