Question

Find the solubility of $\mathbf{Cr}{\left(\mathbf{OH}\right)}_{\mathbf{3}}$in a buffer of pH 13.0 [ ${\text{K}}_{\text{sp}}$of ${\text{Cr(OH)}}_{\text{3}}{\text{=6.3×10}}^{-31}$;${\text{K}}_{\text{f}}$of $\text{Cr(OH}{{\text{)}}_{\text{4}}}^{\text{-}}{\text{=8.0×10}}^{29}$].

Short answer

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

Step-by-step solution

Concept Introduction

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.

Calculation of Solubility

Consider the given information,

$\text{pH}=13.0$

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

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

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

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

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

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

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:

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

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

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

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