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Ionic Equilibrum question

2016 · Q99
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Ionic Equilibrum question

2016 · Q99

NEETChemistryIonic EquilibrumMCQ+4 / −1
The solubility of AgCl(s) with solubility product 1.6 ×\times× 10−-−10 in 0.1 M NaCl solution would be
  1. A
    1.26 ×\times× 10−-−5 M
  2. B
    1.6 ×\times× 10−-−9 M
  3. C
    1.6 ×\times× 10−-−11 M
  4. D
    zero
View written solutionFree

Correct answer: B

AgCl ⇌ Ag+ + Cl–
   s           s     s + 0.1

Concentration of Cl– is (s + 0.1) mol L–1 because s mol L–1 from ionization of AgCl and 0.1 mol L–1 from ionization of 0.1 M NaCl.

Now, Ksp = [Ag+][Cl– ]

⇒\Rightarrow⇒ 1.6 × 10–10 = s (s + 0.1)

⇒\Rightarrow⇒ 1.6 × 10–10 = s (0.1) {∵\because∵ s << 0.1}

⇒\Rightarrow⇒ s = 1.6 × 10–9 M

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