MHT CET Medical202624 April 2026Evening ShiftChemistryIonic EquilibriumActual
Calculate the solubility of AgCl in mol/dm ^3 at 25 ^ C . Given: K_ sp ( AgCl ) = 1.8 10⁻¹⁰
Options
- A1.8 10⁻⁵ M
- B1.34 10⁻⁵ M
- C1.8 10⁻¹⁰ M
- D3.6 10⁻⁵ M
Correct answer
B. 1.34 10⁻⁵ M
Step-by-step solution
The dissociation reaction of AgCl is given by: AgCl(s) Ag ^+( aq ) + Cl ^-( aq ) Let the solubility of AgCl be s . The solubility product constant is: K_ sp = [ Ag ^+][ Cl ^-] = s s = s^2 Given K_ sp = 1.8 10⁻¹⁰ , substituting this value: s^2 = 1.8 10⁻¹⁰ s = 1.8 10⁻¹⁰ s = 1.34 10⁻⁵ mol/dm ^3 Answer: 1.34 10⁻⁵ M