
Solubility of the alkaline earth’s metal sulphates in water decreases in the sequence:
A) $Sr > Ca > Mg > Ba$
B) $Ba > Mg > Sr > Ca$
C) $Mg > Ca > Sr > Ba$
D) $Ca > Sr > Ba > Mg$
Answer
516.9k+ views
Hint: Solubility depends upon the two factors and these are: Lattice enthalpy and Hydration enthalpy. On moving down a group, size increases. So, the size of cation will increase from Be to Ba (group 2 elements, also called alkaline earth metals). Hydration enthalpy decreases as size of the cation increases.
Complete answer:
Alkaline earth metals are two group 2 elements, comprising beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra).
The general chemical formula of sulphates of alkaline earth metals is $MS{O_4}$ where, anion is $SO_4^ - $ (sulphate) and cation is ${M^ + }$ (alkaline earth metals). Now, solubility of alkaline earth’s metal sulphates can be explained as:
Solubility of any substance depends on the following two factors:
- Lattice enthalpy
- Hydration enthalpy
The magnitude of lattice enthalpy of the alkaline earth’s metal sulphates remains almost constant as we move down the group. This is because sulphate ions are so big and the small increase in the size of cations on moving down the group doesn’t make any difference in the lattice enthalpy of the alkaline earth’s metal sulphates.
However, the hydration enthalpy decreases from $B{e^{2 + }}$ to $B{a^{ + 2}}$significantly. This is because the hydration enthalpy depends on the size of cation as:
${\text{Hydration enthalpy}} \propto \dfrac{1}{{{\text{Size of cation}}}}$
As the size of cation increases from Be to Ba, therefore hydration enthalpy decreases of alkaline earth metal sulphates. The greater hydration enthalpies overcome the lattice enthalpy factor.
Thus, solubility of alkaline earth’s metal sulphates decreases in the sequence: $Be > Mg > Ca > Sr > Ba$. Hence, the option, $Mg > Ca > Sr > Ba$ is among the sequence.
Thus, option C is correct.
Note: The alkaline earth’s metal sulphates are all white solids and stable to heat. They are prepared by the action of sulphuric acid on metals, metals oxides, hydroxides and carbonates.
$M + {H_2}S{O_4} \to MS{O_4} + {H_2}$
$MO + {H_2}S{O_4} \to MS{O_4} + {H_2}O$
$M{(OH)_2} + {H_2}S{O_4} \to MS{O_4} + 2{H_2}O$
$MC{O_3} + {H_2}S{O_4} \to MS{O_4} + {H_2}O + C{O_2}$
Complete answer:
Alkaline earth metals are two group 2 elements, comprising beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra).
The general chemical formula of sulphates of alkaline earth metals is $MS{O_4}$ where, anion is $SO_4^ - $ (sulphate) and cation is ${M^ + }$ (alkaline earth metals). Now, solubility of alkaline earth’s metal sulphates can be explained as:
Solubility of any substance depends on the following two factors:
- Lattice enthalpy
- Hydration enthalpy
The magnitude of lattice enthalpy of the alkaline earth’s metal sulphates remains almost constant as we move down the group. This is because sulphate ions are so big and the small increase in the size of cations on moving down the group doesn’t make any difference in the lattice enthalpy of the alkaline earth’s metal sulphates.
However, the hydration enthalpy decreases from $B{e^{2 + }}$ to $B{a^{ + 2}}$significantly. This is because the hydration enthalpy depends on the size of cation as:
${\text{Hydration enthalpy}} \propto \dfrac{1}{{{\text{Size of cation}}}}$
As the size of cation increases from Be to Ba, therefore hydration enthalpy decreases of alkaline earth metal sulphates. The greater hydration enthalpies overcome the lattice enthalpy factor.
Thus, solubility of alkaline earth’s metal sulphates decreases in the sequence: $Be > Mg > Ca > Sr > Ba$. Hence, the option, $Mg > Ca > Sr > Ba$ is among the sequence.
Thus, option C is correct.
Note: The alkaline earth’s metal sulphates are all white solids and stable to heat. They are prepared by the action of sulphuric acid on metals, metals oxides, hydroxides and carbonates.
$M + {H_2}S{O_4} \to MS{O_4} + {H_2}$
$MO + {H_2}S{O_4} \to MS{O_4} + {H_2}O$
$M{(OH)_2} + {H_2}S{O_4} \to MS{O_4} + 2{H_2}O$
$MC{O_3} + {H_2}S{O_4} \to MS{O_4} + {H_2}O + C{O_2}$
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