9.4 pH and pOH

Last updated
Save as PDF

Page ID: 218433

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

If we wish to find the hydronium ion concentration ([H₃O⁺]) and the pH of a solution, we need to know both the strength of the acid (or base) and the concentration of the acid (or base). We will find that we need to treat strong acids (and bases) differently than weak acids (and bases) based on the extent to which they react with water.

Strong Acids

By definition, strong acids are those acids with a \(K_a \geq 1\). Using this definition, we assume that strong acids will react completely with water, so that every molecule of acid reacts with a molecule of water to produce a hydronium ion and the conjugate base. (This assumption is, of course, not possible because there must always be at least a few particles of each reactant and each product at equilibrium, but our assumption is a valid one when \(K_a\) gets to be quite a bit larger than 1.):

\[HA_{(aq)} \; + \; H_2O_{(l)} → H_3O^+_{(aq)} \; + \; A^-_{(aq)} \label {8.5.1} \]

Based on our assumption, \([H_3O^+]_{eq} = [HA]_{initial} \), so we can carry out the following calculation:

\[ pH = -log([H_3O^+]_{eq}) = -log([HA]_{initial}) \label {8.5.2}\]

Example \(\PageIndex{1}\)

What is the pH of a 0.175 M aqueous solution of HBr?

Solution

HBr is a strong acid (\(K_a > 1\)), so \([H_3O^+]_{eq} = [HBr]_{initial} \). Thus, pH = -log (0.175) = 0.757

Exercise \(\PageIndex{1}\)

What is the pH of a 0.0043 M aqueous solution of HNO₃?

Answer: HNO₃ is a strong acid (\(K_a > 1\)), so \([H_3O^+]_{eq} = [HNO_3]_{initial} \). Thus, pH = -log (0.0043) = 2.37

Strong Bases

By definition, strong bases are those bases with a \(K_b \geq 1\). using this definition, we assume that strong bases will react completely with water, so that every molecule of base reacts with a molecule of water to produce a hydroxide ion and the conjugate acid. (This assumption is, of course, not possible because there must always be at least a few particles of each reactant and each product at equilibrium, but our assumption is a valid one when \(K_b\) gets to be quite a bit larger than 1.):

\[B_{(aq)} \; + \; H_2O_{(l)} → OH^-_{(aq)} \; + \; HB^+_{(aq)} \label {8.5.3} \]

Based on our assumption, \([OH^-]_{eq} = [B]_{initial} \), so we can carry out the following calculation:

\[ pOH = -log([OH^-]_{eq}) = -log([B]_{initial}) \label {8.5.4}\]

To find the pH, we then subtract the pOH from 14.

Example \(\PageIndex{2}\)

What is the pH of a 0.175 M aqueous solution of NaNH₂ ?

Solution

NH₂^- is a strong base (\(K_b > 1\)), so \([OH^-]_{eq} = [NH_2^- ]_{initial} \). Thus, pOH = -log (0.175) = 0.757, and pH = 14.000 - 0.757 = 13.243

Exercise \(\PageIndex{2}\)

What is the pH of a 0.0043 M aqueous solution of KOH?

Answer: KOH is a strong base (\(K_b = 1\)), so \([OH^-]_{eq} = [OH^-]_{initial} \). Thus, pOH = -log (0.0043) = 2.37, and pH = 14.00 - 2.37 = 11.63

Weak Acids and Weak Bases

If the acid or base you are working with is weak (\(K_a < 1 or K_b< 1\)), you must carry out amore complicated math problem to determine the \( [H_3O^+] \) because you cannot tell how much \(H_3O^+ \) will form in solution simply by looking at the intial \([HA] \). We will not learn these problem in this course.