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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?

  • $\begingroup$ Now that I think about it, does this have something to do with weak acid and bases? $\endgroup$ – Noaki Sato Mar 27 at 1:45
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    $\begingroup$ This is what happens to Na+: nothing. That's why it is not there in the equations. $\endgroup$ – Ivan Neretin Mar 27 at 7:49
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    $\begingroup$ 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. $\endgroup$ – 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.


Sodium ion is the weak conjugate acid of strong base i.e. NaOH. So the show which one ot of the two ions furnished undergoes hydrolysis to a greater extent the Na+(which does not react) is crossed and hydrolysis of stronger base C2H3O2- to produce OH- ions is shown.


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