# Hydrolysis of sodium carbonate [closed]

This is the equation given by my textbook for hydrolysis of sodium carbonate:

$$\ce{Na2CO3 + 2 H2O -> H2CO3 + 2 Na+ + 2 OH-}$$

and it mentions that sodium ion $$(\ce{Na+})$$ does not tend to combine with the hydroxide ion $$(\ce{OH-})$$ and I was wondering what prevents them from combining together to form $$\ce{NaOH}.$$

• What do you mean by "combine", exactly? A precipitation? If so, you might want to look up solubility of NaOH in water and read about solvation. Dec 7 '19 at 15:03
• @andselisk i mean what prevents them from reacting together to form NaOH Dec 7 '19 at 15:06
• You didn't answer the question. There is NaOH all right, it's dissolved in water. Also note that NaOH is a formula unit for ionic solid, so there is no reason to expect NaOH molecules, if that's what you mean. Dec 7 '19 at 15:11
• Additionally, the textbook is wrong, as carbonate hydrolysis to bicarbonate. $$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$ Carbonic acid is formed just in traces $$\ce{HCO3- + H2O <=> H2CO3 + OH-}$$, as the latter reaction is in alkalic environment pushed strongly toward the left side. Dec 7 '19 at 15:29

Apparently you believe that $$\ce{NaOH}$$ may be an independent molecule. It is not in this case.
$$\ce{NaOH}$$ does not exist in a solution. $$\ce{NaOH}$$ does not exist in the solid state either. $$\ce{NaOH}$$ does not exist in the solid state as a molecular compound. In the solid state, it is made of a huge pile of $$\ce{Na+}$$ and $$\ce{OH-}$$ ions, exactly like a salt grain is made of a pile of $$\ce{Na+}$$ and $$\ce{Cl-}$$ ions. On the average, there is the same amount of $$\ce{Na+}$$ and $$\ce{OH-}$$ ions. But $$\ce{NaOH}$$ does not exist as a separate molecule.