NEETChemistryIonic Equilibrium
Determine the concentration of AgNO ₃ required to reduce the solubility of Ag ₂ CrO ₄ to 1 10⁻⁸ M . Given that the solubility product ( K_ sp ) of Ag ₂ CrO ₄ is 4 10⁻¹² .
Options
- A4 10⁻⁴ M
- B1 10⁻² M
- C2 10⁻⁴ M
- D2 10⁻² M
Correct answer
D. 2 10⁻² M
Step-by-step solution
The target solubility of Ag ₂ CrO ₄ is s = 1 10⁻⁸ M . The dissociation of the sparingly soluble salt is: Ag ₂ CrO ₄(s) 2 Ag ^+(aq) + CrO ₄²⁻(aq) From the stoichiometry, the concentration of chromate ions is equal to the solubility: [ CrO ₄²⁻] = s = 1 10⁻⁸ M Let the required concentration of AgNO ₃ be c . Since AgNO ₃ is a strong electrolyte, it completely dissociates to give [ Ag ^+] c (neglecting the very small amount of Ag ^+ from Ag ₂ CrO ₄ ). The solubility product expression is: K_ sp = [ Ag ^+]^2[ CrO ₄²⁻] Su