The solubility product of AgCl is 1.4 x 10-4 at 100°C. Calculate the solubility of AgCl in boiling water.

AgCl=Ag+ + Cl-

1.4×10^-4 =x × x

x²=1.4×10^-4

x=√(1.4×10^-4)

x=0.012

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explanation:  

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