"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".