Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: BHAVESH on January 29, 2020, 06:43:36 AM
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Hi Everyone,
I found an equation from Google about Ksp of CaSO4 which is as follows
Ksp CaSO4 = 10^[1/(-0.0222XLog(Ionic Strength of Water)+2.2954)2-0.2478]
I just wanted to know how these numeric figures have been derived in above equation.
Thanks....
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Did it give a reference? Can you look up that reference?
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Just by eye it looks like a empirical equation, determined by fitting to a series of experimental data points. If that's the case, the terms and coefficients probably don't have any specific scientific meaning.
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They have just mentioned - (Dupont 1992)
They have mentioned - The solubility products for sparingly soluble salts are based on expressions derived from published data on solubility as a function of the ionic strength of the solution at a given temperature. The equation for calculating Ksp for CaSO4 at 25 Degree C.
And when I checked in Dupont document for Ksp, they have given a graph, it seems it must have been derived from that.
So, still the question remains is that how to calculate Ksp of CaSO4, BaSO4, SrSO4 at any given temp.? - Can I apply Van't Hoff equation here as well???
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So, still the question remains is that how to calculate Ksp of CaSO4, BaSO4, SrSO4 at any given temp.? - Can I apply Van't Hoff equation here as well???
Sure... bearing in mind sources of error.
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Hi Corribus,
I didn't understand about sources of error. Can you please explain.
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For one, the accuracy and presumed temperature independence of thermodynamic quantities (e.g., standard enthalpy change).
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Ksp of CaSO4 @ 25°C is 4.93 X 10-5. When calculated using Van't Hoff Equation @ 40°C, surprisingly reduction in solubility is 4.52 X 10-17 using standard enthalpy of CaSO4 (-1433 KJ mol-1) at 25°C.
Such a drastic drop in solubility!!!
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All that means is that you are doing something wrong.
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Check units in general but especially on your literature value for standard enthalpy of calcium sulfate. That seems way too high (like, by several orders of magnitude).
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What is the value of ΔH that you must use in the van't Hoff equation? Is that what you have used?
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Yes I used -1433 KJ mol-1. This I took from Wikipedia Data Standard Enthalpy. Can you please help me correct my mistake?
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However, it has been observed that when Temp. increases, solubility of CaSO4 increases. In this case, I feel that I am doing some GROSS MISTAKE in calculating solubility using Van't Hoff Equation.
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-1433 kJ/mol is the enthalpy of formation of solid CaSO4. Is that what you need to use in the van't Hoff equation for the solubility equilibrium?
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yes, this is because I want to calculate the prediction of CaSO4 precipitation at given concentration in water. I mean when water gets concentrated by means of either evaporation or by reverse osmosis process, CaSO4 tends to precipitate depending upon concentration of individual ions and comes out as solid from water.
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No - what is the definition of enthalpy of formation? You either need to look up the enthalpy of dissolution, or calculate it from the enthalpies of formation of the solid sulfate and the dissolved ions.
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you mean to say based on Lattice & Hydration Energy should I calculate? As mentioned below
[1ΔHf(Ca+2 (aq)) + 1ΔHf(SO4-2 (aq))] - [1ΔHf(CaSO4 (s))]
[1(-542.83) + 1(-909.27)] - [1(-1434.5)] = -17.5999999999999 kJ
-17.60 kJ (exothermic)
And use this as ΔH° KJ mol1-
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Yes. I have told you before - the ΔH you should use is that of the equilibrium reaction under consideration.. In this case it is
CaSO4(s) ::equil:: Ca2+(aq) + SO42-(aq)
This is not the formation reaction of CaSO4, so ΔH°f(CaSO4) is not the thing to use.
Generally ΔH°reaction = ΣΔH°f(products) - ΣΔH°f(reactants).
Next time you have a van't Hoff equation problem, write down the equilibrium reaction and determine what ΔH you need to use.
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Dear mjc123,
Thank you very much for your guidance. However, the reaction ΔH°reaction = ΣΔH°f(products) - ΣΔH°f(reactants) you have given but the product is CaSO4 and reactants are Ca2+ & SO42- in saturated water.
when it is calculated it comes to 17.6 KJ/mol (Endothermic)??
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No, in the equilibrium for the dissolution of CaSO4, which is what Ksp refers to, CaSO4(s) is the reactant and Ca2+(aq) + SO42-(aq) are the products, as I wrote the equation above.
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Thank you very much dear for making me understand (sometimes even silly things!!!) the conceptual thing. :) :) :)
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How to calculate Lattice Enthalpy of Formation of individual element like Ca2+, Ba2+ etc...
Normally, that we use in enthalpy calculation. For an example we use -537.6 kj mol-1 for Ba2+. So how do we calculate this???
probably, the reaction is Ba(s)<----> Ba2+ + 2e-
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There is no such thing as lattice enthalpy of formation. Lattice energy and enthalpy of formation are different things.
-537.6 kJ/mol is the enthalpy of formation of Ba2+(aq). This is not calculated; it is derived from experiment. It can't be directly measured, but it can be derived from things that can be measured, by Hess's law, and some necessary conventions (e.g. that ΔHf° of H+(aq) ≡ 0).
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Thank you very much. One Lesson over!!!