Which compound in each pair is more soluble in water?

(a) Strontium sulphate or barium chromate

(b) Calcium carbonate or copper(II)carbonate

(c) Barium iodate or silver chromate

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).

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

The 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 the dissolution of some compound

(a)

To compare the solubility of each compound, just compare their K_spvalues.

The K_sp value of Strontium Sulphate ${\text{SrSO}}_4$is –

${\mathrm{K}}_{\mathrm{sp}}=3.2\times {10}^{-7}$

The K_sp value of Barium Chromate ${\text{BaCrO}}_{\text{4}}$is –

${\mathrm{K}}_{\mathrm{sp}}=2.1\times {10}^{-10}$

It can be observed that the K_spof localid="1663359796559" ${\text{SrSO}}_4$> K_sp of ${\text{BaCrO}}_{\text{4}}$.

Therefore, ${\text{SrSO}}_4$is more soluble in water.

To compare the solubility of each compound, just compare their K_spvalues.

The K_sp value of Calcium Carbonate ${\text{CaCO}}_{\text{3}}$ is –

role="math" localid="1663358942215" ${\mathrm{K}}_{\mathrm{sp}}=3.3\times {10}^{-9}$

The K_sp value of Copper (II) Carbonate ${\text{CuCO}}_{\text{3}}$is –

role="math" localid="1663358987059" ${\mathrm{K}}_{\mathrm{sp}}=3\times {10}^{-12}$

It can be observed that the K_sp of ${\text{CaCO}}_{\text{3}}\text{>}$K_sp of ${\text{CuCO}}_{\text{3}}$.

Therefore, ${\text{CaCO}}_{\text{3}}$is more soluble in water.

To compare the solubility of each compound, just compare their K_spvalues.

The K_sp value of Barium iodate ${\text{Ba(IO}}_3{)}_2$ is –

${\mathrm{K}}_{\mathrm{sp}}=1.5\times {10}^{-9}$

The K_sp value of Silver chromate ${\text{Ag}}_2{\mathrm{CrO}}_4$is –

${\mathrm{K}}_{\mathrm{sp}}=2.6\times {10}^{-12}$

It can be observed that the K_sp of ${\text{Ba(IO}}_3{)}_2$> K_sp of localid="1663359575789" ${\text{Ag}}_2{\mathrm{CrO}}_4$.

Therefore, ${\text{Ba(IO}}_3{)}_2$is more soluble in water.