Skip to main content
Chemistry LibreTexts

10.1.1: Solutions of Solids Dissolved in Water

  • Page ID
    367844
    • Anonymous
    • LibreTexts
    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Learning Objectives
    • Explain how temperature influences solubility of solids.

    The solubility of a substance is the amount of that substance that is required to form a saturated solution in a given amount of solvent at a specified temperature. Solubility is often measured as the grams of solute per \(100 \: \text{g}\) of solvent. The solubility of sodium chloride in water is \(36.0 \: \text{g}\) per \(100 \: \text{g}\) water at \(20^\text{o} \text{C}\). The temperature must be specified because solubility varies with temperature. For gases, the pressure must also be specified. Solubility is specific for a particular solvent. We will consider solubility of material in water as solvent.

    The solubility of the majority of solid substances increases as the temperature increases. However, the effect is difficult to predict and varies widely from one solute to another. The temperature dependence of solubility can be visualized with the help of a solubility curve, a graph of the solubility vs. temperature (Figure \(\PageIndex{4}\)).

    alt
    Figure \(\PageIndex{4}\): Solubility curves for several compounds.

    Notice how the temperature dependence of \(\ce{NaCl}\) is fairly flat, meaning that an increase in temperature has relatively little effect on the solubility of \(\ce{NaCl}\). The curve for \(\ce{KNO_3}\), on the other hand, is very steep and so an increase in temperature dramatically increases the solubility of \(\ce{KNO_3}\).

    The trends for gas solubility in aqueous solution are often different than those for solids. You may notice on the graph that there are several substances—\(\ce{HCl}\), \(\ce{NH_3}\), and \(\ce{SO_2}\)— which have solubility that decreases as temperature increases. They are all gases at standard pressure. Additionally, changes in pressure can affect solubility of gases whereas they generally have little or no effect on solubility of solids. Trends of gas solubility will be discussed in the next subsection.

    Solubility curves can be used to determine if a given solution is saturated or unsaturated. Suppose that \(80 \: \text{g}\) of \(\ce{KNO_3}\) is added to \(100 \: \text{g}\) of water at \(30^\text{o} \text{C}\). According to the solubility curve, approximately \(48 \: \text{g}\) of \(\ce{KNO_3}\) will dissolve at \(30^\text{o} \text{C}\). This means that the solution will be saturated since \(48 \: \text{g}\) is less than \(80 \: \text{g}\). We can also determine that there will be \(80 - 48 = 32 \: \text{g}\) of undissolved \(\ce{KNO_3}\) remaining at the bottom of the container. Now suppose that this saturated solution is heated to \(60^\text{o} \text{C}\). According to the curve, the solubility of \(\ce{KNO_3}\) at \(60^\text{o} \text{C}\) is about \(107 \: \text{g}\). Now the solution is unsaturated since it contains only the original \(80 \: \text{g}\) of dissolved solute. Now suppose the solution is cooled all the way down to \(0^\text{o} \text{C}\). The solubility at \(0^\text{o} \text{C}\) is about \(14 \: \text{g}\), meaning that \(80 - 14 = 66 \: \text{g}\) of the \(\ce{KNO_3}\) will re-crystallize.

    Summary

    • The solubility of a solid in water usually increases with an increase in temperature.

    Contributions & Attributions

    This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:


    This page titled 10.1.1: Solutions of Solids Dissolved in Water is shared under a mixed license and was authored, remixed, and/or curated by Anonymous.

    • Was this article helpful?