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Chemistry

Solubility product and common ion effect

In Class 11 Chemistry, this topic from the Equilibrium chapter explains how the solubility product (Ksp) describes the dissolution equilibrium of sparingly soluble salts. Students learn to write Ksp expressions, relate them to molar solubility, and use ion concentrations to predict whether precipitation will occur. The common ion effect shows how adding an ion already present in the equilibrium reduces solubility. The topic also develops understanding of selective precipitation and equilibrium-based calculations.

Medium · Level 1 · 8 questions

TOPIC PRACTICE

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8 questions

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  1. It decreases solubility
  2. It increases solubility
  3. It makes solubility infinite
  4. It remains unrelated to solubility
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  1. Solubility of the salt will decrease
  2. Solubility of the salt will increase
  3. The salt will completely evaporate
  4. The solvent mass will become zero
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  1. The salt is sparingly soluble
  2. The salt is highly soluble
  3. The salt is only a gas
  4. The salt forms no ions
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  1. Precipitate will form
  2. The solution will remain unsaturated
  3. The salt will become gas completely
  4. The solvent will be destroyed
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  1. Unsaturated
  2. Saturated
  3. Supersaturated with precipitate
  4. Completely solid
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  1. Saturated
  2. Unsaturated
  3. Necessarily precipitated
  4. Completely gaseous
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  1. Keep ionic product less than solubility product
  2. Make ionic product very high
  3. Treat solubility product as negative
  4. Remove the solvent completely
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  1. Comparison of ionic product and solubility product
  2. Comparison of mass and colour
  3. Comparison of temperature and container height
  4. Comparison of solvent name and smell

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