Q.

Solubility product of AgCl is 1.8×10-10 M2. Molar solubility of AgCl in 0.1 M NaCl solution will be

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a

1.8×10-11

b

1.8×10-9

c

1.8×10-12

d

1.8×10-6

answer is B.

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Detailed Solution

AgCls⇌Ag+Saq+Cl-Saq ;  NaClaq→Na+0.1 M+Cl-0.1 M ;  Ksp=Ag+Cl-1.8×10-10=ss+0.1M         s<<<0.1M1.8×10-10=s0.1M ; s=1.8×10-9 M ;
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Solubility product of AgCl is 1.8×10-10 M2. Molar solubility of AgCl in 0.1 M NaCl solution will be