The solubility of AgCI in 0.2 M NaCl solution isKsp of AgCl=1.20×10−10 - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
${6.0×10}^{- 10} M$
B
$0 .2 M$
C
$1 .2 \times 10^{- 10} M$
D
$0 .2 \times 10^{- 10} M$

Solution:

Given, concentration of NaCl = 0.2 M
$K_{sp} (AgCl) = 1 .20 \times 10^{- 10} Let the solubility of AgCl in NaCl = x AgCl ⟶ {\underset{X}{Ag}}^{+} + {\underset{X}{Cl}}^{-} Solubility \underset{0 .2}{NaCl} ⟶ \underset{0 .2}{{Na}^{+}} + {\underset{0 .2}{Cl}}^{-} ∴ [{Ag}^{+}] = x and [{Cl}^{-}] = (x + 0 .2) ∴ K_{sp} (AgCl) = [{Ag}^{+}] [{Cl}^{-}] = x (x + 0 .2) = x^{2} + 0 .2 x ∴ K_{sp} = 0 .2 x (x^{2} << 1) or 1 .2 \times 10^{- 10} = 0 .2 x ∴ x = 6 \times 10^{- 10}$

The solubility of AgCI in 0.2 M NaCl solution is $(K_{sp} of AgCl = 1 .20 \times 10^{- 10})$

Solution:

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