Reaction of aqueous sodium carbonate with aluminum foil

Aqueous sodium hydroxide is known to react with aluminum foil in [an exothermic, hydrogen-gas releasing reaction like](https://en.wikipedia.org/wiki/Sodium_aluminate#Reaction_of_aluminium_metal_and_alkali) $$\ce{2NaOH(aq) + 2Al(s) + 2H2O(\ell) → 2NaAlO2(aq) + 3H2(g)}$$


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 $$\ce{Na2CO3(aq) + H2O(\ell) \rightleftharpoons NaHCO3(aq) + NaOH(aq) }$$ 




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](https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation?wprov=sfla1)


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


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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