Question

A sample of Cd(OH)2 is added to pure water and allowed to come to equilibrium at 25ºC. The concentration of Cd +2 = 1.7 x 10 -5M at equilibrium. Calculate Ksp.

Answer 1

Answer:

$Ksp=2.0x10^{-14}$

Explanation:

Hello there!

In this case, given the solubilization of cadmium (II) hydroxide:

$Cd(OH)_2(s)\rightleftharpoons Cd^{2+}(aq)+2OH^-(aq)$

The solubility product can be set up as follows:

$Ksp=[Cd^{2+}][OH^-]^2$

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:

$Ksp=(1.7x10^{-5})(3.4x10^{-5})^2\\\\Ksp=2.0x10^{-14}$

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