Solubility Product

K_sp

The solubility product constant, K_sp is another equilibrium constant used to describe the solubility of a solid and is associated with the following process

\[\mathrm{A}(s) \rightleftharpoons \mathrm{B}(aq) + \mathrm{C}(aq)+ \cdots\]

Solubility products are temperature dependent. Reported K_sp values will be at 25 °C unless otherwise noted.

The higher the K_sp, the more soluble the compound is in the solution. In the reaction above, A can be a salt and can dissociate into a number of aqueous components. One clear example is the dissociation of the highly soluble sodium chloride in water

\[\mathrm{NaCl}(s) \rightleftharpoons \mathrm{Na^+}(aq) + \mathrm{Cl^-}(aq) \qquad K_{\mathrm{sp}} = [\mathrm{Na^+}][\mathrm{Cl^-}] >> 1\]

Sodium chloride is considered to be soluble in water (K_sp >> 1).

An example of a slightly soluble compound in water is lead(II) sulfate (PbSO₄)

\[\mathrm{PbSO_4}(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + \mathrm{SO_4^{2-}}(aq) \qquad K_{\mathrm{sp}} = [\mathrm{Pb^{2+}}][\mathrm{SO_4^{2-}}] = 1.3\times 10^{-8} \]

Notice the relatively small K_sp value for PbSO₄ (it is much less than 1).

An example of an insoluble compound in water is Mg(OH₂)

\[\mathrm{Mg(OH)_2}(s) \rightleftharpoons \mathrm{Mg^{2+}}(aq) + \mathrm{2OH^{-}}(aq) \qquad K_{\mathrm{sp}} = [\mathrm{Mg^{2+}}][\mathrm{OH^{-}}]^2 = 8.9\times 10^{-12} \]

Here, the K_sp for Mg(OH)₂ is not zero but incredibly small.

The qualitative description of solubility is coupled to the quantitative measurement of K_sp. Appendix J of the textbook contains a table of K_sp values for many compounds.

K_sp to mass

\[K_{\mathrm{sp}} \longrightarrow [~]_{\mathrm{eq}} \longrightarrow \rho \longrightarrow m\]

We can determine how many grams of a substance is present in a saturated solution from the K_sp. Suppose we want to know how much AgCl (by mass) is dissolved in a 10.0 L saturated aqueous solution. The K_sp for AgCl is 1.77 × 10^–10.

\[\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag^+}(aq) + \mathrm{Cl^-}(aq)\]

Begin by setting up an ICE table.

Write out the equilibrium expression and solve for x.

\[\begin{align*} [\mathrm{Ag^+}][\mathrm{Cl^-}] &= K_{\mathrm{sp}}\\ (x)(x) &= 1.77\times 10^{-10}\\ x^2 &= 1.77\times 10^{-10}\\ x &= 1.33\times 10^{-5}\\ [\mathrm{Ag^+}]_{\mathrm{eq}} &= [\mathrm{Cl^-}]_{\mathrm{eq}} = 1.33\times 10^{-5}~M \end{align*}\]

Get mass concentration (g L^–1) of AgCl by using the molar mass (143.32 g mol^–1).

\[\begin{align*} \rho_{\mathrm{AgCl}} &= \dfrac{1.33\times 10^{-5}~\mathrm{mol~Ag^+}}{\mathrm{L}} \left ( \dfrac{1~\mathrm{mol~AgCl}}{1~\mathrm{mol~Ag^+}} \right ) \left ( \dfrac{143.32~\mathrm{g~mol^{-1}}}{\mathrm{mol~AgCl}} \right ) \\ &= \dfrac{1.91\times 10^{-3}~\mathrm{g~AgCl}}{\mathrm{L}} \end{align*}\]

Get mass of dissolved AgCl by multiplying by the volume (in L).

\[m_{\mathrm{AgCl}} =\left (1.91\times 10^{-3}~\mathrm{g~L^{-1}}\right ) \left ( 10.0~\mathrm{L} \right ) = 1.91\times 10^{-2}~\mathrm{g}\]

Solutions with a common ion

We can reexamine the mass of a solute in an aqueous solution containing a common ion.
Suppose we want to know how much AgCl (by mass) can be dissolved in a 10.0 L 0.100 M NaCl aqueous solution. The K_sp for AgCl is 1.77 × 10^–10. Note that the solution contains a common ion (Cl^–) with AgCl.

Begin by setting up an ICE table.

Write out the equilibrium expression and solve for x.

\[\begin{align*} [\mathrm{Ag^+}][\mathrm{Cl^-}] &= K_{\mathrm{sp}}\\ x(0.100\textcolor{red}{+x}) &= 1.77\times 10^{-10}\\ 0.100x &= 1.77\times 10^{-10}\\ x &= 1.77\times 10^{-9}\\ [\mathrm{Ag^+}]_{\mathrm{eq}} &= 1.77\times 10^{-9}~M \end{align*}\]

Note: You should “test x” here and make sure it is less than 5%.

Get mass concentration (g L^–1) of AgCl by using the molar mass (143.32 g mol^–1).

\[\begin{align*} \rho_{\mathrm{AgCl}} &= \dfrac{1.77\times 10^{-9}~\mathrm{mol~Ag^+}}{\mathrm{L}} \left ( \dfrac{1~\mathrm{mol~AgCl}}{1~\mathrm{mol~Ag^+}} \right ) \left ( \dfrac{143.32~\mathrm{g~mol^{-1}}}{\mathrm{mol~AgCl}} \right ) \\ &= \dfrac{2.57\times 10^{-7}~\mathrm{g}}{\mathrm{L}} \end{align*}\]

Get mass of dissolved AgCl by multiplying by the volume (in L).

\[m_{\mathrm{AgCl}} =\left (2.57\times 10^{-7}~\mathrm{g~L^{-1}}\right ) \left ( 10.0~\mathrm{L} \right ) = 2.57\times 10^{-6}~\mathrm{g~dissolved}\]

Much less AgCl can be dissolved in NaCl(aq) relative to pure water due to the presence of a common ion.

Practice Problems