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KHCO3 can't be prepared by solvay's preocess as it is soluble and can't be separated out. However the same can be done with NaHCO3 becoz it is insoluble. However according to lattice energy effects, a larger HCO3- anion can be more easily soluble with Na+ than with K+ becoz K+ is a larger cation...


How to solve this contradiction.   

3 years ago

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Answers : (7)

                                        

Hi Swapnil,


 


Ok. What you are talking of is Fahjans Rule.


That a bigger cation has more of a covalent character when compared to that with a smaller cation.


Well that is perfectly right. The covalent character of an ionic compound.... higher for KHCO3 than for NaHCO3 (perfect)


 


But what are we looking at - Solubility.


So we are in effect looking at hydration enthalpies of NaHCO3 and KHCO3 in an aqueous solution, along with the Bond Dissociation energy.


 


Solubility proportional to (|Hydration enthalpy| - |BDE|)


Now you can answer this question i believe.


Which among KHCO3 and NaHCO3 will have a higher bracket term value?


 


Hint: Higher that value in the bracket, higher the solubility.


 


Best Regards,


Ashwin (IIT Madras).

3 years ago
                                        

what is bracket term value?

3 years ago
                                        

Hi,


 


The bracket term value is the value of the expression in the bracket which i had mentioned above.


That is (|Hydration Enthalpy| - |Bond Dissociation Energy|).


 


Regards,


Ashwin (IIT Madras).

3 years ago
                                        

Well Im not talking about fazans rule.


Though if i apply fazans rule, Na2CO3 is more covalent than K2CO3 (Smaller cation has a larger polarising power and than larger one) . So , I dont understand ur explantaion of fazans rule.


Im taking about Larger cation Larger anion lattice energy effects.According to this rule,


When anion is large, lattice energy remain almost constant (relatively small decrease on moving down the group ) throughout the group however hydration enthalpy decreases on going down. As such negative enthalpy of dissolution should decrease on going down the group. As such the solubitlity of KHCO3 should be less than NaHCO3. But opposite is observed.


My explantion is also based upon enthalpies not fazans rule. 

3 years ago
                                        

Hi Swapnil,


 


Extremely sorry.


Fahjans rule states exactly the opposite as you have correctly mentioned.


NaHCO3 is more covalent than KHCO3. (So that would be a direct implication of solubility). Still it is not a correct way of looking at solubility.


 


NaHCO3 definitely has a higher Hyd Enthalpy.


But BDE of NaHCO3 is also very high compared to that of KHCO3, in aqueous solution - (An application of Fahjans rule, more covalent character for NaHCO3)


So (|Hyd Ent| - |BDE|) is less for NaHCO3 as compared to that of KHCO3.


 


So that explains the trend.


 


Regards,


Ashwin (IIT Madras)

3 years ago
                                        Hi Ashwin Sir, 


Ive recently came across an article that defines that due to strong hydrogen bonding in bycarbonates lead to the existence of NaHCO3 in forms of infinite-chain like structure and KHCO3 as dimeric structure due to weak hydrogen bonding due to largen potassium ions.

Do You think this can resolve the deadlock ??? Can u explain how
3 years ago
                                        

Hi Swapnil,


 


Ok. The statement which you have made is new for me. Thanks for that.


The statement which you have made can be very well logically explained.


 


H-Bonding exists for atoms that have high electronegativity. Like that of F,O,N,S.


So H-Bonding (in this example) will be there in HCO3- ion. the presence of the negative charge makes H-Bonding more stronger. So firstly the HCO3- will exist as a long chain with links.


Now we have K+ and Na+. Na+ being much smaller than K+ can fill in the long link of Hydrogen bonded bi-carbonate.


But K+ being bigger in size will break the link to form the ionic bond and can exist only as a dimer.


 


So now that the BDE of NaHCO3 is again more than that for KHCO3. So again that will be a factor (along with its covalent nature). Thus making NaHCO3 practically insoluble.


 


Thanks & Regards,


Ashwin (IIT Madras).

3 years ago

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