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NEET levelSingle correct optionDifficulty 1 of 3ie171

Using $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ , the solubility of $AgCl$ in $0.1 M$ $NaCl$ solution (in $mol L^{- 1}$ ) is approximately

A $1.8 \times 1 0^{- 9} M$
B $1.3 \times 1 0^{- 5} M$
C $1.8 \times 1 0^{- 11} M$
D $4.2 \times 1 0^{- 6} M$

Answer: A

[Cl X^{-}] \approx 0.1 M

since the dissolved

AgCl

adds very little. Then

s = [Ag X^{+}] = K_{s p} / [Cl X^{-}] = 1.8 \times 1 0^{- 10} /0.1 = 1.8 \times 1 0^{- 9} M

.

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