Question

Question #d9f07

Answer 1

`NaOH (aq) + CO_2 (g) -> NaHCO_3`

Explanation:

The reaction depends on the concentration of the alkali solution `NaOH`).

When `NaOH` is concentrated, pH > 10, the reaction produces sodium hydrogen carbonate (`NaHCO_3)`:

`NaOH (aq) + CO_2 (g) -> NaHCO_3`

Answer 2

`OH^-)(aq) + CO_2(g) = HCO_3^-)(aq)`

`CO_2(aq) + 2OH^-)(aq) = CO_3^(2-)(aq) + H_2O(l)`

Explanation:

When pH is very high or of a very basic nature

= `NaOH + CO_2 = NaHCO_3`

The ionic reaction is

First write the ions formed

`Na^+(aq) + OH^-)(aq) + CO_2(g) = Na^+(aq) + HCO_3^-)(aq)`

Where g = gaseous
aq = aqueous

Cut out the same ions

`cancel(Na^+(aq)) + OH^-)(aq) + CO_2(g) = cancel(Na^+(aq)) + HCO_3^-)(aq)`

= `OH^-)(aq) + CO_2(g) = HCO_3^-)(aq)`

This is how we write ionic reactions

Another reaction can be also formed when pH is low or when its more acidic is of a good strength is

`2NaOH + CO_2 = Na_2CO_3 + H_2O`

`2Na^+(aq) +2OH^-)+ CO_2(aq) = 2Na^+(aq) + CO_3^(2-)(aq) + H_2O(l)`

Cut out the same ions

`cancel(2Na^+(aq)) + 2OH^-)(aq) + CO_2(aq) = cancel(2Na^+(aq)) + CO_3^(2-)(aq) + H_2O(l)`

`CO_2(aq) + 2OH^-)(aq) = CO_3^(2-)(aq) + H_2O(l)`