In this lesson, we will learn:

- To recall the expressions for pH and pOH.
- To use the antilog to relate pH and pOH back to aqueous ion concentration.
- How pH and pOH are related to the K
_{w}expression.

__Notes:__- We learned earlier in Introduction to acid-base theory , that pH is defined by the concentration of H
_{3}O^{+}ions in solution:pH = -log[H _{3}O^{+}] - In the same way, pOH can be measured, which is defined by the concentration of OH
^{-}ions in solution:pOH = -log[OH ^{-}]

Be careful with significant figures – with logarithms, only the values in decimal places are considered significant figures. - The reverse of the logarithm is known as the
__antilog__, so the antilog can be used to__convert pH into [H___{3}O^{+}]__and pOH into [OH__^{-}]. The antilog is found by rising 10 to the value for which you are getting the antilog:Antilog (x) = 10 ^{x}__Make sure your calculator gives antilogs in scientific notation, or standard form__. As stated above, the decimal places are the significant figures in a logarithm value. The first digit represents the order of magnitude. For example, log(10) = 2.0 and log(100) = 3.0; 3 is one greater than 2, so 3 as a logarithm is one order of magnitude (10x) greater than 2 as a logarithm.

With this, we can show expressions to find [H_{3}O^{+}] and [OH^{-}] using pH and pOH:[H _{3}O^{+}] = 10^{-pH}[OH ^{-}] = 10^{-pOH} - Because [H
_{3}O^{+}] and [OH^{-}] in aqueous solution at 25^{o}C are related to K_{w}, pH and pOH are related to pK_{w}– which is just the negative log of the K_{w}constant!- pH and pOH give logarithmic expressions of the aqueous ion concentration. Recall that:
K _{w}= [H_{3}O^{+}_{(aq)}] [OH^{-}_{(aq)}] = 1.00 $*$ 10^{-14}at 25^{o}CTaking the negative log of these aqueous ion concentrations, we can determine: pH + pOH = pK _{w}= 14

With these we can relate the four expressions in a ‘grid’ below:

- pH and pOH give logarithmic expressions of the aqueous ion concentration. Recall that: