The solubility of AgCI in 0.2 M NaCl solution isKsp of AgCl=1.20×10−10

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6.0×10−10M

0.2M

1.2×10−10M

0.2×10−10M

answer is A.

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Detailed Solution

Given, concentration of NaCl = 0.2 MKsp(AgCl)=1.20×10−10 Let the solubility of AgCl in NaCl = x AgCl⟶AgX++ClX− Solubility NaCl0.2⟶Na+0.2+Cl0.2− ∴Ag+=x and Cl−=(x+0.2) ∴ Ksp(AgCl)=Ag+Cl−=x(x+0.2)=x2+0.2x ∴ Ksp=0.2xx2<<1 or 1.2×10−10=0.2x ∴ x=6×10−10

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