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SubjectsChemistry
विलेयता
In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn how much of a solute can dissolve in a given amount of solvent under specific conditions. The topic explains saturated, unsaturated and supersaturated solutions, along with the factors that affect solubility, such as the nature of solute and solvent, temperature and pressure. Students also explore why gases behave differently from solids in solutions and apply these ideas to interpret solubility data and related chemical situations.
TOPIC PRACTICE
Up to 6 questions from this page. Select your focus, then start.
At the stated temperature, 45 g is the normal equilibrium limit for 100 g water, but 55 g is present. The solution is therefore supersaturated by about 10 g. A seed crystal provides a nucleation surface, so excess solute may crystallize until roughly the normal limit remains dissolved. The exact separated amount can depend on conditions.
The ionic product is calculated from the current concentrations of the ions. Adding a common ion raises one ion concentration, so the ionic product can become greater than Ksp. When Q exceeds Ksp, the solution is supersaturated with respect to that salt and precipitation proceeds until Q returns to Ksp. Ksp itself is not made infinite by adding an ion.
For salts having the same dissolution stoichiometry, the expression for Ksp has the same powers of solubility for both salts. A larger Ksp then corresponds to a larger equilibrium solubility and hence greater dissolution. This conclusion cannot be applied blindly to salts with different ion ratios, because their Ksp expressions contain different powers and direct numerical comparison becomes misleading.
Dissolving an ionic solid requires separating ions from the crystal lattice, which costs lattice energy. Water molecules then hydrate and stabilise the separated ions, releasing hydration energy. Solubility depends on the overall energy and entropy balance, not on charge alone. If hydration is insufficient to offset lattice forces, the solid may be only slightly soluble.
Using p = K_H x, at a fixed pressure a decrease in gas solubility means a decrease in dissolved mole fraction x. Since K_H = p/x, a smaller x produces a larger K_H. Gas solubility generally decreases with rising temperature, so the Henry constant commonly increases, though the exact temperature dependence is system-specific.
The combination of a gas dissolved in a liquid, its pressure, and mole fraction is the standard clue for Henry’s law. Write p = K_H x, identify the known quantities, and rearrange if necessary. Temperature should not be ignored because K_H depends on it. Lattice energy applies mainly to ionic solids, while colour provides no relevant equation.
QUIZ COMPLETE
