Q19.91P - Chemistry: Molecular Nature Of Matter And Change

Question

A $\mathbf{\text{50.0-mL}}$ volume of  ${\mathbf{\text{0.50 M Fe(NO}}}_{\mathbf{\text{3}}}{\mathbf{\text{)}}}_{\mathbf{\text{3}}}$is mixed with$\mathbf{\text{125 mL}}$ of ${\mathbf{\text{0.25 M Cd(NO}}}_{\mathbf{\text{3}}}{\mathbf{\text{)}}}_{\mathbf{\text{2}}}$. 

(a) If aqueous$\mathbf{\text{NaOH}}$ is added, which ion precipitates first? (See Appendix C.) 

(b) Describe how the metal ions can be separated using . 

(c) Calculate the${\mathbf{\text{[OH}}}^{\mathbf{\text{-}}}\mathbf{\text{]}}$ that will accomplish the separation.

 

Step-by-Step Solution

Answer

(a) If aqueous$NaOH$ is added, the ion which precipitates first is${\text{Fe}}^{\text{3+}}$.

(b) The separation of metal ions using$NaOH$ is done by adding hydroxide ions.

(c) The ${\text{OH}}^{\text{-}}$ions below ${\text{2.01×10}}^{\text{-7}}\text{ M}$will accomplish the separation

Step 1: Concept Introduction

 ${\mathbf{\text{Q}}}_{\mathbf{\text{sp}}}$- ion-product expression; ${\mathbf{\text{Q}}}_{\mathbf{\text{sp}}}$ value is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated  ${\mathbf{\text{Q}}}_{\mathbf{\text{sp}}}$value is called$K_{\mathbf{\text{sp}}}$  value (solubility-product constant).

${\mathbf{\text{MX}}}_{\mathbf{\text{2}}}\mathbf{\to }{\mathbf{\text{M}}}^{\mathbf{\text{2+}}}{\mathbf{\text{+2X}}}^{\mathbf{\text{-}}}$

Solid and liquid state is not included in the$K_{\mathbf{\text{sp}}}$ equations.

${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}{\mathbf{\text{=[M}}}^{\mathbf{\text{2+}}}{\mathbf{\text{][X}}}^{\mathbf{\text{-}}}{\mathbf{\text{]}}}^{\mathbf{\text{2}}}$

S (Molar solubility) is equal to the concentration of one mol of the ion formed by dissolution of some compound.

Step 2: Information Provided

• Volume of  ${\text{Fe(NO}}_3{)}_3$is$\text{50 mL=0.05 L}$: 
• Moles of  ${\text{Fe(NO}}_3{)}_3$is$\text{0.5 M}$: 
• Volume of  ${\text{Cd(NO}}_3{)}_2$is: $\text{125 mL=0.125 L}$
• Moles of${\text{Cd(NO}}_3{)}_2$  is: $\text{0.25 M}$
• The total volume is: $\text{0.05 L+0.125 L=0.175 L}$
• Moles of ${\text{Cd}}^{\text{2+}}$ is${\text{Cd}}^{\text{2+}}\text{=}\frac{\text{0.125 l×0.25 M}}{\text{0.175 l}}\text{=0.179 M}$:
• Moles of ${\text{Fe}}^{\text{3+}}$ is${\text{Fe}}^{\text{3+}}\text{=}\frac{\text{0.05 l×0.5 M}}{\text{0.175 l}}\text{=0.143 M}$:

 

Step 3: Addition of N a O H

(a)

When is $NaOH$added following changes takes place –

$\begin{array}{c}{\text{Cd}}^{\text{2+}}{\text{+2OH}}^{\text{-}}\iff {\text{Cd(OH)}}_{\text{2}}\text{(s)}\\ {\text{Fe}}^{\text{3+}}{\text{+3OH}}^{\text{-}}\iff {\text{Fe(OH)}}_{\text{3}}\text{(s)}\\ {\text{K}}_{\text{sp1}}\text{=}\left[{\text{Cd}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}{\text{=7.2×10}}^{\text{-15}}\\ {\text{K}}_{\text{sp2}}\text{=}\left[{\text{Fe}}^{\text{3+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}{\text{=1.6×10}}^{\text{-39}}\\ {\left[{\text{OH}}^{\text{-}}\right]}_{\text{1}}\text{=}\sqrt{\frac{{\text{7.2×10}}^{\text{-15}}}{\left[{\text{Cd}}^{\text{2+}}\right]}}{\text{=2.01×10}}^{\text{-7}}\text{ M}\\ {\left[{\text{OH}}^{\text{-}}\right]}_{\text{2}}\text{=}\sqrt{\frac{{\text{1.6×10}}^{\text{-39}}}{\left[{\text{Fe}}^{\text{3+}}\right]}}{\text{=2.24×10}}^{\text{-13}}\text{ M}\end{array}$

 

Therefore, it can be seen that needs much lower concentration of ${\text{OH}}^{\text{-}}$ions to begin precipitation, thus,${\text{Fe}}^{\text{3+}}$ precipitatesfirst.

Step 4: Separation of ions using N a O H

(b)

In order to separate this metal cations by using , slowly add hydroxide ions so the precipitation of${\text{Fe}}^{\text{3+}}$ ions can begin, but keep the concentration of ${\text{OH}}^{\text{-}}$lower than${\text{2.01×10}}^{\text{-7}}\text{ M}$ precipitation in order to prevent precipitation.

 

Therefore, slowly add hydroxide ions to carry out separation.

Step 5: Concentration of OH -

(c) 

According to answer(b), the ${\text{OH}}^{\text{-}}$concentration would be just a bit lower than .${\text{2.01×10}}^{\text{-7}}\text{ M}$

 

Therefore, the concentration should be lower than .${\text{2.01×10}}^{\text{-7}}\text{ M}$