I learnt about how ocean acidification can be described in terms of equilibrium. This is summarized in the textbook diagram below:

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However, this diagram doesn't explain to me why the shells should be softening. As more CO2 is put into the atmosphere, this will increase $\ce{CO2_{(aq)}}$. By the second equation this should create more H+ and HCO3-.

People say that the increase in H+ concentration will eat up the CO3 ions in solution and thus the loss of CO3 will shift the dissolution equilibrium right and make shells soft. However, although there is an increase in H+ which eats up the CO3, there is also an increase in the amount of HCO3 so shouldn't that prevent a shift in equilibrium?


1 Answer 1


If we play a little with all the equilibrium systems shown in the diagram we can get:

$$ \require{cancel} \begin{align} \ce{\cancel{\ce{CO2(aq)}} &<=> CO2(g)} \tag{R1} \\ \ce{HCO3^-(aq) + \cancel{\ce{H^+(aq)}} &<=> \cancel{\ce{CO2(aq)}} + H2O(l)} \tag{R2} \\ \ce{HCO3-(aq) &<=> \cancel{\ce{CO3^2-(aq)}} + \cancel{\ce{H^+(aq)}}} \tag{R3} \\ \ce{Ca^2+(aq) + \cancel{\ce{CO3^2-(aq)}} &<=> CaCO3(s)} \tag{R4} \\ \hline \ce{Ca^{2+}(aq) + 2 HCO3^-(aq) &<=> CaCO3(s) + CO2(g) + H2O(l)} \tag{R5} \end{align} $$

From this overall reaction, we can see that increasing the concentration of $\ce{CO2}$ will result in an equilibrium shift to the left according to Le Chatelier, which translates to the decomposition of $\ce{CaCO3}$.

Another explanation is that the solubility of $\ce{CaCO3}$ increases with decreasing pH, and vice-versa. In other words, if the ocean becomes acidic enough, more $\ce{CaCO3}$ will dissolve in the water.


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