Question

Does any solid ${\mathbf{\text{PbCl}}}_{\mathbf{\text{2}}}$form when$\mathbf{\text{3.5 mg}}$of$\mathbf{NaCl}$is dissolved in$\mathbf{\text{0.250 L}}$of${\mathbf{\text{0.12 M Pb(NO}}}_{\mathbf{\text{3}}}{\mathbf{\text{)}}}_{\mathbf{\text{2}}}$?

Short answer

The Q_spvalue is lower than the K_spvalue, thus, the solution is unsaturated and ${\mathbf{\text{PbCl}}}_{\mathbf{2}}$precipitate won't form.

Step-by-step solution

Concept Introduction

Q_sp - ion-product expression; Q_spvalue is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_spvalue is called K_spvalue (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 theequations.

role="math" localid="1663578476456" ${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}{\mathbf{\text{= [M}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{][X}}}^{\mathbf{\text{-}}}{\mathbf{\text{]}}}^{\mathbf{\text{2}}}$

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

If:

$$\begin{gathered} {\mathbf{Q}}_{\mathbf{sp}}\mathbf{>}{\mathbf{K}}_{\mathbf{sp}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{ }\mathbf{precipitation}\mathbf{ }\mathbf{occurs}\mathbf{.} \\ {\mathbf{Q}}_{\mathbf{sp}}\mathbf{<}{\mathbf{K}}_{\mathbf{sp}\mathbf{ }}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{ }\mathbf{No}\mathbf{ }\mathbf{precipitate}\mathbf{ }\mathbf{will}\mathbf{ }\mathbf{be}\mathbf{ }\mathbf{formed}\mathbf{.} \end{gathered}$$

Given Data

Given,

• Molar Mass of NaCl is:$3.5\mathrm{mg}=0.0035\mathrm{g}$
• Moles of data-custom-editor="chemistry" $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$is: $\text{0.12 M}$
• Volume of $\mathrm{Pb}({\mathrm{NO}}_3{)}_2$is:$\text{0.250 L}$

Calculation

First write the dissolution equation for ${\mathrm{PbCl}}_2$–

${\mathrm{PbCl}}_2\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

The K_spvalue is as follows –

${\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{PbCl}}_2\right)=1.7\times {10}^{-5}$

The Q_sp is represented as –

${\text{Q}}_{\text{sp}}{\text{= [Pb}}^{\text{2 +}}{\text{][Cl}}^{\text{-}}{\text{]}}^2$

If Q_spvalue is equal to K_spvalue, than there is saturated solution with no precipitate. If Q_sp value is greater than K_sp value, than there is saturated solution with precipitate. If K_sp value is lower than K_sp value, than there is unsaturated solution without precipitate.

localid="1663598237738" $$\begin{gathered} \left[{\mathrm{Pb}}^{2+}\right]=\left[\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\right] \\ \left[{\mathrm{Pb}}^{2+}\right]=0.12\mathrm{M} \\ \left[{\mathrm{Cl}}^{-}\right]=\frac{3.5\mathrm{mg}\mathrm{NaCl}}{0.250\mathrm{L}\mathrm{solv}.}\times \frac{1\mathrm{g}}{1000\mathrm{mg}}\times \frac{1\mathrm{mol}\mathrm{NaCl}}{88.44\mathrm{g}} \\ \left[{\mathrm{Cl}}^{-}\right]=2.4\times {10}^{-4}\mathrm{M} \\ {\mathrm{Q}}_{\mathrm{sp}}=(0.12\mathrm{M})\times {\left(2.4\times {10}^{-4}\mathrm{M}\right)}^2 \\ {\mathrm{Q}}_{\mathrm{sp}}=6.89\times {10}^{-9} \end{gathered}$$

Therefore, localid="1663591051734" ${\text{Q}}_{\text{sp}}<K_{sp}$ and no solid (precipitate) will be formed.