Difficult
Explain the factors affecting the solubility of salts in a solution.
(N/A) The solubility of ionic solids in water varies significantly.
Some ionic solids,such as $CaF_2$,are highly soluble,while others like $LiF$ (lithium fluoride) have very low solubility and are considered insoluble. The solubility of a salt depends on two primary factors:
$(i)$ Lattice enthalpy of the salt: This is the energy required to break one mole of a solid salt into its constituent ions. $A$ higher lattice enthalpy means more energy is needed to break the crystal lattice.
$(ii)$ Solvation enthalpy of ions: This is the energy released when one mole of a solid salt dissolves in a solvent. For a salt to dissolve,the energy released from ion-solvent interactions (solvation enthalpy) must be sufficient to overcome the strong electrostatic forces holding the ions together in the crystal lattice (lattice enthalpy).
Therefore,a salt is soluble in a solvent only when the solvation enthalpy is greater than the lattice enthalpy. If the solvation enthalpy is less than the lattice enthalpy,the salt remains insoluble.
Some ionic solids,such as $CaF_2$,are highly soluble,while others like $LiF$ (lithium fluoride) have very low solubility and are considered insoluble. The solubility of a salt depends on two primary factors:
$(i)$ Lattice enthalpy of the salt: This is the energy required to break one mole of a solid salt into its constituent ions. $A$ higher lattice enthalpy means more energy is needed to break the crystal lattice.
$(ii)$ Solvation enthalpy of ions: This is the energy released when one mole of a solid salt dissolves in a solvent. For a salt to dissolve,the energy released from ion-solvent interactions (solvation enthalpy) must be sufficient to overcome the strong electrostatic forces holding the ions together in the crystal lattice (lattice enthalpy).
Therefore,a salt is soluble in a solvent only when the solvation enthalpy is greater than the lattice enthalpy. If the solvation enthalpy is less than the lattice enthalpy,the salt remains insoluble.
Explore More
Similar Questions
If $N_2$ gas is bubbled through water at $293 \ K$,how many millimoles of $N_2$ gas would dissolve in $1 \ L$ of water? Assume that $N_2$ exerts a partial pressure of $0.987 \ bar$. Given that Henry's law constant for $N_2$ at $293 \ K$ is $76.48 \ kbar$.
Difficult
View SolutionIf $O_2$ gas is bubbled through water at $303 \; K$,the number of millimoles of $O_2$ gas that dissolve in $1 \; L$ of water is (Nearest Integer). (Given: Henry's Law constant for $O_2$ at $303 \; K$ is $46.82 \; kbar$ and partial pressure of $O_2 = 0.920 \; bar$). (Assume solubility of $O_2$ in water is too small,nearly negligible).
When a bottle of soft drink is opened, which of the following phenomena occur?
$(A)$ Solubility of dissolved gas decreases.
$(B)$ The bottle releases internal pressure.
$(C)$ Effervescence is observed from the bottle.
$(D)$ Increase in external pressure.
Identify the correct choice.
$(A)$ Solubility of dissolved gas decreases.
$(B)$ The bottle releases internal pressure.
$(C)$ Effervescence is observed from the bottle.
$(D)$ Increase in external pressure.
Identify the correct choice.
The partial pressure of ethane over a solution containing $7.14 \times 10^{-3} \ g$ of ethane is $1 \ bar$. If the solution contains $5 \times 10^{-2} \ g$ of ethane,then what shall be the partial pressure of the gas (in $bar$)? (At equal temperature)
Solubility of $N_2$ gas in water at $25\,^{\circ}C$ and $0.78\, atm$ is $5.4 \times 10^{-4}\, M$. Calculate Henry's law constant for $N_2$ in water at $25\,^{\circ}C$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo