J.R. S. answered • 02/20/23
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Not sure density plays a role since Kc is based on molar concentration which is moles / liter of solution, and not moles / liter of solvent. Seems like Kc and Kp would be the same values.
Ca(OH)2(s) <==> Ca2+(aq) + 2OH-(aq)
Ksp = 5.50x10-6 = [Ca2+][OH-]2
Let [Ca2+] = x, then [OH-] = 2x
5.5x10-6 = (x)(2x)2 = 4x3
x3 1.375x10-6
x = 1.11x10-2 M = solubility of Ca(OH)2
Ca(OH)2(s) <==> Ca2+(aq) + 2OH-(aq)
Kc = [Ca2+][OH-]2 / 1
Kc = (1.11x10-2)(2.22x10-2)2
Kc = 5.5x10-6
Suleiman N.
It seems that Ksp is used as the concentration of the products and the reactant is Ca(OH)2(s). The concentration of Ca(OH)2(s) is found by using the density to convert from g/ml to mol/ml and then mol/L which is equivilant to M molarity (mol/L). After the concentration is found then its Kc = Ksp/[Ca(OH)2(s)]
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02/21/23
J.R. S.
tutor
Thanks for the clarification, but to me, that doesn't make sense. Not sure how you can find the concentration of SOLID Ca(OH)2. In the Ksp expression, the [Ca(OH)2] is taken as unity, and thus Ksp = [Ca2+][OH-]^2/1 = [Ca^2+][OH-]. I still can't see how the density plays a role. But thank you anyway for responding. Maybe someone else will 'chime in' to clarify further.
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02/21/23
Suleiman N.
Thank you!02/20/23