Calcium reactor and pH

In a $400$ liter aquarium, whose bicarbonate level is kept constant, it is initially equal to $122\space\text{mg/L}$ and you want to raise the $\text{pH}$ from $8.3$ to $8.48$. For this, a calcium reactor can be used. This equipment uses a media composed predominantly of calcium carbonate and a carbon dioxide stream confined in a cylinder subjected to a pressure of $4.1\space\text{atm}$. Carbon dioxide serves to create a slightly acidic $\text{pH}$, dissolving the calcareous media and producing an aqueous solution of calcium bicarbonate, which is returned to the aquarium. In order to avoid sudden changes in the level of bicarbonate in the system, the aquarist prefers to act within $10$ days, in which the calcium reactor must remain on for $10$ hours a day. Assuming that the partial pressure and solubility of $\text{CO}_{2\text{(g)}}$ will not be changed during the process, give the approximate value of the correct flow of $\text{CO}_{2\text{(g)}}$ over the reactor. Given: $\text{pKa}_1=6, \text{pKa}_2=10, \log2=0.3, \log3=0.48$

Initially, I did

$pH=pKa+\log\frac{[\text{salt}]}{\text{[acid]}}$

So, in $pH=8.3$, we have

$\frac{[CO_3^{2-}]}{[HCO_3^-]}=0.02\\frac{[HCO_3^-]}{[H_2CO_3]}=200$

In $pH=8.48$,

$\frac{[CO_3^{2-}]}{[HCO_3^-]}=0.03\\frac{[HCO_3^-]}{[H_2CO_3]}=300$

The reaction in reactor I think that is

$\ce{CaCO3 + H2O + CO2 <=> Ca^2+ + 2HCO3-}$

But now I don't know how to proceed to find this $CO_2$ flow. Can someone help me?

The answer is ${0,09\space\text{L/h}}$