Question

What is the molar solubility of AgCl in a 0.050 M NaCl solution? The Ksp of AgCl is 1.6 x 10-10. (Assume that the contribution of [Cl-] from AgCl is negligible relative to the [Cl-] from NaCl)

Answer 1

The molar solubility of AgCl:

The molar solubility of AgCl in a 0.050 M NaCl solution is 8 x $10^{-8}$ M $Ag^{+}$

What is solubility?

The solubility is the quantity of reagent required to saturate the solution or bring about the dissociation reaction's equilibrium.

Reaction:

The dissociation reaction of AgCl in water is:

$AgCl$  ⇄ $Ag^{+} + Cl^{-}$

Each mole of AgCl that dissolves in this reaction yields 1 mole of both $Ag^{+}$ and  $Cl^{-}$. The concentration of either the Ag or Cl ions would then be equal to the solubility.

Solubility= [$Ag^{+}$] = [$Cl^{-}$]

Calculation:

in 0.050 M NaCl, the [$Cl^{-}$] = 1 x $10^{-2}$

ksp = [$Ag^{+}$] x [$Cl^{-}$]

1.6 x $10^{-10}$ = [$Ag{+}$] x ( 5 x $10^{-2}$)

[$Ag^{+}$] = 5 x $10^{+2}$ x 1.6 x $10^{-10}$

[$Ag^{+}$] = 5 x 1.6 x $10^{-10+2}$

[$Ag^{+}$] = 8 x $10^{-8}$ M

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