Q19.101P - Chemistry: Molecular Nature Of Matter And Change

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 $\text{Cr(OH}{{\text{)}}_{\text{4}}}^{\text{ - }}{\text{ = 8.0×10}}^{29}$].

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 $0.0504 \text{M.}$