Answer:
[tex]Ksp=2.0x10^{-14}[/tex]
Explanation:
Hello there!
In this case, given the solubilization of cadmium (II) hydroxide:
[tex]Cd(OH)_2(s)\rightleftharpoons Cd^{2+}(aq)+2OH^-(aq)[/tex]
The solubility product can be set up as follows:
[tex]Ksp=[Cd^{2+}][OH^-]^2[/tex]
Now, since we know the concentration of cadmium (II) ions at equilibrium and the mole ratio of these ions to the hydroxide ions is 1:2, we infer that the concentration of the latter at equilibrium is 3.5x10⁻⁵ M. In such a way, the resulting Ksp turns out to be:
[tex]Ksp=(1.7x10^{-5})(3.4x10^{-5})^2\\\\Ksp=2.0x10^{-14}[/tex]
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