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

Q_sp- ion-product expression; Q_spvalue is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_sp value is calledK_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_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.

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.

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.

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.

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 –

role="math" localid="1663365225441" $K_{sp}=2.2\times {10}^{-16}$

It can be observed that the K_spofrole="math" localid="1663365115119" $\text{CuI}$> K_sp of $\mathrm{AgCN}$.

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