Question

The scene below depicts a saturated solution of ${\mathbf{\text{MCl}}}_{\mathbf{\text{2}}}\mathbf{\text{(s)}}$ in the presence of dilute aqueous $\mathbf{\text{NaCl}}$ ;each sphere represents: $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{mol}$of ion, and the volume is $\mathbf{\text{250.0 ml}}$(solid MCl₂ is shown as green chunks; Na⁺ ions and water molecules are not shown).

1. Calculate the K_sp of ${\mathbf{MCl}}_{\mathbf{2}}$.
2. If $\mathbf{M}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}$ is added, there is increase, decrease or no change in Cl^- particle ? In the K_sp ? In the mass of ${\mathbf{MCl}}_{\mathbf{2}}$?

Short answer

1. The required value is ${\mathrm{K}}_{\mathrm{sp}}=1.9\times {10}^{-14}$.
2. Concentration of Cl^- will decrease.

K_sp is temperature dependent and so it does not change.

Mass of will increase.

Step-by-step solution

Le Chatelier’s principle

According to the Le Chatelier’s principle, if a system in equilibrium is disturbed, the equilibrium moves in such a direction so as to nullify the effect of disturbance

Calculating the amount of each spherical moles

(a)

To begin, we may write the dissolving equation for MCl₂ as follows:

${\text{MCl}}_2\left(s\right)\leftrightharpoons {\text{M}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$

role="math" localid="1663695132988" ${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{M}}^{2+}\right]{\left[{\text{Cl}}^{-}\right]}^2$

We now need to calculate $\left[{\mathrm{M}}^{2+}\right]\left[{\text{Cl}}^{-}\right]$ by calculating the amount of each spherical moles and counting them:

$\left[{\mathrm{M}}^{2+}\right]=\frac{3\times 1\times {10}^{-6}}{0.250\mathrm{L}}=0.000012\mathrm{M}$

$$\begin{gathered} \left[{\mathrm{Cl}}^{-}\right]=\frac{10\times 1\times {10}^{-6}}{0.25\mathrm{L}}=0.00004\mathrm{M} \\ {\mathrm{K}}_{\mathrm{sp}}=1.2\times {10}^{-5}\times {\left(4\times {10}^{-5}\right)}^2=1.9\times {10}^{-14} \end{gathered}$$

Therefore, the required value of ${\mathrm{K}}_{\mathrm{sp}} \mathrm{is} 1.9\times {10}^{-14}$.

Effects of adding  M(NO3)2

(b)

According to the Le Chatelier’s principle, if a system in equilibrium is disturbed, the equilibrium moves in such a direction so as to nullify the effect of disturbance.

So, when $\text{M}{\left({\text{NO}}_3\right)}_2$ is added, the concentration of M⁺² ion increases. Thus, the reaction will move towards the right and the concentration of Cl^- will decrease.

K_sp is temperature dependent and so it does not change.

Since, the backward reaction is favored, more MCl₂ will be formed and its mass will increase.