Answer: The solubility of this compound in pure water is 0.012 M
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as 
The equation for the ionization of the is given as:
By stoichiometry of the reaction:
1 mole of 
gives 1 mole of 
and 2 mole of 
When the solubility of is S moles/liter, then the solubility of will be S moles\liter and solubility of will be 2S moles/liter.
![K_{sp}=[Ca^{2+}][OH^{-}]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BOH%5E%7B-%7D%5D%5E2)
![6.5\times 10^{-6}=[S][2S]^2](https://tex.z-dn.net/?f=6.5%5Ctimes%2010%5E%7B-6%7D%3D%5BS%5D%5B2S%5D%5E2)
Thus solubility of this compound in pure water is 0.012 M
A 1 molar solution is the molecular weight in grams in 1 litre of water, so a 3.5 molar solution would be 58.44g multiplied by 3.5, which is 204.54g in 1L
The overall order of a reaction is directly
related to the reaction mechanism of the reaction. The reaction mechanism is
defined by the reaction rate equation. In this case, we are given by the
equation, Rate=k. The order of this reaction is zero since the rate is
not dependent on the concentration of reactants or products.
Answer:
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