Short FL #5, P/C, Q.21

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Short FL #5, P/C, Q.21

Post by jbrostrom » Sun Aug 02, 2020 1:22 pm

I am confused about Ksp. My initial understanding of Ksp is that it is effectively a Keq value for a solid ionic compound; a measure of the amount of a solid ionic solute that can be dissolved in a saturated solution. I thought that the reason the reactant was not included in calculating Ksp is because the ionic cmpd is a solid, and solids are not included in Keq calculations.

In the Chem/Physics section of shortened FL#5, Ksp is given for Fe(OH)2 (aq) and the question asks you to solve for molar solubility. The equilibrium for the dissolution is given by the equ:
Fe(OH)2 (aq) + H2O (l) <--> Fe^2+(aq) + 2OH^-(aq)
Since Fe(OH)2 is aquesous, I assumed it would be included in the calculation of Ksp

Ksp = ([Fe2+][OH-]^2)/[Fe(OH)2]

How come Fe(OH)2 (aq) is not included in the calculation of Ksp.
Additionally, can Ksp be given for other non-solid compounds, and if it is given would you include the solute in the denominator of the equation? For example, could you be given a Ksp of glucose?
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Joined: Fri May 29, 2020 11:43 pm

Re: Short FL #5, P/C, Q.21

Post by NS_Tutor_Yuqi » Tue Aug 04, 2020 1:25 am

Hi! The reason why Fe(OH)2 is not included in the Ksp equation is because the passage says that it is poorly soluble. You are right that Ksp usually pertains to the dissociation of solids, but there are instances where gases and liquids are also relevant.
Since Ksp only pertains to ionic compounds, you can't find the Ksp of sugars like glucose.
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