I know sodium metal reacts violently with water, but what about aqueous solutions (like sodium bicarbonate). Also, what would the reaction kinetics be?

I have watched YouTube videos were a small piece of elementary sodium reacts completely with water in a matter of seconds. Would that also be the case of elementary sodium in aqueous solutions (for example sodium bicarbonate)?

  • 1
    $\begingroup$ Yes, it would be pretty much the same as with water. $\endgroup$ Aug 13, 2016 at 6:35
  • $\begingroup$ what its the solution is very concentrated (like 6M sodium bicarbonate) would it still behave like it will with water? $\endgroup$
    – user510
    Aug 13, 2016 at 15:25
  • 1
    $\begingroup$ You can't make sodium bicarbonate 6M. Water simply wouldn't dissolve that much. $\endgroup$ Aug 13, 2016 at 18:21
  • $\begingroup$ This video shows what happened in 1947 when the U.S. Army dumped 20 thousand pounds of sodium (in 3500 pound containers) into frozen alkaline Lake Lenore, WA, USA: youtube.com/watch?v=HY7mTCMvpEM . Not sure how alkaline the lake was prior to the sodium addition. $\endgroup$
    – Ed V
    Aug 7, 2019 at 20:20

1 Answer 1


It helps to know what the reaction actually is. When sodium metal reacts with water, it is oxidised and the hydrogen in water is reduced according to the following equation:

$$\ce{2Na (s) + 2 H2O (l) -> 2 Na+ (aq) + 2 OH- (aq) + H2 ^ (g)}\tag{1}$$

Note that no other components play a role; the only thing that does is the $\mathrm{pH}$ value of the water (lower $\mathrm{pH}$ would mean a more violent reaction) — this is a spoiler and later shown in equation $(8)$. Since this is a redox reaction, we can use standard half-cell theory to analyse it. Most importantly, thanks to equation $(2)$, only the effective cell potential influences $\Delta G$ ($z$ is the number of electrons transferred while $F$ is Faraday’s constant) which in turn can be separated into two half-cells according to equation $(3)$ which can then be analysed by the Nernst equation $(4)$.

$$\Delta G_\text{cell} = - n_{\ce{e-}}FE_\text{cell}\tag{2}$$ $$E_\text{cell} = E_\text{cathode} - E_\text{anode}\tag{3}$$ $$E = E^0 + \frac{RT}{zF} \ln \frac{a_\text{Ox}}{a_\text{red}} = E^0 + \frac{0.059~\mathrm{V}}{z} \lg \frac{a_\text{Ox}}{a_\text{Red}}\tag{4}$$

The half-cell reactions are shown in equations $(5)$ (oxidation) and $(6)$ (reduction) respectively, along with their standard potentials $E^0$ (under standard conditions, i.e. most importantly $\mathrm{pH\ 0}$).

$$\ce{Na+ + e- <=> Na (s)} \qquad E^0 = -2.71~\mathrm{V}\tag{5}$$ $$\ce{2 H+ + 2 e- <=> H2 (g)} \qquad E^0 = 0.000~\mathrm{V}\tag{6}$$

Let’s take a look at the anodic oxidation first. The initial concentration of sodium ions could be, say, $3~\mathrm{M}$. Remember that the activity is defined as $1$ for solids. With the assumption that $a \approx c$, that would mean our oxidation potential is $-2.68~\mathrm{V}$ as shown in equation $(7)$.

$$E_\mathrm{anode} = -2.71~\mathrm{V} + \frac{0.059~\mathrm{V}}{1} \lg \frac{3}{1} = -2.71~\mathrm{V} + 2.82 \cdot 10^{-2}~\mathrm{V} = -2.68~\mathrm{V}\tag{7}$$

The corresponding calculation for the reductive side, equation $(8)$, is a little harder since we have no real starting value for the partial pressure (and thus activity) of hydrogen gas.

$$E_\text{cathode} = \frac{0.059~\mathrm{V}}{2} \lg \frac{a(\ce{H3O+})}{p(\ce{H2})} = -0.0295~\mathrm{V} \times \mathrm{pH} - 0.0295~\mathrm{V} \times \lg p(\ce{H2})\tag{8}$$

But we do know that a reaction will happen if $E_\text{cell} > 0~\mathrm{V}$ and we have a good idea of the anode’s potential. Thus, if we want no reaction we must set the $E_\text{cell} < 0$ (equation $(9)$):

$$\begin{align}0 &> E_\text{cathode} - E_\text{anode}\\ - 2.68~\mathrm{V} &> E_\text{cathode}\\ - 2.68~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH} - 0.0295~\mathrm{V} \times \lg p(\ce{H2})\end{align}\tag{9}$$

The second summand is negative if $p(\ce{H2}) > 1~\mathrm{bar}$. (The equation itself requires a dimensionless pressure which is typically defined to have a the same value as if it were in $\mathrm{bar}$). However, that basically requires performing the reaction under a hydrogen atmosphere. This setup is rather hard and not part of the question. Instead, let’s take the natural concentration of hydrogen which is $1~\mathrm{ppm}$ according to Wikipedia. That results in $0.177~\mathrm{V}$ for the second summand (equation $(10)$).

$$-0.0295~\mathrm{V} \times \lg 10^{-6} = (-6) \times -0.0295~\mathrm{V} = 0.177~\mathrm{V}\tag{10}$$

At least we didn’t lose too much. But there is still the first summand, which is also negative by sign. We just need a high enough $\mathrm{pH}$ value to compensate. The calculation is in equation $(11)$.

$$\begin{align}-2.68~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH} + 0.177~\mathrm{V}\\ -2.86~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH}\\ 96.9 &< \mathrm{pH}\end{align}\tag{11}$$

So we would need a $\mathrm{pH}$ value of almost $97$ to prevent the reaction from happening due to too basic conditions. Of course, this value is unobtainable in water or practically any solvent known to us. Even tert-butyllithium, one of the most basic species known has a $\mathrm{p}K_\mathrm{a}$ value of only $\approx 50$.

You could now try to argue that you would just need to increase the hydrogen gas partial pressure appropriately. So how does the value change if the reaction is performed under hydrogen atmosphere? See equation $(12)$ for the answer.

$$\begin{align}-2.86~\mathrm{V} &> -0.0295~\mathrm{V} \mathrm{pH}\\ 90.8 &< \mathrm{pH}\end{align}\tag{12}$$

As you may or may not have guessed, a logarithmic difference of $6$ for the partial pressure results in a lowering of the $\mathrm{pH}$ by $6$ units. So to arrive at achieveable $\mathrm{pH}$ values, your reaction would need to happen at a hydrogen pressure of $10^{75}~\mathrm{bar}$ — go figure.

All things considered, the reaction is so violently endergonic that there is basically no way to inhibit it whatsoever. If metallic sodium sees a sufficient amount of water, it will start reacting, end of discussion.


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