When 10.0 mL of a 0.1 M Cu+ solution are mixed with 1.0 mL of a 1.0 M Cl- solution, a precipitate of CuCl is formed: Cu+(aq) + Cl-(aq) = CuCl(s). This equilibrium is characterised at ambient...

When 10.0 mL of a 0.1 M Cu+ solution are mixed with 1.0 mL of a 1.0 M Cl- solution, a precipitate of CuCl is formed: Cu+(aq) + Cl-(aq) = CuCl(s). 
This equilibrium is characterised at ambient temperature by pKsp = 6.49. What is the final concentration of chloride anions that remain in solution?

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t-nez | High School Teacher | (Level 3) Associate Educator

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According to the stoichiometry of this reaction the solubility product constant is:

`K_s_p = [Cu^+][Cl^-]` = `10^-^6.49 = 3.24x10^-^7`

This is the maximum amount of CuCl that will remain dissolved in solution. Any additional will precipitate. Since this is the product of Cu+ and Cl-, the Cl- concentration is given by the square root of the Ksp:

``

`sqrt(3.24x10^-^7) = 5.69x10^-^4` 

Now let's determine if the moles of ions in the final solution exceeds the maximum solubility:

moles Cu+ = (0.010 L)(0.1 M) = 0.001 moles

moles Cl- = (0.0010 L)(1.0 M) = 0.0010 moles

There are equal moles of the two ions in 11.0 ml (0.0110 L) of solution, giving concentrations of each ion equal to (0.001 moles)/(0.0110 L) = 0.091 M

This exceeds the maximum solubility so the concentration of Cl- dissolved will equal the maximum solubility of `5.69x10^-^4M`

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