Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
I realized that the $\ce{-OH}$ group when it reacts with some metal like $\ce{Na}$ or $\ce{K}$ will replace the $\ce{H}$ element to form $\ce{-OK}$ or $\ce{-ONa}$:
\begin{align} \ce{H2O + Na &-> NaOH + H2}\\ \ce{CH3OH + Na &-> CH3ONa + H2}\\ \ce{H2O + K &-> KOH + H2}\\ \end{align}
So I am just wondering is that always true? Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$? Does $\ce{NaOH + Na}$ form $\ce{Na2O}$?
The question: Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$?
This is a redox reaction regardless of what species $\ce{K}$ would react with. The answer is two-fold: (a) Having potassium added to an aqueous solution of sodium hydroxide; and (b) Having potassium added to molten sodium hydroxide, therefore no water present (or trace amount).
Answer to situation (a) is potassium would essentially react with water (other ions in the solution are $\ce{Na+}$ and $\ce{OH-}$) based on standard potentials of following possible half-reactions.
$$ \begin{align} \ce{K+ + e- &<=> K} &E^\circ &=\pu{-2.931V}&\\ \ce{Na+ + e- &<=> Na} &E^\circ &= \pu{-2.71V}&\\ \ce{2H2O + 2e- &<=> H2 + 2OH-} &E^\circ &=\pu{-0.8277V}&\\ \end{align}$$
Therefore, the only ionic species in the solution would be $(\ce{Na+ + K+ + OH-})$ in $\ce{H2O}$.
Answer to situation (b) is a little complicated than this. Nowadays, in electrochemical research, use of molten $\ce{NaOH}$ is the choice of the electrolyte. The choice of molten $\ce{NaOH}$ over molten $\ce{Na2CO3}$ is based on a number of advantages of the molten hydroxide electrolyte over others. The first advantage is its higher electrical (ionic) conductivity, which is important in the question at hand. The ionic conductivity of molten $\ce{NaOH}$ exceeds that of molten sodium carbonate by a factor of 1.5 [1]. Thus, it is safe to assume that molten $\ce{NaOH}$ functions as $\ce{Na+}$ and $\ce{OH-}$ ions in molten state. Additionally, $\ce{Ni}$ has already been shown to be a chemically stable material in a molten $\ce{NaOH}$ environment, and hence, nickel containers are used to produce anhydrous $\ce{NaOH}$. Concentrating an aqueous $\ce{NaOH}$ solution in $\ce{Ni}$ containers by heating up to $\pu{500 ^\circ C}$ will produce the desired $\ce{NaOH}$ melt.
According to [1], molecular oxygen is extremely oxidizing in fused hydroxides and reacts with hydroxide ions to produce superoxide and peroxide ions. The concentration of these ions is determined by the following equilibria (Note also that increasing the water vapor pressure would reduce the peroxide and superoxide ion content by favoring the equilibria to LHS, but it's not our concern here):
$$ \begin{align} \ce{O2 + 4OH- &<=> 2O_2^{2-} &+& 2H2O }\\ \ce{3O2 + 4OH- &<=> 4O_2^{-} &+& 2H2O }\\ \ce{2OH- &<=> 4O^{2-} &+& H2O} \end{align}$$
For example, see following Abstract:
A thermodynamical discussion of the $\ce{Pt/O2}$ electrode in molten $\ce{NaOH}$ shows that the redox potential depends on $p_\mathrm{O_2}$, $p_\mathrm{H_2O}$, $a_\mathrm{O_2^{2-}}$ and $a_\mathrm{O^{2-}}$. Potential measurements have shown that the reaction $\ce{O2 + 2e− = O2^{2−}}$ is potential-controlling. From the electrode materials investigated—$\ce{Pt}$, $\ce{Au}$, $\ce{Ag}$, $\ce{Ni}$ and $\ce{Fe}$—the last one only exhibits a positive potential shift by comparison to the other metals; this shift has been interpreted in terms of an $\ce{O2^{2-}}$ impoverishment following corrosion.
The peroxide which is required for the corrosion reactions can be formed only by chemical reaction of the melt with $\ce{O2}$ , not, however, by anodic oxidation, because peroxide oxidation essentially precedes the oxidation of hydroxyl ions yielding oxygen and water vapour. With anodic and cathodic polarization, peroxide is reacted with limiting current flow, the anodic reaction being less inhibited. Water vapour is cathodically reduced to $\ce{H2}$ and $\ce{OH-}$ —with limiting current flow—before $\ce{NaOH}$ is reduced. $\ce{Pt}$ corrosion is anodically enhanced by water vapour; in this context, current fluctuations are observed. In the pure $\ce{NaOH}$ melt the cathidic reduction may be assumed to involve formation of $\ce{H2}$ and $\ce{Na2O}$ ; not, however, $\ce{Na}$ deposition. [2]
Thus, in a molten hydroxide, one can expect ionic species such as metal cation ($\ce{Na+}$ in this case), $\ce{OH-}$, $\ce{O^{2-}}$, $\ce{O_2^{2-}}$, and $\ce{O_2^{-}}$ anions, and molecular $\ce{O2}$ present in this so-called basic melt in equilibria. Thus, when $\ce{K}$ reacts with any of these (most probably with $\ce{H2O}$ and $\ce{OH-}$ based on their concentrations at equilibrium) to oxidize to $\ce{K+}$, it will stay as mixture of $\ce{KOH}$, $\ce{K2O}$, $\ce{K2O2}$, and $\ce{KO2}$.
Example for ionic nature in molten salt solution molten $\ce{NaCl}$:
References:
Direct electrochemical power generation from carbon in fuel cells with molten sodium hydroxide: Chemical Engineering Communications, 2005, 192 (12), 1655-1670. DOI: 10.1080/009864490896241.
Elektrochemische messungen in NaOH-schmelzen: Electrochimica Acta, 1968, 13 (3), 625-643. DOI: 10.1016/0013-4686(68)87031-8.
The last question (Does $\ce{NaOH + Na}$ form $\ce{Na2O}$?) is answered in Wikipedia: yes. Similarly, $\ce{KOH + molten K}$ gives $\ce{K2O}$ (Wikipedia again).
The first question (Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$?) suggests two products, but the real question is whether any reaction will occur. If a reaction occurs, either hydrogen will be released or sodium metal will be released.
If the reaction proceeds by hydrogen evolution, the product will be $\ce{NaOK}$ or what is essentially a mixture of $\ce{Na2O + K2O}$.
If sodium metal is released, it will be in contact with some of the original $\ce{NaOH}$, which has been demonstrated to react to form $\ce{Na2O}$. If the reaction proceeds in this manner, at the halfway point it will be: $\ce{2 K + 2 NaOH -> K + KOH + NaOH + Na}$. Then the $\ce{Na}$ can react with $\ce{NaOH}$ to form $\ce{Na2O}$ and the $\ce{K}$ can react with the $\ce{KOH}$ to form $\ce{K2O}$, giving a mixture of $\ce{Na2O + K2O}$, or what could be represented as $\ce{NaOK}$. (Interestingly, at the halfway point, you have $\ce{NaK}$ alloy, which is extremely reactive!)
The essential reaction is elimination of hydrogen, which is not energetically favored, but which drives the reaction by eliminating the reverse reaction ($\ce{H2 + Na2O -> 2 NaOH}$ or $\ce{H2 + K2O -> 2 KOH}$). From data in the CRC Handbook, for the reaction: $\ce{2 NaOH + 2 K -> Na2O + K2O +H2}$, the heats are $2 \times (-102) + 0 < -99 -86 + 0$, so entropy of mixing $\ce{Na2O}$ or $\ce{K2O}$ must be factored in, or just consider that removing hydrogen, however slow, will drive the reaction to the mixture of oxides.
The comment about using molten potassium to react with $\ce{KOH}$ (and presumably liquid sodium to react with $\ce{NaOH}$) suggests that simply pushing a solid hydroxide against a solid metal might not give a rapid reaction. So, a little heat might be needed, but the reaction will almost certainly give the mixed oxide.
Two good references to address this question are:
DETERMINATION OF THE SOLUBILITY OF OXYGEN BEARING IMPURITIES IN SODIUM, POTASSIUM AND THEIR ALLOYS J. Phys. Chem., 1959, 63(1), pp 68–70;
and
US Patent 1872611 Process of making potassium metal or sodium-potassium metal alloy
According to the journal article:
At $\ce{400 ^\circ C}$, under vacuum distillation conditions:
$\ce{NaOH + Na -> Na2O + NaH}$
$\ce{ 2NaH -> 2Na + H2}$
It is also stated:
The equilibrium oxide in a sodium-potassium ($\ce{NaK}$)-oxygen system is sodium monoxide.
According to the patent:
Sodium metal can be reacted with $\ce{KOH}$ to yield a $\ce{NaK}$ alloy plus a mixture of $\ce{NaOH}$ and $\ce{KOK}$. This is presented as a cost-effective way to make the more desirable potassium metal from the less desirable sodium metal.
That replacement takes place when oxygen in the hydroxyl group does not have any charge, but in a covalent compound.
Take ethanol for instance. The hydroxyl hydrogen will be replaced by $\ce{Na}$ or $\ce{K}$, if you add corresponding metals on pure ethanol. Same through for water, at first, oxygen has two covalent bonds with two hydrogens. Alkali metal comes in, oxygen gets one electron from its outer shell and displaces the hydrogen homolytically. Then there will be a negatively charged hydroxyl group left and plus charged alkali metal, they will have an ionic association in solution. Say it $\ce{M-OH}$, this will be $\ce{M+ + OH-}$ in solution, and adding another alkali metal will not have any effect on them, $\ce{OH-}$ is quite stable in this condition, likewise $\ce{M+}$.
This is inherently a redox reaction, the word "replacement" instead is confusing. In short, I expect sodium displacement by potassium, not another oxygen reduction. Potassium ionic radii ($\pu{280 pm}$) and sodium ionic radii ($\pu{227 pm}$) seem quite different. It makes unlikely to see its solid as $\ce{NaKO}$ or $\ce{NaOK}$, and there would be no such product in liquid reaction if lattice energy does not stabilize possible end product.
Having both sodium hydroxide and potassium in aqueous solution would result in the following reaction:
$$ \begin{align} \ce{NaOH + H2O &-> Na+ + OH- + H2O}\\ \ce{K + H2O + OH- &-> K+ + 2OH- + H2} \end{align} $$
Essentially, the potassium is going to react with the water long before anything else. The only way I can think of to get your reaction to go forward would be to have the molten forms of $\ce{NaOH}$ and $\ce{K}$, which would react like this:
$$\ce{NaOH + K -> KOH + Na}$$
$\ce{NaOH}$ reacts with $\ce{K}$ to form $\ce{KOH}$ there is no compound which is called as NaOK
Does $\ce{NaOH + Na}$ form $\ce{Na2O}$? Yes of course
$\ce{NaOH + Na -> Na2O + ½ H2}$
$\ce{Na2O}$ is Sodium oxide, which is like anhydrous form of $\ce{NaOH}$. When you add water to Sodium oxide it will give you back $\ce{NaOH}$.
Note - Answer edited completely
KOH because K is highly reactive in the metal reactive series and metals don't react with metals as both are elctropostive
