. At a certain temperature, the solubility product of AgCl at 373 K. The
solubility product of Agch is 3.4 x10".
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We have to find the solubility of AgCl when solubility product of AgCl is 3.4 × 10¯⁴.
solution : dissociation reaction of AgCl is ..
AgCl ⇔ Ag⁺ + Cl¯
S S
so, Ksp = [Ag⁺][Cl¯] = S × S
⇒Ksp = S²
⇒3.4 × 10¯⁴ = S²
⇒S = √(3.4) × 10¯² M
⇒S = 1.84 × 10¯² M
Therefore the solubility of AgCl would be 1.84 × 10¯² M.
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