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Q.

The silver ion in a solution is precipitated by addition of chloride ion. The final volume of solution is 500 mL. What should be the concentration of Cl^-ion if not more than 0.10 mg of Ag⁺ ion remains unprecipitated?
( K_sp(AgCl) = 1.0 x 10^-10 )

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a

$5 .4 \times 10^{- 5} M$

b

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

c

$5 .4 \times 10^{- 4} M$

d

$1 .9 \times 10^{- 7} M$

answer is A.

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Detailed Solution

$[{Ag}^{+}] = 0 .1 mg = (0 .1 \times 10^{- 3} g) in 500 {mLH}_{2} O = \frac{0 .1 \times 10^{- 3}}{108} mol in 500 {mLH}_{2} O = \frac{0 .1 \times 10^{- 3} \times 2}{108} mol in 1 {LH}_{2} O = 1 .85 \times 10^{- 6} M [{Ag}^{+}] [{Cl}^{-}] = 1 .0 \times 10^{- 10} ∴ [{Cl}^{-}] = \frac{1 .0 \times 10^{- 10}}{1 .85 \times 10^{- 6}} = 5 .4 \times 10^{- 5} M$

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