Chemistry, asked by shallungmorandangsha, 8 months ago

The solubility of AgCl in water at 25°C is 1×10^-5M. calculate the solubility products of AgCl at this temperature

Answers

Answered by abhilasha69

0

Answer:

Let AgCl solubility be S

Then Ksp=S

2

Given, S=1.06×10

−5

Ksp=1.1236×10

−10

Explanation:

hope it is useful to you..!!

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