Question

What is the molarity of NaCl in which AgCl has a molar solubility of 2.38 x 10-9 mol /L? The Ksp for Silver Chloride is: 1.83 x 10-10.

Answer 1

The concentration of this NaCl solution is 0.0769M

Step-by-step explanation:

Hello there!

In this case, since we are dealing with a solubility equilibrium in which it is desired to compute the concentration of a common-ion effect to chloride solution in the form of NaCl, we can set up the equilibrium reaction and expression:

`AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)\\\\Ksp=[Ag^+][Cl^-]\\`

It is possible to insert the concentration of starting chloride ions provided by the ionization of NaCl and in terms of the molar solubility, s, equal to 2.38x10^{-9}M:

`Ksp=(s)(C+s)\\\\1.83x10^(-10)=(2.38x10^(-9))(M+2.38x10^(-9))`

Thus, solving for M, we obtain:

`(1.83x10^(-10))/(2.38x10^(-9)) =M+2.38x10^(-9)\\\\M=(1.83x10^(-10))/(2.38x10^(-9)) -2.38x10^(-9)\\\\M=0.0769M`

Thus, the concentration of this NaCl solution is 0.0769M.

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