# Why is sodium carbonate less soluble in water than sodium bicarbonate?

Why is sodium carbonate less soluble in water than sodium bicarbonate? If you think about their structures, the only thing that is different is that sodium carbonate has two sodium atoms, while sodium bicarbonate has one sodium atom and one hydrogen atom instead of the second sodium atom. Sodium and hydrogen have different electronegativity values, giving sodium bicarbonate polar character.

Does the difference in solubility have to do with the dipole moment of the molecules? If so, how can it be explained in more detail?

• May 8, 2017 at 20:27
• I suspect that it's a matter of the carbonate ion having a greater formal charge, causing it to be better solvated than the singly charged bicarbonate ion. Keep in mind that if I were certain of this, I would have written a formal answer ;) May 8, 2017 at 23:20
• It isnt, as another answer has stated. To my knowledge this is only true for Mg and Ca (bi)carbonates. The trend in the alkali metals are oppostie: carbonates will be always more soluble than bicarbonates. Jun 22, 2020 at 8:16

Oops! The question contains an incorrect assumption. But chemists are amazing - they can still answer such a question:

Sodium carbonate is way more soluble than sodium bicarbonate at all temperatures above about freezing. $$Na2CO3$$ has several crystalline hydrates that are temperature sensitive and make the curve less smooth than most other salts. (I suppose the caffeine is in there to wake us up.)

So, I woke up to the fact that when we calculate the weight of "sodium carbonate" which dissolves at 0 C, it's the decahydrate, whereas sodium bicarbonate is anhydrous, so at 0 C, the solubility of sodium bicarbonate is 0.833 moles per liter, while the same weight of sodium carbonate decahydrate dissolved in a liter is only 0.245 moles.

Now I'm more puzzled by why sodium carbonate has such a complex hydrate system, and sodium bicarbonate doesn't.

Likely, this is due to hydrogen bonding. $\ce{NaHCO3}$ has one hydrogen bond donor site, but $\ce{Na2CO3}$ has none.

For an organic compound, one hydrogen bond for each four-five carbons may be enough to make it water soluble.

According to the Born–Landé equation, the lattice energy is proportional to Z+Z-. Therefore in $\ce{Na2CO3}$ the Z+Z- value is 2 where as in $\ce{NaHCO3}$ the value of Z+Z- value is one. So the lattice energy is higher in $\ce{Na2CO3}$ than $\ce{NaHCO3}$. In addition to that the size of bicarbonate ion is higher than carbonate. Therefore packing efficiency will be higher in sodium carbonate than sodium bicarbonate. Thus more energy will require to disrupt $\ce{Na2CO3}$ than $\ce{NaHCO3}$. Therefore $\ce{Na2CO3}$ is less soluble in water than $\ce{NaHCO3}$.