# 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.

• NaOH + NaHCO3 -> Na2CO3 + H2O, with sodium carbonate being still caustic, even if less than NaOH. For neutralization of residues after water washing, it is adviced 2% NaHCO3 for acids and 2% acetic acid for bases, which are advised to kept handy as stock solutions. Dec 10, 2022 at 6:49

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.

• Sorry once again but could you explain how you got $K_n=K_a(V_a+V_b)$ thanks :) Dec 12, 2022 at 11:09
• $$K_a=\frac{C_C^c\;C_D^d}{C_A^a}=\frac{\left(\frac{n_C}{V}\right)^c\;\left(\frac{n_D}{V}\right)^d}{\left(\frac{n_A}{V}\right)^a}=\frac{n_C^c\;n_D^d}{n_A^a}\left(\frac{1}{V}\right)^{\Delta n}=\;K_n\left(\frac{1}{V}\right)^{\Delta n}$$ $$\Delta n=1$$ $$V=V_A+V_B$$ Dec 12, 2022 at 16:22
• Sorry I am really struggling with the notation, just note I'm still a highschool chemistry student. What is $\Delta n$ and what chemicals are subscript $C, D$ and $A$? Could you perhaps write working as $K_a=\frac{\ce{[H3O^+][NaCO3^-]}}{[\ce{NaHCO3}]}$ or something like that thank you! Dec 12, 2022 at 23:18
• I just represented the reaction: $\ce{HCO3- <=>H+ +CO3^{2-}}$ as: $$\ce{A<=>C + D}$$ $\Delta n$ is the difference between the sum of product molar coefficients and reactant molar coefficients of the reaction: $$\Delta n=c+d-a=1+1-1=1$$ Dec 12, 2022 at 23:33
• Thanks got it and finally, where does $K_n=\frac{x^2}{C_aoV_a-x}$ come from? What does $x$ represent and how did you derive the expression? (Sorry for all the questions but I feel like it'll be so beneficial for me to understand) Dec 13, 2022 at 1:58

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, 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.

• I would like to ask, for me it appears the current answer attempts to find how much bicarb is needed for the pH to be 7, whereas for the NaOH + NaHCO3 reaction, the salt will be basic, meaning, more bicarb will be needed for a neutral pH than just what is required to complete the reaction, hence the calculation. What are your thoughts on this, what calculations would you make for a neutral pH? Jan 1 at 5:35
• Upvoted. The pH of a typical baking soda solution is in the neighborhood of 8.3 or 8.4. Adding sodium hydroxide raises the pH: the pH has to be intermediate between that of the baking soda and NaOH. It cannot be 7 unless vast quantities of pH 7 water are used to dilute the baking soda and NaOH to ultra-trace levels.
– Ed V
Jan 1 at 16:28
• @Quippy What Ed said. You won't ever get to pH 7 by adding bicarb to sodium hydroxide, because bicarbonate is itself slightly alkaline. Jan 1 at 20:41