# Solubility Product

Solubility rules are generally covered in Chemistry I in a qualitative way. Below are some solubility rules of some ionic compounds in water.

 Soluble Ions Except Alkali metals (Group I) Li+, Na+, K+, etc. none Ammonium Ions NH4+ none Nitrates, acetates, chlorates, and perchlorate NO3–, C2H3O2–, ClO3–, ClO4– none Binary compounds of halogens (X) with metals (M) MCl, MBr, MI, etc. F–, Ag+, Pb2+, Hg2+ Sulfates SO42– Ba2+, Sr2+, Ca2+, Pb2+, Ag+, and Hg2+ Slightly Soluble Ions Except Sulfates of lead, silver, and mercury SO42– with Pb2+, Ag+, and Hg2+ none Hydroxides of alkaline earth metals (Group II) OH– with Ca2+, Sr2+, etc. Ba2+ Insoluble Ions Except Sulfides S2– Ca2+, Ba2+, Sr2+, Mg2+, Na+, K+, and NH4+ Hydroxides OH– Alkali metals (Group I), transition metals, Al3+, and NH4+ Carbonates, oxalates, chromates, and phosphates CO32–, C2O42–, CrO42–, and PO43– Alkali metals (Group I) and NH4+

## Ksp

The solubility product constant, Ksp 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 Ksp values will be at 25 °C unless otherwise noted.

The higher the Ksp, the more soluble the compuond 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 (Ksp >> 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 Ksp 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 Ksp for Mg(OH)2 is not zero but incredibly small.

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

## Ksp 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 Ksp. Suppose we want to know how much AgCl (by mass) is dissolved in a 10.0 L saturated aqueous solution. The Ksp for AgCl is 1.77 × 10–10.

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

Begin by setting up an ICE table.

 AgCl(s) $$\rightleftharpoons$$ Ag+(aq) + Cl– I 0 0 C +x +x E x x

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 conaining 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 Ksp 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.

 AgCl(s) $$\rightleftharpoons$$ Ag+(aq) + Cl– I 0 0.100 C +x +x E x x

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

### Problem 1

Fe(OH)2 is dissolved in water to make a saturated solution. How much Fe(OH)2 (in g) can be dissolved if the final volume of the solution was 100.0 L?

$\mathrm{Fe(OH)_2}(s) \rightleftharpoons \mathrm{Fe^2+}(aq) + \mathrm{2OH^-}(aq) \qquad K_{\mathrm{sp}} = 1.80\times 10^{-15}$

Solution

Set up an ICE table.

Fe(OH)2(s) $$\rightleftharpoons$$ Fe2+(aq) + 2OH
I 0 0
C +x +2x
E x 2x

Write out the equilibrium expression and solve for x.

\begin{align*} [\mathrm{Fe^{2+}}][\mathrm{OH^-}]^2 &= K_{\mathrm{sp}}\\ x(2x)^2 &= 1.80\times 10^{-15}\\ 4x^3 &= 1.80\times 10^{-15}\\ x &= 7.66\times 10^{-6}\\[2ex] [\mathrm{Fe^{2+}}] &= 7.66\times 10^{-6}~M\\ [\mathrm{OH^{-}}] &= 1.53\times 10^{-5}~M \end{align*}

Using stoichiometry and molar mass (89.86 g mol–1), determine the mass concentration of Fe(OH)2 dissolved.

\begin{align*} \rho_{\mathrm{Fe(OH)_2}} &= \dfrac{7.66\times 10^{-6}~\mathrm{mol~Fe^{2+}}}{\mathrm{L}} \left ( \dfrac{1~\mathrm{mol~Fe(OH)_2}}{1~\mathrm{mol~Fe^{2+}}} \right ) \left ( \dfrac{89.86~\mathrm{g~mol^{-1}}}{\mathrm{mol~Fe(OH)_2}} \right ) \\ &= \dfrac{6.88\times 10^{-4}~\mathrm{g}}{\mathrm{L}} \end{align*}

Use the total volume of solution to determine the mass of Fe(OH)2 dissolved.

$m_{\mathrm{Fe(OH)_2}} =\left (6.88\times 10^{-4}~\mathrm{g~L^{-1}}\right ) \left ( 100.0~\mathrm{L} \right ) = 6.88\times 10^{-2}~\mathrm{g~dissolved}$

Previous
Next