Chemistry, asked by srivastavapranjal737, 4 months ago


. 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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Answered by iTzRiYaNsH
3

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Answered by abhi178
3

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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