# If the Cr_3^+ concentration in a saturation solution of Cr(OH)_3 is 4.0xx10^-6 M. How do you calculate the Ksp?

Jun 9, 2016

${K}_{s p} = 6.9 \cdot {10}^{- 21}$

#### Explanation:

Chromium(III) hydroxide, "Cr"("OH")_3, is made up of chromium(III) cations, ${\text{Cr}}^{3 +}$, and hydroxide anions, ${\text{OH}}^{-}$.

Chromium(III) hydroxide is considered to be insoluble in water, which means that when the salt dissolves, only very, very small amounts will actually dissociate to produce ions in solution.

${\text{Cr"("OH")_ (color(red)(3)(s)) rightleftharpoons "Cr"_ ((aq))^(3+) + color(red)(3)"OH}}_{\left(a q\right)}^{-}$

Notice that for every mole of chromium(III) hydroxide that dissociates you get $1$ mole of chromium cations and $\textcolor{red}{3}$ moles of hydroxide anions.

This means that at equilibrium, an aqueous solution of chromium(III) hydroxide will contain three times as many moles of hydroxide anions than of chromium cations.

Therefore, for any concentration of chromium(III) cations you have in solution, you will also have

$\left[{\text{OH"^(-)] = color(red)(3) xx ["Cr}}^{3 +}\right]$

In your case, you know that

["Cr"^(3+)] = 4.0 * 10^(-6)"M"

which means that this saturated solution contains

["OH"^(-)] = color(red)(3) xx 4.0 * 10^(-6)"M" = 1.2 * 10^(-5)"M"

Now, the solubility product constant for this dissociation equilibrium is equal to

${K}_{s p} = {\left[{\text{Cr"^(3+)] * ["OH}}^{-}\right]}^{\textcolor{red}{3}}$

Plug in your values to find

K_(sp) = 4.0 * 10^(-6)"M" * (1.2 * 10^(-5)"M")^color(red)(3)

${K}_{s p} = 6.9 \cdot {10}^{- 21} {\text{M}}^{4}$

You'll usually see the ${K}_{s p}$ expressed without added units, so you can say that you have

${K}_{s p} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{6.9 \cdot {10}^{- 21}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.