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5.4: pH Calculations for Weak Acids

  • Page ID
    364672
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    Learning Objectives
    • Carry out equilibrium calculations for weak acid systems
    • Determine the pH of a weak acid solution based on acid concentration and the acid ionization constant
    • Calculate percent ionization for an acid

    Acid Equilibrium Calculations

    Acid equilibrium problems are one type of aqueous equilibrium reaction, and we can use the acid ionization constants ( \(\ce{Ka}\) ) to quantitatively determine the concentrations of the acid, hydronium ions, and the conjugate base. Example \(\PageIndex{1}\) demonstrates how an ICE table can be used for find equilibrium concentrations. To determine the pH of the solution at equilibrium we can use the equilibrium proton concentration, \( \ce{[H3O+]} \), from the ICE table.

    Example \(\PageIndex{1}\): Equilibrium Concentrations in a Solution of a Weak Acid

    Formic acid, HCO2H, is the irritant that causes the body’s reaction to ant stings.

    A photograph is shown of a large black ant on the end of a human finger.
    The pain of an ant’s sting is caused by formic acid. (credit: John Tann)

    What is the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid?

    \[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−4} \nonumber \]

    Solution

    1. Determine x and equilibrium concentrations. The equilibrium expression is:

    \[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \nonumber \]

    Because water is the solvent, it has a fixed activity equal to 1. Any small amount of water produced or used up during the reaction will not change water's role as the solvent, so the value of its activity remains equal to 1 throughout the reaction so we do not need to consider it when setting up the ICE table.

    The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color):

    Example 14.3.6.png

    2. Solve for \(x\) and the equilibrium concentrations. At equilibrium:

    \[\begin{align*} K_\ce{a} &=1.8×10^{−4}=\ce{\dfrac{[H3O+][HCO2- ]}{[HCO2H]}} \\[4pt] &=\dfrac{(x)(x)}{0.534−x}=1.8×10^{−4} \end{align*}\]

    Now solve for \(x\). Because the initial concentration of acid is reasonably large and \(K_a\) is very small, we assume that \(x << 0.534\), which permits us to simplify the denominator term as \((0.534 − x) = 0.534\). This gives:

    \[K_\ce{a}=1.8×10^{−4}=\dfrac{x^{2}}{0.534} \nonumber\]

    Solve for \(x\) as follows:

    \[\begin{align*} x^2 &=0.534×(1.8×10^{−4}) \\[4pt] &=9.6×10^{−5} \\[4pt] x &=\sqrt{9.6×10^{−5}} \\[4pt] &=9.8×10^{−3} \end{align*}\]

    To check the assumption that \(x\) is small compared to 0.534, we calculate:

    \[\begin{align*} \dfrac{x}{0.534} &=\dfrac{9.8×10^{−3}}{0.534} \\[4pt] &=1.8×10^{−2} \, \textrm{(1.8% of 0.534)} \end{align*}\]

    \(x\) is less than 5% of the initial concentration; the assumption is valid.

    We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table:

    \[\begin{align*} \ce{[H3O+]} &=~0+x=0+9.8×10^{−3}\:M. \\[4pt] &=9.8×10^{−3}\:M \end{align*}\]

    The pH of the solution can be found by taking the negative log of the \(\ce{[H3O+]}\), so:

    \[pH = −\log(9.8×10^{−3})=2.01 \nonumber\]

    Exercise \(\PageIndex{1}\): acetic acid

    Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH3CO2H?

    \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5} \nonumber\]

    Hint

    Determine \(\ce{[CH3CO2- ]}\) at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or \(\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]_{initial}}}×100\).

    Answer

    percent ionization = 1.3%

    Some weak acids ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation.

    Example \(\PageIndex{2}\): Equilibrium Concentrations in a Solution of a Weak Acid

    Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the \(\ce{HSO4-}\) ion, a weak acid. What is the pH of a 0.50-M solution of \(\ce{HSO4-}\)?

    \[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{a}=1.2×10^{−2} \nonumber \]

    Solution

    We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of \(\ce{HSO4-}\) so that we can use \(\ce{[H3O+]}\) to determine the pH. As in the previous examples, we can approach the solution by the following steps:

    Example 14.3.8A.png

    1. Determine \(x\) and equilibrium concentrations. This table shows the changes and concentrations:

    Example 14.3.8B.png

    2. Solve for \(x\) and the concentrations.

    As we begin solving for \(x\), we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of \(x\).

    At equilibrium:

    \[K_\ce{a}=1.2×10^{−2}=\ce{\dfrac{[H3O+][SO4^2- ]}{[HSO4- ]}}=\dfrac{(x)(x)}{0.50−x} \nonumber\]

    If we assume that x is small and approximate (0.50 − x) as 0.50, we find:

    \[x=7.7×10^{−2} \nonumber \]

    When we check the assumption, we confirm:

    \[\dfrac{x}{\mathrm{[HSO_4^- ]_i}} \overset{?}{\le} 0.05 \nonumber\]

    which for this system is

    \[\dfrac{x}{0.50}=\dfrac{7.7×10^{−2}}{0.50}=0.15(15\%) \nonumber\]

    The value of \(x\) is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find \(x\).

    The equation:

    \[K_\ce{a}=1.2×10^{−2}=\dfrac{(x)(x)}{0.50−x}\nonumber \]

    gives

    \[6.0×10^{−3}−1.2×10^{−2}x=x^{2+} \nonumber\]

    or

    \[x^{2+}+1.2×10^{−2}x−6.0×10^{−3}=0 \nonumber \]

    This equation can be solved using the quadratic formula. For an equation of the form

    \[ax^{2+} + bx + c=0, \nonumber\]

    \(x\) is given by the quadratic equation:

    \[x=\dfrac{−b±\sqrt{b^{2+}−4ac}}{2a} \nonumber\]

    In this problem, \(a = 1\), \(b = 1.2 × 10^{−3}\), and \(c = −6.0 × 10^{−3}\).

    Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:

    \[x=7.2×10^{−2} \nonumber\]

    Now determine the hydronium ion concentration and the pH:

    \[\begin{align*} \ce{[H3O+]} &=~0+x=0+7.2×10^{−2}\:M \\[4pt] &=7.2×10^{−2}\:M \end{align*}\]

    The pH of this solution is:

    \[\mathrm{pH=−log[H_3O^+]=−log7.2×10^{−2}=1.14} \nonumber\]

    Percent Ionization of Acids

    Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:

    \[\% \:\ce{ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\% \label{PercentIon} \]

    Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.

    Example \(\PageIndex{3}\): Calculation of Percent Ionization from pH

    Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.

    Solution

    The percent ionization for an acid is:

    \[\ce{\dfrac{[H3O+]_{eq}}{[HNO2]_0}}×100 \nonumber\]

    The chemical equation for the dissociation of the nitrous acid is:

    \[\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{NO2-}(aq)+\ce{H3O+}(aq). \nonumber\]

    Since \(10^{−pH} = \ce{[H3O+]}\), we find that \(10^{−2.09} = 8.1 \times 10^{−3}\, M\), so that percent ionization (Equation \ref{PercentIon}) is:

    \[\dfrac{8.1×10^{−3}}{0.125}×100=6.5\% \nonumber \]

    Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.

    Exercise \(\PageIndex{3}\)

    Calculate the percent ionization of a 0.10 M solution of acetic acid with a pH of 2.89.

    Answer

    1.3% ionized

    Summary

    The strengths of Brønsted-Lowry acids in aqueous solutions can be determined by their acid ionization constants. Weak acids are only partially ionized in water. The pH of an acid solution can be determined using the concentration of the acid and the its acid ionization constant.

    Key Equations

    • \(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\)
    • \(\textrm{Percent ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\)

    Glossary

    acid ionization constant (Ka)
    equilibrium constant for the ionization of a weak acid
    percent ionization
    ratio of the concentration of the ionized acid to the initial acid concentration, times 100

    This page titled 5.4: pH Calculations for Weak Acids is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.