-1
$\begingroup$

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 $\ce{NaOH}.$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – andselisk Dec 7 '19 at 15:03
  • $\begingroup$ @andselisk i mean what prevents them from reacting together to form NaOH $\endgroup$ – AmirWG Dec 7 '19 at 15:06
  • 1
    $\begingroup$ 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. $\endgroup$ – andselisk Dec 7 '19 at 15:11
  • 3
    $\begingroup$ 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. $\endgroup$ – Poutnik Dec 7 '19 at 15:29
1
$\begingroup$

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.

When dissolved in water, the water molecules are going to separate the ions, which become independent from one another.

| improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.