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).