Question

Does any solid ${\mathbf{\text{Ag}}}_{\mathbf{\text{2}}}{\mathbf{\text{CrO}}}_{\mathbf{\text{4}}}$form when $\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{g}$of ${\mathbf{\text{AgNO}}}_{\mathbf{3}}$is dissolved in $\mathbf{\text{15.0 mL}}$of $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{M}{\mathbf{K}}_{\mathbf{2}}{\mathbf{CrO}}_{\mathbf{4}}$?

Short answer

The Q_spvalue is lower than the K_spvalue, thus, the solution is unsaturated and${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ precipitate won't form.

Step-by-step solution

Ionic product

Q_spis the ionic product expression; Q_sp value is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_sp value is called K_sp value (solubility-product constant).

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

Solid and liquid state is not included in theK_spequations.

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

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

$$\begin{gathered} {\mathit{Q}}_{\mathbf{s}\mathbf{p}}\mathbf{>}{\mathit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{ }\mathit{p}\mathit{r}\mathit{e}\mathit{c}\mathit{i}\mathit{p}\mathit{i}\mathit{t}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{o}\mathit{c}\mathit{c}\mathit{u}\mathit{r}\mathit{s}\mathbf{.} \\ {\mathit{Q}}_{\mathbf{s}\mathbf{p}}\mathbf{<}{\mathit{K}}_{\mathbf{s}\mathbf{p}\mathbf{ }}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{ }\mathit{N}\mathit{o}\mathbf{ }\mathit{p}\mathit{r}\mathit{e}\mathit{c}\mathit{i}\mathit{p}\mathit{i}\mathit{t}\mathit{a}\mathit{t}\mathit{e}\mathbf{ }\mathit{w}\mathit{i}\mathit{l}\mathit{l}\mathbf{ }\mathit{b}\mathit{e}\mathbf{ }\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{e}\mathit{d}\mathbf{.} \end{gathered}$$

Given Data

Given,

• Molar Mass of ${\mathrm{AgNO}}_3$is:
• Moles of${\text{K}}_2{\mathrm{CrO}}_4$is:$2.7\times {10}^{-5}\mathrm{g}$
• Volume of ${\text{K}}_2{\mathrm{CrO}}_4$is:$\text{15 mL}$

Calculation

First write the dissolution equation for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$–

${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{(s)}\iff {\text{2Ag}}^{\text{+}}{\text{(aq) + CrO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

The K_sp value is as follows –

${\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{Ag}}_2{\mathrm{CrO}}_4\right)=2.6\times {10}^{-12}$

The Q_spis represented as –

${\mathrm{Q}}_{\mathrm{sp}}=[{\mathrm{Ag}}^{+}{]}^2\left[{\mathrm{CrO}}_4^{2-}\right]$

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

$$\begin{gathered} \left[{\mathrm{CrO}}_4^{2-}\right]=\left[{\mathrm{K}}_2{\mathrm{CrO}}_4\right] \\ \left[{\mathrm{CrO}}_4^{2-}\right]=4\times {10}^{-4}\mathrm{M} \\ \left[{\mathrm{Ag}}^{+}\right]=\frac{2.7\times {10}^{-5}\mathrm{g}{\mathrm{Ag}}_2{\mathrm{CrO}}_4}{15\mathrm{mL}\mathrm{solv}.}\times \frac{1000\mathrm{mL}}{1\mathrm{L}}\times \frac{1\mathrm{mol}{\mathrm{Ag}}_2{\mathrm{CrO}}_4}{169.87\mathrm{g}} \\ \left[{\mathrm{Ag}}^{+}\right]=1.1\times {10}^{-6}\mathrm{M} \end{gathered}$$

localid="1663604041397" $$\begin{gathered} {\mathrm{Q}}_{\mathrm{sp}}=(1.1\times {10}^{-6}\mathrm{M}{)}^2\times \left(4\times {10}^{-4}\mathrm{M}\right) \\ =4.84\times {10}^{-16} \end{gathered}$$

Therefore, Q_sp value is lower and no solid will be formed.