6.24: pH and pOH, pKa and pKb

pH and pOH

The concentration of ions in solution can be quite high- and can differ from solution to solution by many orders of magnitude. For this reason, it is useful to refer to a logarithmic scale. We will use an operator \(\mathrm{p}\) defined to be the negative \(\log\) (base 10) of the value it precedes: for example, \(\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]\).

The \(\mathrm{pH}\) of a solution is a measure of how acidic it is: a \(\mathrm{pH}\) of 7 indicates the solution is neutral, while lower values are acidic and higher values are basic. The compliment to \(\mathrm{pH}\) is \(\mathrm{pOH}\), a measure of the concentration of hydroxide ions: \(\mathrm{pOH}=-\log _{10}\left[\mathrm{OH}^{-}\right]\). A high \(\mathrm{pOH}\) corresponds to an acidic solution, while a low \(\mathrm{pOH}\) indicates a basic solution. \(\mathrm{pH}\) and \(\mathrm{pOH}\) are related as

\(p H+p O H=14\)

Example: The \(\mathrm{K}_a\) for nitrous acid, \(\mathrm{HNO}_2\), is \(4.0 \times 10^{-4}\). Determine the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of \(0.1 \mathrm{M}\) nitrous acid. 

Answer

First, let's write an expression for the dissociation of nitrous acid:

\[\mathrm{HNO}_2(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{NO}_2^{-}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q) \nonumber\]

Then, we can write an expression for the acid dissociation constant:

\[K_a=\dfrac{\left[\mathrm{NO}_2^{-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{HNO}_2\right]} \nonumber\]

To find the \(\mathrm{pH}\), we need to determine the concentration of \(\mathrm{H}^{+}\) ion (equivalently \(\mathrm{H}_3 \mathrm{O}^{+}\)). The initial concentration of nitrous acid is \(0.1 \mathrm{M}\): we can set up an ICE table to solve for the final concentrations:

\(\begin{array}{c|c|c|c} 
& {\left[\mathrm{HNO}_2\right]} & {\left[\mathrm{NO}_2^{-}\right]} & {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]} \\
\hline \hline \text { Initial } & 0.1 & 0 \mathrm{M} & 0 \mathrm{M} \\
\text { Change } & -x \mathrm{M} & +x \mathrm{M} & +x \mathrm{M} \\
\text { Final } & 0.1-x \mathrm{M} & x \mathrm{M} & x \mathrm{M}
\end{array}\)

Plugging in the final values to the expression for \(\mathrm{K}_a\), we can solve for \(x\) to find the concentration of hydrogen ions:

\begin{gathered}
K_a=\dfrac{\left[\mathrm{NO}_2^{-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{HNO}_2\right]}=\frac{x^2}{0.1-x}=4.0 \times 10^{-4} \\
x^2=4.0 \times 10^{-5}-x\left(4.0 \times 10^{-4}\right) \\
x^2+x\left(4.0 \times 10^{-4}\right)-4.0 \times 10^{-5}=0 \\
{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=x \approx 0.006}
\end{gathered}

Finally, we can solve for \(\mathrm{pH}\) :

\[p H=-\log \left[H^{+}\right]=-\log \left[H_3 O^{+}\right]=-\log (0.006)=2.21 \nonumber\]

The value of \(\mathrm{pOH}\) follows using the relation that \(\mathrm{pH}+\mathrm{pOH}=14\) :

\[p O H=14-p H=11.79 \nonumber\]

Example: What is the \(\mathrm{pH}\) of \(0.04 \mathrm{M} \mathrm{HI}\), which is a strong acid?

Answer

The key phrase to solving this problem is 'strong acid.' When we have a strong acid, we don't need an acid dissociation constant because we know it fully dissociates! Here, the concentration of hydrogen ion simply equals the initial concentration of acid:

\[p H=-\log \left[H^{+}\right]=-\log (0.04)=1.4 \nonumber\]

\(\mathbf{p K}_a\) and \(\mathbf{p K}_b\)

In the same way that it is convenient to express the concentration of hydronium or hydroxide ions as a logarithmic value, it can be useful to express the acid and base dissociation constants as \(\mathrm{pK}_a\) and \(\mathrm{pK}_b\), since they can again differ by many orders of magnitude between species. The definition of \(\mathrm{p}\) as an operator remains the same, so \(\mathrm{pK}_a=-\log _{10}\left(\mathrm{~K}_a\right)\) and \(\mathrm{pK}_b=-\log _{10}\left(\mathrm{~K}_b\right)\).

Example: For an \(0.2 \mathrm{M}\) solution of pyridine \(\left(\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}\right)\), a base with a \(\mathrm{pH}\) of \(9.28\), determine the \(\mathrm{pK}_b\).

Answer

First, let's write a dissociation equation for pyridine:

\[C_5 H_5 N(a q)+H_2 O(l) \rightarrow C_5 H_5 N^{+}(a q)+O H^{-} \nonumber\]

Then, we can convert the \(\mathrm{pH}\) to a \(\mathrm{pOH}\), since pyridine is a base:

\[p O H=14-p H=14-9.28=4.72 \nonumber\]

Next, we can convert the \(\mathrm{pOH}\) into a concentration of hydroxide ions:

\begin{gathered}
p O H=-\log _{10}\left[O H^{-}\right] \\
{\left[O H^{-}\right]=10^{-p O H}=10^{-4.72}=1.89 \times 10^{-5}}
\end{gathered}

Then, we can write an expression for the base dissociation constant:

\[K_b=\dfrac{\left[C_5 H_5 \mathrm{NH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[C_5 \mathrm{H}_5 \mathrm{~N}\right]} \nonumber\]

And write our ICE table: Plugging these final values into the expression from above for \(\mathrm{K}_b\):

\(\begin{array}{c|c|c|c} 
& {\left[\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}\right]} & {\left[\mathrm{C}_5 \mathrm{H}_5 \mathrm{NH}^{+}\right]} & {\left[\mathrm{OH}^{-}\right]} \\
\hline \hline \text { Initial } & 0.2 & 0 \mathrm{M} & 0 \mathrm{M} \\
\text { Change } & -x \mathrm{M} & +x \mathrm{M} & +x \mathrm{M} \\
\text { Final } & 0.2-x \mathrm{M} & x \mathrm{M} & x \mathrm{M}
\end{array}\)

\[K_b=\dfrac{x^2}{0.2-x} \nonumber\]

Above, we determined the value of \(x=\left[\mathrm{OH}^{-}\right]=1.89 \times 10^{-5}\), so we can simply plug in to solve for \(\mathrm{K}_b\) :

\[K_b=\dfrac{\left(1.89 \times 10^{-5}\right)^2}{0.2-1.89 \times 10^{-5}}=1.8 \times 10^{-9} \nonumber\]

Finally,

\[p K_b=-\log \left(K_b\right)=8.74 \nonumber\]