What is the relationship between solubility s and solubility product K_sp for a sparingly soluble salt of type AB2 (like PbCl2)? (Hard)
Answer and explanation
Correct answer: B. K_sp = 4 s^3
- Step 1: Dissociation equilibrium: AB2(s) <=> A^(2+) + 2 B^(-). Step 2. [A^(2+)] = s, [B^(-)] = 2s. K_sp = [A^(2+)][B^(-)]^2 = (s)(2s)^2 = 4 s^3. Answer: K_sp = 4 s^3. Why not the others: s^2 is for 1:1 salts (AB); 27 s^4 is for 1:3 salts (AB3); 2 s^2 forgets squaring concentration.
Practise more Chemical Equilibrium, Energetics and Environmental Chemistry.
Free questions with an explanation for every one, and full timed Higher Secondary Certificate Examination (Bangladesh) mock tests.
More Chemical Equilibrium, Energetics and Environmental Chemistry questions
- What is the relation between equilibrium constants K_p and K_c for a general gaseous reaction?
- For the reaction N2(g) + 3H2(g) <=> 2NH3(g), what is the value of Delta n?
- According to Le Chatelier's Principle, how does increasing total pressure affect the equilibrium position of N2(g) + 3H2(g) <=> 2NH3(g)?
- How does increasing temperature affect the equilibrium of an exothermic reaction (Delta H < 0)?
- What is the pH of a 0.01 M aqueous solution of Hydrochloric Acid (HCl)?
- What is the pH of a 0.001 M aqueous solution of Sodium Hydroxide (NaOH) at 25 degrees Celsius?