Q19.122CP - Chemistry: Molecular Nature Of Matter And Change

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 ${\mathbf{MCl}}_{\mathbf{2}}$ ?

Step-by-Step Solution

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. 

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:

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

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

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 ( N O 3 ) 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.