91Ó°ÊÓ

In polar liquids, the ionic material dissociates into its ions in ionic equilibrium. The ions generated in the solution are always in equilibrium with the undissociated solute.

Consider the given information,

$\text{pH}=13.0$

First, we can write the dissolution equation for ${\text{Cr(OH)}}_{\text{3}}$

$$\begin{gathered} {\text{Cr(OH)}}_{\text{3}}\text{(s)}⇌{\text{Cr}}^{\text{3 +}}{\text{(aq) + 3OH}}^{\text{-}}\text{(aq)} \\ \text{Ksp =}\left[{\text{Cr}}^{\text{3 +}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}                                      \left(1\right) \end{gathered}$$

Now we can write the reaction for formation of ${\text{Cr(OH)}}_{\text{4}}^{\text{-}}$

$$\begin{gathered} {\text{Cr}}^{\text{3 +}}{\text{+ 4OH}}^{\text{-}}⇌{\text{Cr(OH)}}_{\text{4}}^{\text{-}} \\ {\text{K}}_{\text{f}}\text{=}\frac{{\displaystyle \left[{\text{Cr(OH)}}_{\text{4}}^{\text{-}}\right]}}{{\displaystyle \left[{\text{Cr}}^{\text{3 +}}\right]\text{×}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{4}}}}\text{(2)} \end{gathered}$$

When we sum up those two equilibrium reactions we get the final reaction:

${\text{Cr(OH)}}_{\text{3}}{\text{+ OH}}^{\text{-}}⇔{\text{Cr(OH)}}_{\text{4}}^{\text{-}}$

And when we multiply those two constants $({\text{K}}_{\text{sp}} \text{and}  {\text{K}}_{\text{f}})$we get the overall constant.

$\text{K =} \frac{\left[{\text{Cr(OH)}}_{\text{4}}^{\text{-}}\right]}{\left[{\text{OH}}^{\text{-}}\right]}                          \left(3\right)$

We can find $\left[{\text{OH}}^{\text{-}}\right]$from value:

$$\begin{gathered} \text{pH + pOH = 14} \\ ⇒\text{pOH = 14 - pH} \\ ⇒\text{pOH = 14 - 13 = 1} \\ ⇒\left[{\text{OH}}^{\text{-}}\right]\text{= 0.1M} \end{gathered}$$

Now we can find $\left[{\text{Cr(OH)}}_{\text{4}}^{\text{-}}\right]$using equation (3):

$\begin{array}{rcl}\left[\mathrm{Cr}{\left(\mathrm{OH}\right)}_4^{-}\right]& =& \mathrm{K}×\left[{\mathrm{OH}}^{-}\right]\\ & =& {\mathrm{K}}_{\mathrm{sp}}×{\mathrm{K}}_{\mathrm{p}}×\left[{\mathrm{OH}}^{-}\right]\\ & =& 6.3×{10}^{-31}×8.0×{10}^{29}×0.1\mathrm{M}\\ & =& 0.0504\mathrm{M}\end{array}$

Thus, the solubility of ${\text{Cr(OH)}}_{\text{4}}^{\text{-}}$is$0.0504 \text{M.}$