91Ó°ÊÓ

Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

(a)

First we can write the dissolution equation for MZ:

$\begin{array}{l}\mathrm{MZ}\left(\mathrm{s}\right)\mathrm{C}{\mathrm{M}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{Z}}^{2-}\left(\mathrm{aq}\right)\\ \mathrm{Ksp}=\left[{\mathrm{M}}^{2+}\right]\left[{\mathrm{Z}}^{2-}\right]\end{array}$

Simply look for the picture in which the concentrations of ${\mathrm{M}}^{2+}$and${\mathrm{Z}}^{2-}$ are equal, which is picture B.

Therefore, the required answer is picture B.

(b)

Considering the given information:

$\mathrm{c}\left(\mathrm{eachion}\right)=2.5×{10}^{-6}\mathrm{M}$

We can see that each species has four ions, so ${\text{K}}_{\text{sp}}$expression will look like this.

as an example:

$\begin{array}{c}\mathrm{Ksp}=\left[{\mathrm{M}}^{2+}\right]\left[{\mathrm{Z}}^{2-}\right]\\ =4×2.5×{10}^{-6}×4×2.5×{10}^{-6}\\ =1×{10}^{-10}\end{array}$

Therefore, the required value is $1×{10}^{-10}$.

(c)

If we add ${\mathrm{Na}}_2\mathrm{Z}$, the concentration of ${\mathrm{Z}}^{2+}$increases and the concentration of ${\mathrm{M}}^{2+}$decreases, so that would be picture $C$.

(d)

A lower pH indicates that there is less base in the solution. It is said that ${\mathrm{Z}}^{2-}$represents carbonate ion, which is basic, so picture A would have lowered pH .