The solubility of copper(I) chloride is 3.91mg per 100.0mL of solution.?

Calculate Ksp for CuCl .

CuCl —-> Cu+ + Cl-
… s ………. s …… s ……
where s is the molar solubility
Ksp = [Cu+]*[Cl-] = s^2
moles of CuCl = grams / MM = 0.00391 / 98.99 = 3.95*10^-5 mol
molar solubility = moles / V = 3.95*10^-5 / 0.100 = 3.95*10^-4 M
Ksp = s^2 = 1.56*10^-7

Hi man,
So you know that a maximum of 39.1mg of CuCl will dissolve into 1L of solution giving you a 0.000394949495M CuCl solution. That is, 0.0003949 moles of CuCl dissociate 1:1 into .0003949moles of Cu+ and .0003949moles of Cl- in 1L of water.
Ksp = (.003949)(.003949) = 1.55985104 × 10-7
But be careful, when an ionic compound does not dissovle into 1:1 proportions, you cannot simply square the molar concentration of the ionic compound to get Ksp.
Take BaF2 for example. When it dissolves in water, it yields 2 ions of F- for every Ba in a 1:2 ratio. The Ksp for this would be written:
Ksp = (x)(2x)^2
For reasons why the 2x was raised to the 2nd power, see his cool tutorial:
http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Ca…

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