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Aqueous sodium hydroxide is known to react with aluminum foil in an exothermic, hydrogen-gas releasing reaction like $$\ce{2NaOH(aq) + 2Al(s) + 2H2O(\ell) → 2NaAlO2(aq) + 3H2(g)}$$

However, on youtube, it looks like someone was able to substitute NaOH with Na2CO3 and get a vigorous similar-looking reaction as well: https://www.youtube.com/watch?v=ggPWdwcsyqM

Which makes me wonder — did they also form $\ce{NaAl(CO3)2}$ as well? E.g., via $$\ce{2 Na2CO3(aq) + 4Al(s) + 6H2O(\ell) -> NaAl(CO3)2(aq) + 3NaAlO2(aq) + 6H2(g)}$$

Or was sodium carbonate merely used as a source of NaOH, from $$\ce{Na2CO3(aq) + H2O(\ell) \rightleftharpoons NaHCO3(aq) + NaOH(aq) }$$

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    $\begingroup$ I don't think there is something like NaAl(CO3)2. The closest I can get is dihydroxy aluminium sodium carbonate. pubchem.ncbi.nlm.nih.gov/compound/… $\endgroup$
    – Nilay Ghosh
    Commented Sep 18, 2019 at 7:03
  • $\begingroup$ @Nilay Ghosh Indeed, mostly likely something like "sodium aluminum carbonate" would probably be rather unstable as well, especially in aqueous solution. Even the Wikipedia article en.wikipedia.org/wiki/Aluminium_carbonate is pretty doubtful of the existence of any (non-hydroxylated) "aluminum carbonates" too... $\endgroup$
    – ManRow
    Commented Sep 20, 2019 at 13:06

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It is rather:

$$\ce{2 Al + 6 H2O + 2 OH- -> 2 [Al(OH)4]^- + 3 H2 ^}$$where $\ce{OH-}$ comes either from hydroxide dissociation, either from carbonate hydrolysis.

$$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$

The reaction with carbonate would gradually slow down as the carbonate/bicarbonate buffer will kick in.

Initial $\mathrm{pH}$ is ( see notes below line ):

$$\mathrm{pH_{init}}=14 - 0.5 ( \mathrm{p} K_\mathrm{b2} - \log{ [ \ce{Na2CO3} ]} )\\= 7 + 0.5 ( \mathrm{p} K_\mathrm{a2} + \log{ [ \ce{Na2CO3} ]} ) $$

because $\mathrm{p} K_\mathrm{a2} + \mathrm{p} K_\mathrm{b2} = \mathrm{p} K_\mathrm{w} = 14 $ ( at $\mathrm{ 25\ ^{°}C}$ )

As $ \mathrm{p} K_\mathrm{a2} = 10.329$:

$$\mathrm{pH_{init}}=12.17 + 0.5 \cdot \log{ [ \ce{Na2CO3} ]} $$

..and when some $\ce{OH-}$ is spent and $\ce{HCO3^-}$ is formed:

$$\mathrm{pH}=10.329 + \log \frac {[ \ce{CO3^2-} ]} {[ \ce{HCO3^-} ]}$$

From Henderson–Hasselbalch equation

$$\ce{pH} = \ce{p}K_\ce{a} + \log \left( \frac{[\ce{A^-}]}{[\ce{HA}]} \right) $$

If there is supposed all $[\ce{H+}]$ comes from acid dissociation and majority of acid is not dissociated, then :

$$\ce{pH} \overset{cca}= \ce{p}K_\ce{a} - \ce{pH} - \log {[\ce{HA}]_\ce{total}} $$

$$\ce{pH} \overset{cca}= \frac12 ( \ce{p}K_\ce{a} - \log {[\ce{HA}]_\ce{total}} )$$

Alternatively

$$\ce{pOH} = \ce{p}K_\ce{b} + \log \left( \frac{[\ce{BH+}]}{[\ce{B}]} \right) \\ \overset{cca}= \ce{p}K_\ce{b} - \ce{pOH} - \log {[\ce{B}]_\ce{total}} $$

Then

$$\ce{pOH} \overset{cca}= \frac12 ( \ce{p}K_\ce{b} - \log {[\ce{B}]_\ce{total}} )$$

$$14 - \ce{pH} \overset{cca}= \frac12 ( 14 - \ce{p}K_\ce{a} - \log {[\ce{B}]_\ce{total}} )$$

$$\ce{pH} \overset{cca}= 7 + \frac12 ( \ce{p}K_\ce{a} + \log {[\ce{B}]_\ce{total}} )$$

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  • $\begingroup$ How to derive the following equation : $$\mathrm{pH_{init}}=7 + 0.5 ( \mathrm{p} K_\mathrm{a2} + \log{ [ \ce{Na2CO3} ]} )$$ $\endgroup$
    – Adnan AL-Amleh
    Commented Sep 19, 2019 at 3:50
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    $\begingroup$ @Adnan AL-Amleh From analogous equation pH=0.5(pKa - log c ) for solution of a weak acid, that is derived with a simplification from definition of an acid dissociation constant. All is the basic stuff from chemical textbooks. $\endgroup$
    – Poutnik
    Commented Sep 19, 2019 at 3:58
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    $\begingroup$ @Adnan AL-Amleh Perhaps you are looking for en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation ? $\endgroup$
    – ManRow
    Commented Sep 19, 2019 at 23:20

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