Ionic EquilibriumHard

Question

The solubility of AgCl in water is 0.001435 g per litre at 15o C. The solubility product of AgCl is

(Ag = 108, Cl = 35.3)

Options

A.10−5
B.10−10
C.2 × 10−10
D.10−9

Solution

$S = \frac{0.001435}{143.5} = 10^{- 5}\text{ M}$

∴ Solubility product, Ksp = S2 = 10–10

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