(Another) question about Le Chatelier's principle (1 Viewer)

[hungryhippoos]

Hi everyone,

I'm hoping someone could clarify this for me.

In the CO2/H2CO3 equilibrium scenario: CO2(g) + H2O(l) ⇌ H2CO3(aq), what would be the effect of adding water to the reaction? My head tells me that the equilibrium would shift to the right (increased formation of H2CO3 to counteract the change, similar to adding CO2, according to LCP), but I'm having difficulty imagining how this would play out. If this was a real life scenario involving a carbonated drink, would the drink therefore become more sour and fizzy (due to the increased presence of H2CO3)?

Thanks in advance for your help.

 

[captainhelium]

I'm not fully sure of this but it might be because although you're increasing the formation of H₂CO₃, you're also diluting the solution at the same which I think decreases the overall concentration of aqueous stuff. So in essence, I don't think the increase in H₂CO₃ becomes very noticeable (in terms of taste and appearance) due to dilution.

 

[jazz519]

The CO2 equilibrium can be described by these three equations:
CO2(g) ⇌ CO2(aq)
CO2(g) + H2O(l) ⇌ H2CO3(aq)
H2CO3 + H2O(l) ⇌ H3O+(aq) + HCO3-(aq) (all exothermic)

The effect of adding water can be described by the last two equations. You are correct in stating that the equilibrium will shift to the right to minimise this change. However, we must note that LCP never fully minimises the change i.e the initial change in concentration may have been a decrease by lets say 1.0M and LCP might only increase the concentrations after by like 0.5M. This means that overall there is still a decrease in concentration from before the addition of water by 0.5M.

So essentially, the drink would still taste more diluted than before and this is also useful in explaining why dilution of weak acids can't be explained by pH=-log(H+) as you may dilute a 100mL solution to 1000mL which you would expect would result in a one increase in pH however it actually is more like 0.5. As LCP shifts the equilibrium right in the last equation making more H3O+ to minimise the change. However, once again it cannot fully minimise the change.

 

[hungryhippoos]

Sorry, I forgot about this! Thank you very much for the detailed responses, everyone!