# Question d9f07

Mar 20, 2017

$N a O H \left(a q\right) + C {O}_{2} \left(g\right) \to N a H C {O}_{3}$

#### Explanation:

The reaction depends on the concentration of the alkali solution $N a O H$).

When $N a O H$ is concentrated, pH > 10, the reaction produces sodium hydrogen carbonate (NaHCO_3):

$N a O H \left(a q\right) + C {O}_{2} \left(g\right) \to N a H C {O}_{3}$

Mar 20, 2017

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

= $N a O H + C {O}_{2} = N a H C {O}_{3}$

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

$2 N a O H + C {O}_{2} = N {a}_{2} C {O}_{3} + {H}_{2} O$

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)#