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Question
Ksp of AgCl is 1.8 × 10−10. Calculate molar solubility in 1 M AgNO3.
Solution
\[\ce{AgCl_{(s)} ⇌ Ag^+_{( aq)} + Cl^-_{( aq)}}\]
x = solubility of AgCI in 1 M AgNO3
\[\ce{AgNO3_{(aq)} ⇌ \underset{1 M}{Ag^+_{( aq)}} + \underset{1 M}{NO^-_{3(aq)}}}\]
[Ag+] = x + 1 ≃ 1 M ..........(∵ x << 1)
[Cl–] = x
Ksp = [Ag+] [Cl–]
1.8 × 10−10 = (1) (x)
x = 1.8 × 10−10 M
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