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7.14: Calculating pH of Strong Acid and Base Solutions

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    Learning Objectives

    • Give the names and formulas of some strong acids and bases.
    • Explain the pH scale, and convert pH and concentration of hydronium ions.
    • Evaluate solution pH and pOH of strong acids or bases.

    Acids and bases that are completely ionized when dissolved in water are called strong acids and strong bases There are only a few strong acids and bases, and everyone should know their names and properties. These acids are often used in industry and everyday life. The concentrations of acids and bases are often expressed in terms of pH, and as an educated person, you should have the skill to convert concentrations into pH and pOH. The pH is an indication of the hydrogen ion concentration, \(\ce{[H+]}\).

    Strong Acids

    Strong acids are acids that are completely or nearly 100% ionized in their solutions; Table \(\PageIndex{1}\) includes some common strong acids. Hence, the ionization in Equation \(\ref{gen ion}\) for a strong acid HA can be represented with a single arrow:

    \[\ce{HA(aq) + H2O(l) \rightarrow H3O^{+}(aq) + A^{-}(aq)} \label{gen ion}\]

    Water is the base that reacts with the acid \(\ce{HA}\), \(\ce{A^{−}}\) is the conjugate base of the acid HA, and the hydronium ion is the conjugate acid of water. By definition, a strong acid yields 100% of \(\ce{H3O+}\) and \(\ce{A^{−}}\) when the acid ionizes in water. Table \(\PageIndex{1}\) lists several strong acids.

    Table \(\PageIndex{1}\): Some of the common strong acids and bases are listed here.
    Strong Acids Strong Bases
    perchloric acid (\(\ce{HClO4}\)) lithium hydroxide (\(\ce{LiOH}\))
    hydrochloric acid (\(\ce{HCl}\)) sodium hydroxide (\(\ce{NaOH}\))
    hydrobromic acid (\(\ce{HBr}\)) potassium hydroxide (\(\ce{KOH}\))
    hydroiodic acid (\(\ce{Hl}\)) calcium hydroxide (\(\ce{Ca(OH)2}\))
    nitric acid (\(\ce{HNO3}\)) strontium hydroxide (\(\ce{Sr(OH)2}\))
    sulfuric acid (\(\ce{H2SO4}\)) barium hydroxide (\(\ce{Br(OH)2}\))

    For a strong acid, \(\ce{[H+]}\) = \(\ce{[A^{-}]}\) = concentration of acid if the concentration is much higher than \(1 \times 10^{-7}\, M\). However, for a very dilute strong acid solution with concentration less than \(1 \times 10^{-7}\, M\), the pH is dominated by the autoionization of water

    \[\ce{H2O \rightleftharpoons H+ + OH-}\]

    Example \(\PageIndex{1}\)

    Calculate the pH of a solution with \(1.2345 \times 10^{-4}\; M \ce{HCl}\), a strong acid.

    Solution

    The solution of a strong acid is completely ionized. That is, this equation goes to completion

    \[\ce{HCl(aq) -> H(aq) + Cl^{-}(aq)} \nonumber\]

    Thus, \(\ce{[H+]} = 1.2345 \times 10^{-4}\).

    \[\ce{pH} = -\log(1.2345 \times 10^{-4}) = 3.90851 \nonumber\]

    Exercise \(\PageIndex{1}\)

    What is the pH for a solution containing 0.234 M \(\ce{[HCl]}\)?

    Answer

    pH = 0.63

     

    Strong Bases

    Strong bases are completely ionized in solution. Table \(\PageIndex{1}\) includes some common strong bases. For example, \(\ce{KOH}\) dissolves in water in the reaction

    \[\ce{KOH \rightarrow K+ + OH-} \nonumber\]

    Relative to the number of strong acids, there are fewer number of strong bases and most are alkali hydroxides. Calcium hydroxide is considered a strong base, because it is completely, almost completely, ionized. However, the solubility of calcium hydroxide is very low. When \(\ce{Ca(OH)2}\) dissolves in water, the ionization reaction is as follows:

    \[\ce{Ca(OH)2 \rightarrow Ca^2+ + 2 OH-} \nonumber\]

    Because of the stoichiometry of calcium hydroxide, upon dissociation, the concentration of \(\ce{OH-}\) will be twice the concentration of \(\ce{Ca^2+}\):

    \[\mathrm{[OH^-] = 2 [Ca^{2+}]} \nonumber\]

    Example \(\PageIndex{4}\)

    Calculate the pH of a solution containing \(1.2345 \times 10^{-4}\; M \; \ce{Ca(OH)2}\).

    Solution

    Based on complete ionization of

    \[\ce{Ca(OH)2 \rightarrow Ca^{+2} + 2OH-} \nonumber\]

    \[\begin{align*} \ce{[OH^{-}]} &= 2 \times 1.2345 \times 10^{-4} \\[4pt] &= 2.4690 \times 10^{-4}\; M \\[4pt] \ce{pOH} &= -\log( 2.4690 \times 10^{-4})\\[4pt] &= 3.6074 \end{align*}\]

    \[pH=14-pOH\]

    \[pH=14-3.6074\]

    \[pH=10.39\]

    Exercise \(\PageIndex{4}\)

    The molar solubility of calcium hydroxide is 0.013 M \(\ce{Ca(OH)2}\). Calculate the pH.

    Answer

    pOH = 12.42

    Questions

    1. What is the pH of a solution containing 0.01 M \(\ce{HNO3}\)?
    2. What is the pH of a solution containing 0.0220 M \(\ce{Ba(OH)2}\)? Give 3 significant figures.
    3. Exactly 1.00 L solution was made by dissolving 0.80 g of \(\ce{NaOH}\) in water. What is \(\ce{[H+]}\)? (Atomic mass: \(\ce{Na}\), 23.0; \(\ce{O}\), 16.0; \(\ce{H}\), 1.0)
    4. What is the pH for a solution which is 0.050 M \(\ce{HCl}\)?
    5. Which of the following is usually referred to as strong acid in water solution?

      \(\ce{HF}\), \(\ce{HNO2}\), \(\ce{H2CO3}\), \(\ce{H2S}\), \(\ce{HSO4-}\), \(\ce{Cl-}\), \(\ce{HNO3}\), \(\ce{HCN}\)

    Solutions

    1. Answer 2
      Hint...
      You do not need a calculator to evaluate \(-\log (0.01) = 2\)
    2. Answer 12.64
      Hint...
      \(\ce{Ba(OH)2 \rightarrow Ba^2+ + 2 OH-}\)
    3. Answer \(5.0\times 10^{-13}\)

      Hint...
      \(\mathrm{[OH^-] = \dfrac{0.80}{40} = 0.020\: M}\); \([H^+] = \dfrac{1.0 \times 10^{-14}}{0.020} = 5\times 10^{-13} M\). The pH is 12.30.

    4. Answer 1.3
      Hint...
      This solution contains 1.83 g of \(\ce{HCl}\) per liter. \(\mathrm{[H^+] = 0.050}\).
    5. Answer \(\ce{HNO3}\)
      Consider...
      All others are weak acids

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