Q19.79P - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

Answer

(a) Strontium sulphate is more soluble in water than Barium chromate.

(b) Calcium carbonate is more soluble in water than Copper (II)   carbonate.

(c) Barium iodate is more soluble in water than Silver chromate.

Concept Introduction

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

Compound More Soluble between SrSO 4 and role="math" localid="1663358088758" BaCrO 4

(a)

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

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_sp of ${\text{SrSO}}_4$ > K_sp  of ${\text{BaCrO}}_{\text{4}}$. 

 

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

Compound More Soluble between CaCO 3 and CuCO 3

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

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

  ${\mathrm{K}}_{\mathrm{sp}}=3.3\times {10}^{-9}$

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

${\mathrm{K}}_{\mathrm{sp}}=3\times {10}^{-12}$ 

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.

Compound More Soluble between Ba(IO 3 ) 2 and Ag 2 CrO 4

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

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 ${\text{Ag}}_2{\mathrm{CrO}}_4$.

 

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