Q19.81P - Chemistry: Molecular Nature Of Matter And Change

Question

Which compound in each pair is more soluble in water? 

(a) Manganese (II) hydroxide or calcium iodate 

(b) Strontium carbonate or cadmium sulfide 

(c) Silver cyanide or copper (I) iodide

Step-by-Step Solution

Answer

a) Calcium iodate is more soluble in water than Manganese (II)  hydroxide.

(b) Strontium carbonate is more soluble in water than Calcium sulphide.

(c) Copper (I) iodide is more soluble in water than Silver cyanide.

 

Solubility product

 Q_sp - ion-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).

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

Solid and liquid state is not included in the K_sp equations.

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

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

The solubility of salts can be compared by suing their   value. The salt with the greater  value will have a high solubility than the one with lower  value.

Subpart (a)

To compare the solubility of each compound, just compare their   K_sp values.

The K_sp value of Manganese (II) Hydroxide ${\text{Mn(OH)}}_{\text{2}}$ is –

${\mathrm{K}}_{\mathrm{sp}}=1.6\times {10}^{-13}$ 

The K_sp value of Calcium Iodate ${\text{Ca(IO}}_{\text{3}}{\text{)}}_{\text{2}}$ is –

 ${\mathrm{K}}_{\mathrm{sp}}=7.1\times {10}^{-7}$

It can be observed that the K_sp of ${\text{Ca(IO}}_{\text{3}}{\text{)}}_{\text{2}}$ > K_sp of ${\text{Mn(OH)}}_{\text{2}}$. 

 

Therefore, ${\text{Ca(IO}}_{\text{3}}{\text{)}}_{\text{2}}$  is more soluble in water.

Subpart (b)

To compare the solubility of each compound, just compare their   K_sp values.

The K_sp value of Strontium Carbonate ${\text{SrCO}}_3$ is –

${\mathrm{K}}_{\mathrm{sp}}=5.4\times {10}^{-10}$ 

The K_sp value of Cadmium Sulphide $\mathrm{CdS}$ is –

 ${\mathrm{K}}_{\mathrm{sp}}=1\times {10}^{-24}$

It can be observed that the K_sp of ${\text{SrCO}}_3$ > K_sp of $\mathrm{CdS}$.

 

Therefore, ${\text{SrCO}}_3$  is more soluble in water.

Subpart (c)

To compare the solubility of each compound, just compare their   K_sp values.

The K_sp value of Copper (I) Iodide $\text{CuI}$ is –

$K_{sp}=1\times {10}^{-12}$ 

The K_sp value of Silver Cyanide $\mathrm{AgCN}$ is –

  $K_{sp}=2.2\times {10}^{-16}$

It can be observed that the K_sp of $\text{CuI}$> K_sp of $\mathrm{AgCN}$.

 

Therefore, $\text{CuI}$ is more soluble in water.