Question #d9f07

Question

1954 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-20T05:57:02

$N a O H (a q) + C O_{2} (g) \to N a H C O_{3}$ $NaOH (aq) + CO_2 (g) -> NaHCO_3$

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

When $N a O H$ $NaOH$ is concentrated, pH > 10, the reaction produces sodium hydrogen carbonate ( $N a H C O_{3})$ $NaHCO_3)$ :

$N a O H (a q) + C O_{2} (g) \to N a H C O_{3}$ $NaOH (aq) + CO_2 (g) -> NaHCO_3$

Answer 2 · 2017-03-20T06:18:00

$O H^{-} (a q) + C O_{2} (g) = H C O_{3}^{-} (a q)$ $OH^-)(aq) + CO_2(g) = HCO_3^-)(aq)$

$C O_{2} (a q) + 2 O H^{-} (a q) = C O_{3}^{2 -} (a q) + H_{2} O (l)$ $CO_2(aq) + 2OH^-)(aq) = CO_3^(2-)(aq) + H_2O(l)$

When pH is very high or of a very basic nature

= $N a O H + C O_{2} = N a H C O_{3}$ $NaOH + CO_2 = NaHCO_3$

The ionic reaction is

First write the ions formed

$N a^{+} (a q) + O H^{-} (a q) + C O_{2} (g) = N a^{+} (a q) + H C O_{3}^{-} (a q)$ $Na^+(aq) + OH^-)(aq) + CO_2(g) = Na^+(aq) + HCO_3^-)(aq)$

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

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

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