Question #d9f07
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#NaOH (aq) + CO_2 (g) -> NaHCO_3#
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#
#OH^-)(aq) + CO_2(g) = HCO_3^-)(aq)#
#CO_2(aq) + 2OH^-)(aq) = CO_3^(2-)(aq) + H_2O(l)#
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)#
