At a given temperature Ksp of AgCl is 4.83 x 10-10. Calculate the molar solubility of AgCl at this temperature. -...

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Answer 1

Answer #1

Answeg) a) Ksp Of Agde = 4.83x101 Agce - Ag + ce Ksp - [Ag+] [u] Ksp = [5] x [s] 4.83x10-10 = [s] S = V4.83x1070 S = 2.197 X1

Answer 2

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