# 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

Solubility product of AgCl = K

_{sp}

AgCl (s) = Ag

^{+}+ Cl

^{-}

KCl = K

^{+}+ Cl

^{-}

So, [Cl

^{-}] = x M

At equilibrium,

[Ag

^{+}] = a ; [Cl

^{-}] = a + x

Equating the two we get -

K

_{sp}= a x (a + x) = a

^{2}+ ax

After approximation, we get -

ax = K

_{sp}

So, a = $\frac{{K}_{sp}}{x}$

Here, a = moles of AgCl dissolved = Solubility of AgCl = $\frac{{K}_{sp}}{x}$

So, option (d) is correct.

Regards

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