How do you calculate Ksp from molar solubility?

Aug 16, 2016

Here's how you can do that.

Explanation:

Let's take a generic dissociation equilibrium to work with here

${\text{X"_ n "Y"_m rightleftharpoons color(blue)(n) * "X"^(m+) + color(purple)(m) * "Y}}^{n -}$

Now, the molar solubility of this generic salt ${\text{X"_n"Y}}_{m}$ tells you the number of moles of salt that can be dissolved in one liter of solution to form a saturated solution.

Let's assume that you are given a molar solubility equal to $s$ ${\text{mol L}}^{- 1}$ for this salt in water at room temperature. This tells you that you can dissolve $s$ moles of ${\text{X"_n"Y}}_{m}$ per liter of solution at this temperature.

Now, notice that every mole of ${\text{X"_n"Y}}_{m}$ that dissolves produces $\textcolor{b l u e}{n}$ moles of ${\text{X}}^{m +}$ cations and $\textcolor{p u r p \le}{m}$ moles of ${\text{Y}}^{n -}$ anions.

This means that the saturated solution will contain

$\left[{\text{X}}^{m +}\right] = \textcolor{b l u e}{n} \cdot s$

$\left[{\text{Y}}^{n -}\right] = \textcolor{p u r p \le}{m} \cdot s$

The solubility product constant, ${K}_{s p}$, for this dissociation equilibrium looks like this

${K}_{s p} = {\left[{\text{X"^(m+)]^color(blue)(n) * ["Y}}^{n -}\right]}^{\textcolor{p u r p \le}{m}}$

Plug in the expressions you have for the concentrations of the two ions in terms of $s$ to find

${K}_{s p} = {\left(\textcolor{b l u e}{n} \cdot s\right)}^{\textcolor{b l u e}{n}} \cdot {\left(\textcolor{p u r p \le}{m} \cdot s\right)}^{\textcolor{p u r p \le}{m}}$

${K}_{s p} = \textcolor{b l u e}{{n}^{n}} \cdot {s}^{\textcolor{b l u e}{n}} \cdot \textcolor{p u r p \le}{{m}^{m}} \cdot {s}^{\textcolor{p u r p \le}{m}}$

This is equivalent to

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{K}_{s p} = \textcolor{b l u e}{{n}^{n}} \cdot \textcolor{p u r p \le}{{m}^{m}} \cdot {s}^{\left(\textcolor{b l u e}{n} + \textcolor{p u r p \le}{m}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}}$

Let's take a numerical example to test out this expression.

Magnesium hydroxide, "Mg"("OH")_2, has a molar solubility of $1.44 \cdot {10}^{- 4} \text{M}$ in pure water at room temperature. What is the ${K}_{s p}$ of the salt?

The first thing to do is identify the values of $n$ and $m$ by writing the dissociation equilibrium for magnesium hydroxide

${\text{Mg"("OH")_ (2(s)) rightleftharpoons "Mg"_ ((aq))^(2+) + 2"OH}}_{\left(a q\right)}^{-}$

As you can see, you have

$\left\{\begin{matrix}n = 1 \\ m = 2\end{matrix}\right.$

This means that the ${K}_{p s}$ of magnesium hydroxide is

${K}_{s p} = {1}^{1} \cdot {2}^{2} \cdot {\left(1.44 \cdot {10}^{- 4} \text{M}\right)}^{\left(1 + 2\right)}$

${K}_{s p} = 1.2 \cdot {10}^{- 11} {\text{M}}^{3}$

The solubility product constant is usually given without added units, so you'd have

${K}_{s p} = 1.2 \cdot {10}^{- 11}$