Question

A $\mathbf{\text{50.0-mL}}$volume of${\mathbf{\text{0.50MFe(NO}}}_{\mathbf{\text{3}}}{\mathbf{\text{)}}}_{\mathbf{\text{3}}}$is mixedwith$\mathbf{\text{125mL}}$ of${\mathbf{\text{0.25MCd(NO}}}_{\mathbf{\text{3}}}{\mathbf{\text{)}}}_{\mathbf{\text{2}}}$.

(a) Ifaqueous$\mathbf{\text{NaOH}}$ isadded, which ion precipitates first? (See Appendix C.)

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

(c) Calculatethe${\mathbf{\text{[OH}}}^{\mathbf{\text{-}}}\mathbf{\text{]}}$ thatwill accomplish the separation.

Short 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-by-step solution

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 inthe$K_{\mathbf{\text{sp}}}$ 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.

Information Provided

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

Addition ofNaOH

(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.

Separation of ions usingNaOH

(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 ofOH-

(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 .${\text{2.01×10}}^{\text{-7}}\text{M}$