How do you calculate something on a pH scale?

1 Answer

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

Explanation:

"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 "pH" from the concentration of "H"_3"O"^+ (or just simply "H"^+) you need to use the formula:

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

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

If you have the concentration of "OH"^-, however, simply find the "pOH" from the expression:

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

After you get this value, you use the formula:

"pH= 14-pOH"

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

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

M="moles"/"liters"

So whenever I say concentration, I mean molarity.

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

"HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^-

… where "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 "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 "H"_3"O"^+ from the K_a and your equation, simply plug the numbers that you have into this expression:

K_a=(["A"^-]["H"_3"O"^+])/(["HA"])

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

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