The solubility of AgCl(s) with solubility product 1.6×10−10 is in 0.1 M NaCl solution would be

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The solubility of $AgCl$ (s) with solubility product $1.6 \times 10^{- 10}$ is in $0.1 M NaCl$ solution would be

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Zero

$1 .26 \times 10^{- 5} M$

$1 .6 \times 10^{- 9} M$

$1 .6 \times 10^{- 11} M$

answer is C.

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Detailed Solution

AgCl ⇌ Ag⁺+ Cl⁻
a 0 0
a-S S S+0.1
ksp = 1.6×10⁻¹⁰= [Ag⁺] [Cl⁻]

=S(0.1+S)

ksp is value seems to be very small
S value can be ignored, with respect to 0.1 M

1.6×10⁻¹⁰=S×0.1

S=1.6×10^-9M

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