# What happened to the Na+ ion? [closed] I was in the middle of my acid and base schools questions when I came to a stop at this question. I have posted the answer to tell you what I am confused about. My teacher did not explain this very well, and my friends are not sure either. In the picture I posted, the teacher cancels out the Na from the chemical compound NaC2H2O2. I'm not sure why he did this, and then he proceeded to solve the problem via a separate equation using a hydrolysis reaction with H2O and C2H3O2-. Could someone explain to me why this happens, and would this happen with other acid and bases?

• Now that I think about it, does this have something to do with weak acid and bases? – Noaki Sato Mar 27 at 1:45
• This is what happens to Na+: nothing. That's why it is not there in the equations. – Ivan Neretin Mar 27 at 7:49
• Na+ does not participate in the reaction, i.e. it is a spectator ion. So, you can ignore it for the acid base part of calculation. – S R Maiti Mar 28 at 20:29

The $$\ce{Na+}$$ ion is a spectator ion in this case. It doesn't participate in the chemical reaction. It is found unchanged on both the reagent and product sides (red top-right and green bottom-right reaction formulas), so it can be canceled out to get the net reaction (red left). From there, he did the usual acid-base calculation to get a formula for $$\ce{Kb}$$. If you actually know $$\ce{Kb}$$, you can then find $$x$$, the $$\ce{OH-}$$ concentration, and from there, the $$\ce{pH}$$. Or, if you know the $$\ce{pH}$$ for this reactant concentration, then you can calculate $$\ce{Kb}$$.
The reason $$\ce{Na+}$$ doesn't participate, is as Aditya wrote: it is the conjugate acid of a strong base ($$\ce{NaOH}$$), so it is a weak acid (weak enough to be ignored). It can't abstract a proton from water, and has no protons of its own to donate. At most, it can combine with $$\ce{OH-}$$ to form $$\ce{NaOH}$$, but $$\ce{NaOH}$$ is very soluble, so it will break back apart right away. In solution, $$\ce{NaOH}$$ is just separate $$\ce{Na+}$$ and $$\ce{OH-}$$ ions.