Skip to main content
Chemistry LibreTexts

10.22: Conversion of Solubility to \(K_\text{sp}\)

  • Page ID
    289586
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Baking soda (sodium bicarbonate) is prepared by bubbling carbon dioxide gas through a solution of ammonia and sodium chloride. Ammonium carbonate is first formed, which then reacts with the \(\ce{NaCl}\) to form sodium bicarbonate and ammonium chloride. The sodium bicarbonate is less soluble than the other materials, so it will precipitate out of solution.

    Conversion of Solubility to \(K_\text{sp}\)

    Solubility is normally expressed in \(\text{g/L}\) of saturated solution. However, solubility can also be expressed in moles per liter. Molar solubility is the number of moles of solute in one liter of saturated solution. In other words, the molar solubility of a given compound represents the highest molarity solution that is possible for that compound. The molar mass of a compound is the conversion factor between solubility and molar solubility. Given that the solubility of \(\ce{Zn(OH)_2}\) is \(4.2 \times 10^{-4} \: \text{g/L}\), the molar solubility can be calculated as shown below:

    \[\frac{4.2 \times 10^{-4} \: \cancel{\text{g}}}{\text{L}} \times \frac{1 \: \text{mol}}{99.41 \: \cancel{\text{g}}} = 4.2 \times 10^{-6} \: \text{mol/L} \: \left( \text{M} \right)\nonumber \]

    Solubility data can be used to calculate the \(K_\text{sp}\) for a given compound. The following steps need to be taken.

    1. Convert from solubility to molar solubility.
    2. Use the dissociation equation to determine the concentration of each of the ions in \(\text{mol/L}\).
    3. Apply the \(K_\text{sp}\) equation.

    Example \(\PageIndex{1}\)

    The solubility of lead (II) fluoride is found experimentally to be \(0.533 \: \text{g/L}\). Calculate the \(K_\text{sp}\) for lead (II) fluoride.

    Solution

    Step 1: List the known quantities and plan the problem.

    Known

    • Solubility of \(\ce{PbF_2} = 0.533 \: \text{g/L}\)
    • Molar mass \(= 245.20 \: \text{g/mol}\)

    Unknown

    The dissociation equation for \(\ce{PbF_2}\) and the corresponding \(K_\text{sp}\) expression:

    \[\ce{PbF_2} \left( s \right) \rightleftharpoons \ce{Pb^{2+}} \left( aq \right) + 2 \ce{F^-} \left( aq \right) \: \: \: K_\text{sp} = \left[ \ce{Pb^{2+}} \right] \left[ \ce{F^-} \right]^2\nonumber \]

    The steps above will be followed to calculate the \(K_\text{sp}\) for \(\ce{PbF_2}\).

    Step 2: Solve.

    \[\text{molar solubility} = \frac{0.533 \: \cancel{\text{g}}}{\text{L}} \times \frac{1 \: \text{mol}}{245.20 \: \cancel{\text{g}}} = 2.17 \times 10^{-3} \: \text{M}\nonumber \]

    The dissociation equation shows that for every mole of \(\ce{PbF_2}\) that dissociates, \(1 \: \text{mol}\) of \(\ce{Pb^{2+}}\) and \(2 \: \text{mol}\) of \(\ce{F^-}\) are produced. Therefore, at equilibrium, the concentrations of the ions are:

    \[\left[ \ce{Pb^{2+}} \right] = 2.17 \times 10^{-3} \: \text{M} \: \: \text{and} \: \: \left[ \ce{F^-} \right] = 2 \times 2.17 \times 10^{-3} = 4.35 \times 10^{-3} \: \text{M}\nonumber \]

    Substitute into the expression and solve for the \(K_\text{sp}\).

    \[K_\text{sp} = \left( 2.17 \times 10^{-3} \right) \left( 4.35 \times 10^{-3} \right)^2 = 4.11 \times 10^{-8}\nonumber \]

    Step 3: Think about your result.

    The solubility product constant is significantly less than 1 for a nearly insoluble compound such as \(\ce{PbF_2}\).

    Summary

    • Molar solubility calculations are described.
    • Calculations of \(K_\text{sp}\) using molar solubility are described.


    10.22: Conversion of Solubility to \(K_\text{sp}\) is shared under a CC BY-NC license and was authored, remixed, and/or curated by LibreTexts.