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:

$\displaystyle \text{pH} = - \log{\left [ \text{H}^{+} \right ]} \qquad \Leftrightarrow \qquad \left [ \text{H}^{+} \right ] = 10^{- \text{pH}}$.

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

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

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

$\displaystyle \text{pOH} = - \log{\left [ \text{OH}^{-} \right ]} \qquad \Leftrightarrow \qquad \left [ \text{OH}^{-} \right ] = 10^{- \text{pOH}}$.

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

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

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:

$\displaystyle \left [ \text{H}^{+} \right ] = 2 \left [ \text{H}_{2} \text{SO}_{4} \right ] = 2 \left ( 0.042 \, \text{mol/L} \right ) = 0.084 \, \text{mol/L}$.

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

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

and then we can calculate pOH with

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

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

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