#"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"#.
