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NEETChemistryIonic Equilibrium

The solubility product ( K_ sp ) of AgCl is 1.6 10⁻¹⁰ . What is the molar solubility of AgCl in a 0.1 M aqueous solution of NaCl ?

Options

  1. A1.6 10⁻⁹ mol/L
  2. B1.26 10⁻⁵ mol/L
  3. C1.6 10⁻¹¹ mol/L
  4. D1.6 10⁻¹⁰ mol/L

Correct answer

A. 1.6 10⁻⁹ mol/L

Step-by-step solution

In a 0.1 M NaCl solution, the concentration of chloride ions is [ Cl ^-] = 0.1 M (due to complete dissociation of NaCl ). Let the molar solubility of AgCl in this solution be s . The dissociation of AgCl is: AgCl (s) Ag ^+(aq) + Cl ^-(aq) At equilibrium: [ Ag ^+] = s [ Cl ^-] = 0.1 + s 0.1 M (since s is very small compared to 0.1 ) The solubility product expression is: K_ sp = [ Ag ^+][ Cl ^-] 1.6 10⁻¹⁰ = (s)(0.1) s = 1.6 10⁻¹⁰ 0.1 = 1.6 10⁻⁹ mol/L Thus, the molar solubility is 1.6 10⁻⁹ mol/L . Answer: 1.6 10⁻⁹ mol

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