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Umnica [9.8K]
3 years ago
15

Calcium hydroxide, Ca(OH)2, is an ionic compound with a solubility product constant, Ksp, of 6.5×10–6. Calculate the solubility

of this compound in pure water.
Chemistry
1 answer:
kari74 [83]3 years ago
5 0

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 K_{sp}

The equation for the ionization of the  is given as:

Ca(OH)_2\rightarrow Ca^{2+}+2OH^-  

By stoichiometry of the reaction:

1 mole of  Ca(OH)_2 gives 1 mole of Ca^{2+} and 2 mole of OH^-

When the solubility of  Ca(OH)_2 is S moles/liter, then the solubility of Ca^{2+}  will be S moles\liter and solubility of OH^- will be 2S moles/liter.

K_{sp}=[Ca^{2+}][OH^{-}]^2

6.5\times 10^{-6}=[S][2S]^2

S=0.012M

Thus solubility of this compound in pure water is 0.012 M

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The formula for estimating the ratio of population in 1st excited state to the ground state can be computed as:

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From the above equation:

Δ E = energy difference =  3.13 × 10⁻¹⁸  J

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Thus:

0.05 =e^{^{ -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}

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