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Ksp of AgCl is 1 * 10-10. Its solubility in 0.1 M KNO3 will be..?

Thakshayini , 8 Years ago
Grade 11
anser 1 Answers
Rituraj Tiwari
Calculate the solubility of AgCl (Ksp = 1.8 x10-10) in 0.1 M, KCl

So we can calculate it as follows;

AgCl ---------> Ag+ + Cl-

ksp = S²

S = √1.8 x10-10

S = 1.34 x 10^-5

So the solubility in 0.1 M KCl

determine the new ksp = 1.34 x 10^-5 x (0.1) = 1.34 x 10^-6

So the new solubility is = √1.34 x 10^-6

= 1.16 x 10^-3 M
Last Activity: 5 Years ago
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