MY CART (5)

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price:

There are no items in this cart.
Continue Shopping
                   

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

Share

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

Post Your Answer

More Questions On Inorganic Chemistry

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details
which of the following scientist is not associated with the periodic table? newlands b.rutherford c.lother meyer d.prout’s
 
 
Not only Newland is associated. We have lother meyer curve which shows that atomic volume = atomic mass/density of elements. This is also a contribution to the periodic table as we can study...
 
Naga Himanshu one month ago
 
actually there is a not in the ques. newland is associated with the periodic table..
 
mansi dabriwal one month ago
 
in ques thr is a NOT ….
 
mansi dabriwal one month ago
Which one of the following is a covalent hydride? A.) BeH2 B.) AsH3
 
 
its BeH2 because of brillyium’s anamolus properties
 
akshat 6 days ago
 
BeH2 is covalent hydride as it form as dimer
 
Aditya Sharma 7 days ago
 
Yes..@akshat
 
Aditya Sharma 5 days ago
silver metal crystalline with a face centred cubic lattice the lenght of unit cell in found tobethe value 4.077x10 -8 m.ca;culate the atomic radius and density of silver {atomic mass of...
 
 
“silver metal crystalline with a face centred cubic lattice the lenght of unit cell in found tobethe value 4.077x10-8m.ca;culate the atomic radius and density of silver {atomic mass of...
  img
Sunil Kumar FP 8 months ago
 
we have density of the fcc=z*M/a^3*Na z=no of atom fro fcc it is 4 M= molecular mass Na=avagadro number a=edge length for face centered cubic unit cell=sqrt2...
  img
Sunil Kumar FP 8 months ago
what osmotic preaaure would be of 1.25m sucrose solution exhibit at 25C ? The density of th solution is 1.34g/ml.?
 
 
sir kindly give the solution
 
Shriya Mehrotra 6 months ago
 
28.6 atm
 
himateja 6 months ago
does ethanamide give bromoform as well as amine when reacted with Naobr?
 
 
ethanamide cannot give the reaction because of no free methyl group
 
akshat 5 days ago
 
please tell the type of amine 1 deg 2 deg or 3 deg
 
akshat 5 days ago
 
It cannot give Naobr.
 
Adarsh 3 days ago
How do the vanderwaals constants a, b vary for different gases? The value of a is higher for ethane than that for ammonia?
 
 
Hello Student Vander Waals constant for attraction (a) and volume (b) are characteristic for a given gas. Some salient feature of a & b are: 1. For a given gas Vander Waal’s...
  img
Pankaj 4 months ago
 
Extent of attraction and repulsion depends on the nature of molecule. So, different molecules have different values of a and b.
  img
Pankaj 22 days ago
 
a actually depends on te extent of forces of attraction b/w te molecules in a gas. these forces are more in ammonia (H-Bonds) tan in ethane.But the actual a value is more for etane.
 
Himaja 21 days ago
View all Questions »