pH & pOH

pH is a unitless number used to convey the acidity of a substance without having to deal with numbers in scientific notation. You must use an extra significant digit when calculating it (and drop a digit when you are going from pH to concentration). You can go between pH and concentration with these equivalent formulae:

$\text{pH}=-\log [{\text{H}}^{+}]⟺[{\text{H}}^{+}]={10}^{-\text{pH}}\text{.}$

The pH value tells us about the acidity of a solution:

• $\text{pH}=7⟹$ neutral
• $\text{pH}<7⟹$ acidic
• $\text{pH}>7⟹$ basic

pOH is just like pH, except it uses the hydroxide ion concentration:

$\text{pOH}=-\log [{\text{OH}}^{-}]⟺[{\text{OH}}^{-}]={10}^{-\text{pOH}}\text{.}$

If you have one of pH and pOH, you can easily find the other (at SATP):

$\text{pH}+\text{pOH}=14\text{.}$

Example

Calculate the concentrations of H⁺_(aq) and OH⁻_(aq) and the values of pH and pOH for a 0.042 mol/L H₂SO_4(aq) solution.

H₂SO_4(aq) is a strong acid, so finding the H⁺_(aq) concentration is trivial:

$[{\text{H}}^{+}]=2[{\text{H}}_2{\text{SO}}_4]=2(0.042\text{mol/L})=0.084\text{mol/L}\text{.}$

Now we can use this to calculate pH. Remembering to keep an extra significant digit, we have

$\text{pH}=-\log [{\text{H}}^{+}]=-\log (0.084\text{mol/L})=1.08\text{,}$

and then we can calculate pOH with

$\text{pOH}=14-\text{pH}=14-1.08=12.9\text{.}$

From that we can get the hydroxide ion concentration, remembering to drop the extra significant digit:

$[{\text{OH}}^{-}]={10}^{-\text{pOH}}={10}^{-12.9}=1.2\times {10}^{-13}\text{mol/L}\text{.}$