Reaction of aqueous sodium carbonate with aluminum foil
It is rather:

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

$$\ce{CO3^2- + H2O &lt;=&gt; HCO3- + OH-}$$

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

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

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

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