Answer:
The concentration of this NaCl solution is 0.0769M
Explanation:
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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:
[tex]AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)\\\\Ksp=[Ag^+][Cl^-]\\[/tex]
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:
[tex]Ksp=(s)(C+s)\\\\1.83x10^{-10}=(2.38x10^{-9})(M+2.38x10^{-9})[/tex]
Thus, solving for M, we obtain:
[tex]\frac{1.83x10^{-10}}{2.38x10^{-9}} =M+2.38x10^{-9}\\\\M=\frac{1.83x10^{-10}}{2.38x10^{-9}} -2.38x10^{-9}\\\\M=0.0769M[/tex]
Thus, the concentration of this NaCl solution is 0.0769M.
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