The solubility product of AgCl is Ksp. Then the solubility of AgCl in xM KCl is 1.Ksp×x2 2.x/Ksp 3.Ksp/x2 4.Ksp/x

Dear Student

Solubility product of AgCl = Ksp
AgCl (s) = Ag+ + Cl​-

KCl = K+ + Cl- 
So, [Cl-] = x M

At equilibrium, 
[Ag+] = a ;     [Cl-] = a + x

Equating the two we get -
Ksp = a x (a + x) = a2 + ax

After approximation, we get -
ax = Ksp
So, a = Kspx
Here, a = moles of AgCl dissolved = Solubility of AgCl = Kspx
So, option (d) is correct.

Regards
 

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