Why don't we use sodium bicarbonate to neutralise base spills?

According to this document on lab safety (<http://faculty.washington.edu/korshin/Class-486/AEESP-safety-notes.pdf>), it says to not use $\ce{NaHCO3}$ to neutralise specifically base spills, stating "Do not use acetic acid or
sodium bicarbonate to clean a base spill. The sodium bicarbonate will not neutralize the spill, and acetic
acid could react strongly with the base."


Could someone explain why sodium bicarbonate won't neutralise a base spill? I have always learnt that it is amphiprotic and therefore can be used to neutralise BOTH acid AND base spills.




Suppose we have a strong basic aqueous solution $B$ that we attempt to neutralize with $\ce{NaHCO3}$ working as an acid $A$.


The source of excess $\ce{OH-}$ ions comes from the base, and since it's strong, we can consider:


$$n\_{Bo}=n\_{\ce{OH-}}\implies C\_{Bo}{V\_B}=C\_{\ce{OH-}}V\_B$$


The source of excess $\ce{H+}$ ions is bicarbonate, but since it's a weak acid, we have to consider the equilibrium:


$$\ce{HCO3-(aq)<=>H+(aq) + CO3^{2-}(aq)}$$


The equilibrium expression in terms of moles after mixing both solutions is:


$$K\_n=K\_a\;(V\_A+V\_B)=\frac{x^2}{C\_{Ao}V\_A-x}$$


For complete neutralization to take place, both excess $\ce{H+}$ and excess $\ce{OH-}$ ions need to completely react with each other, so:


$$x=C\_{Bo}V\_B$$


Substituting above:


$$K\_a\;(V\_A+V\_B)=\frac{(C\_{Bo}V\_B)^2}{C\_{Ao}V\_A-C\_{Bo}V\_B}$$


For simplicity, let's consider equal molar concentrations of both solutions and a volume of $\pu{0.1L}$ for our basic solution that needs neutralizing:


$$C\_{Ao}=C\_{Bo}=\pu{1mol/L}$$


$$V\_B=\pu{0.1L}$$


The acid dissociation constant of bicarbonate at 25°C is approximately:


$$K\_a=5.012\times10^{-11}$$


Substituting all values:


$$5.012\times10^{-11}\;(V\_A+0.1)=\frac{0.1^2}{V\_A-0.1}$$


Solving for $V\_A$:


$$V\_A=\pu{14125.2L}$$


In other words, bicarbonate is considerably weaker as an acid (about 400 times) than it is as a base, so attempting to neutralize a strong basic solution with it would require an enormous amount.


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I am going to argue that the currently-accepted answer is incorrect.


As the current accepted answer correctly pointed out, the bicarbonate ion dissociates into a proton and a carbonate ion, and that an equilibrium between the bicarbonate ion and the carbonate and hydrogen ion is established.


However, Sam202's calculation assumes that there is no driving force that would shift the equilibrium towards one side or the other. In the case of an acid/base reaction, this is simply not true. Following from [Le Chatelier's principle](https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle), the equilibrium can be shifted to favor the products by removing products as they are formed. In the case of sodium bicarbonate reacting with sodium hydroxide, hydrogen ions formed by the dissociation of bicarbonate are removed by reaction with hydroxide ions to form water, causing the equilibrium to shift to favor greater dissociation of bicarbonate, driving the reaction to completion.


As a sidenote, acid/base titrations assume that the reaction between the acid and base is stoichiometric. If Sam202's answer were correct, titrating a weak acid with a strong base or vice-versa would be impractical.


As to the question at hand, the reaction of sodium bicarbonate with sodium hydroxide would produce water and sodium carbonate as pointed out by commenter Poutnik. While sodium carbonate is a weaker base compared to sodium hydroxide (pKa of sodium carbonate is 10.33 compared to 15.7 for sodium hydroxide, according to Wikipedia) it is still basic (For reference, pKa of ammonia is 9.25, pKa of sodium bicarbonate is 6.3). Thus, the reaction of sodium bicarbonate with sodium hydroxide would still leave a base spill, as Poutnik correctly deduced.


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