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?
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`