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NEET levelSingle correct optionDifficulty 1 of 3ia059

Using $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ , the solubility of $AgCl$ in $0.050 M$ $NaCl$ (in $mol L^{- 1}$ ) is about

A $9.0 \times 1 0^{- 12}$
B $1.3 \times 1 0^{- 5}$
C $7.2 \times 1 0^{- 8}$
D $3.6 \times 1 0^{- 9}$

Answer: D

The common ion fixes

[Cl X^{-}] \approx 0.050 M

, so

s = [Ag X^{+}] = K_{s p} /0.050 = 3.6 \times 1 0^{- 9} M

.

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