Question

A solution is $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M}$in NaF, $\mathit{N}{\mathit{a}}_{\mathbf{2}}\mathit{S}$, and $\mathit{N}{\mathit{a}}_{\mathbf{3}}\mathit{P}{\mathit{O}}_{\mathbf{4}}$. What would be the order of precipitation as a source of $\mathit{P}{\mathit{b}}^{\mathbf{2}\mathbf{+}}$is added gradually to the solution? The relevant ${\mathit{K}}_{\mathbf{sp}}$values are role="math" localid="1657794770089" ${\mathit{K}}_{\mathbf{sp}}\mathbf{\left(}{\mathbf{PbF}}_{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}$, ${\mathit{K}}_{\mathbf{sp}}\mathbf{\left(}\mathbf{PbS}\mathbf{\right)}\mathbf{=}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{29}}$, and ${\mathit{K}}_{\mathbf{sp}}\mathbf{\left[}{\mathbf{Pb}}_{\mathbf{3}}{\left({\mathrm{PO}}_4\right)}_{\mathbf{2}}\mathbf{\right]}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{54}}$.

Short answer

Answer

The order of precipitation as a source of $\mathit{P}{\mathit{b}}^{\mathbf{2}\mathbf{+}}$is added gradually to the solution is $\mathit{P}\mathit{b}\mathit{S}\mathbf{>}\mathit{P}{\mathit{b}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{>}\mathit{P}\mathit{b}{\mathit{F}}_{\mathbf{2}}$.

Step-by-step solution

The given Information.

The concentration can be given as ${\mathit{K}}_{\mathbf{s}\mathbf{p}}\left(Pb{F}_2\right)\mathbf{=}\mathbf{4}\mathbf{\times }{\mathbf{-}}^{\mathbf{8}}$, ${\mathit{K}}_{\mathbf{sp}}\mathbf{\left(}\mathbf{PbS}\mathbf{\right)}\mathbf{=}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{29}}$, and ${\mathit{K}}_{\mathbf{sp}}\mathbf{\left[}{\mathbf{Pb}}_{\mathbf{3}}{\left({\mathrm{PO}}_4\right)}_{\mathbf{2}}\mathbf{\right]}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{54}}$.

$\mathbf{\left[}{\mathbf{F}}^{\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{\left[}{\mathbf{S}}^{\mathbf{2}\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\mathbf{P}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{3}\mathbf{-}}\mathbf{\right]}\mathbf{=}{\mathbf{10}}^{\mathbf{-}}\mathit{M}$

The precipitation takes place when $\mathit{Q}\mathbf{>}{\mathit{K}}_{\mathbf{sp}}$as the solution becomes supersaturated.

Here, Q is the constant at any condition and ${\mathit{K}}_{\mathbf{sp}}$is the constant at equilibrium only. They have generally the similar meaning.

The concentration of Pb2+ in PbF2.

As,

$$\begin{gathered} \mathit{N}\mathit{a}\mathit{F}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M} \\ \mathit{N}\mathit{a}\mathit{F}\mathbf{}\mathbf{}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{}\mathbf{}\mathit{N}{\mathit{a}}^{\mathbf{+}}\mathbf{}\mathbf{}\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{F}}^{\mathbf{-}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M} \end{gathered}$$

Also, for $\mathit{P}\mathit{b}{\mathit{F}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathit{F}$

$$\begin{gathered} {\mathit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{=}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}} \\ \mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{=}\left[P{b}^{2+}\right]{\left[F^{-}\right]}^{\mathbf{2}} \\ \mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\left[P{b}^{2+}\right]\mathbf{\times }{\left({10}^{-4}\right)}^{\mathbf{2}} \end{gathered}$$

Therefore,

$\mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}}\mathbf{\right]}\mathbf{=}\mathbf{4}\mathit{M}$

Thehighest concentration of $\mathit{P}{\mathit{b}}^{\mathbf{2}}$in the solution.

The concentration of  in .

As,

$$\begin{gathered} \mathit{N}{\mathit{a}}_{\mathbf{2}}\mathit{S}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M} \\ \mathit{N}{\mathit{a}}_{\mathbf{2}}\mathit{S}\mathbf{}\mathbf{}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathit{N}{\mathit{a}}^{\mathbf{+}}\mathbf{}\mathbf{}\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{S}}^{\mathbf{2}\mathbf{-}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\left({10}^{-4}M\right)}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M} \end{gathered}$$

Also, for $\mathit{P}\mathit{b}\mathit{S}\mathbf{=}\mathbf{1}{\mathit{S}}^{\mathbf{2}\mathbf{-}}$

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{29}} \\ {\mathit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{=}\mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{F}}^{\mathbf{2}\mathbf{-}}\mathbf{\right]} \\ 7\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{29}}\mathbf{=}\mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}\left[1\times {10}^{-4}\right] \end{gathered}$$

Therefore,

$\mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}\mathbf{=}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{25}}\mathit{M}$

The lowest concentration of $\mathit{P}{\mathit{b}}^{\mathbf{2}\mathbf{+}}$in the solution.

The concentration of Pb2+ in Pb3(PO4)2.

As,

$$\begin{gathered} \mathit{N}{\mathit{a}}_{\mathbf{3}}\mathit{P}{\mathit{O}}_{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M} \\ \mathit{N}{\mathit{a}}_{\mathbf{3}}\mathit{P}{\mathit{O}}_{\mathbf{4}}\mathbf{}\mathbf{}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{}\mathbf{}\mathbf{3}\mathit{N}{\mathit{a}}^{\mathbf{+}}\mathbf{}\mathbf{}\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{S}}_{\mathbf{4}}^{\mathbf{3}\mathbf{-}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{\left(}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{M}\mathbf{\right)}}^{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M} \end{gathered}$$

Also, for $\mathit{P}{\mathit{b}}_{\mathbf{3}}{\left(P{O}_4\right)}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{}\mathit{P}{\mathit{O}}_{\mathbf{4}}^{\mathbf{3}\mathbf{-}}$

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{54}} \\ \mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{54}}\mathbf{=}\mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{F}}^{\mathbf{2}\mathbf{-}}\mathbf{\right]} \\ \mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{54}}\mathbf{=}{\mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}}^{\mathbf{3}}{\left({10}^{-4}\right)}^{\mathbf{2}} \end{gathered}$$

And

$$\begin{gathered} {\mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}}^{\mathbf{3}}\mathbf{=}\frac{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{54}}}{{\mathbf{10}}^{\mathbf{-}\mathbf{8}}}{\mathbf{10}}^{\mathbf{-}\mathbf{46}} \\ \mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}\mathbf{}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{16}}\mathit{M} \end{gathered}$$

As $\mathit{P}{\mathit{b}}^{\mathbf{2}\mathbf{+}}$has the lowest concentration in the solution for PbS, therefore it is least soluble there and has the maximum part precipitated out.

Hence,theorder of precipitation can be found is

$\mathit{P}\mathit{b}\mathit{S}\mathbf{>}\mathit{P}{\mathit{b}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{>}\mathit{P}\mathit{b}{\mathit{F}}_{\mathbf{2}}$