Question

The solubility product for silver chloride is 1.2*10-10 at 293k calculate the solubility of silver chloride at 298k

Answer 1

Solubility product of AgCl is 4*10^-10 at 298 K. Solubility of AgCl in 0.04 M CaCl2 is?

Answer

78

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3 ANSWERS



Nissim Raj Angdembay, A Levels (2016)

Answered Sep 23, 2015

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AgCl(s)AgCl(s) disassociates in water in the following equation:

AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s)⇌Ag+(aq)+Cl−(aq)

That means, 

Ksp=[Ag+(aq)][Cl−(aq)]Ksp=[Ag+(aq)][Cl−(aq)]

4×10−10=[Ag+(aq)][Cl−(aq)]4×10−10=[Ag+(aq)][Cl−(aq)]

Since the same number of ions are produced, you could represent it using 'x':

4×10−10=[x][x]4×10−10=[x][x]

x2=4×10−10x2=4×10−10

x=2×10−5 mol dm−3