## OA not OK part 8: 170 to 1

#### Posted on 21 July 2011 by Doug Mackie

This is post 8 in a series about ocean acidification. Other posts: Introduction , 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, Summary 1 of 2, Summary 2 of 2.

Welcome to the 8^{th} post in our series about the fundamental chemistry of ocean acidification. In this post we show that adding acid to seawater removes carbonate.

In the last post we introduced the concept of **reaction coefficient** to describe the state of a reaction relative to the **equilibrium constant**. For equation 8 we have the following expressions:

The direction in which a reaction responds to a disturbance is given by comparing *Q* to *K*. If *Q* < *K *then the reaction system is *not yet* at equilibrium and favours the right side (products) to reach equilibrium. If *Q *> *K* then the reaction system has 'gone past' equilibrium and favours the left side (reactants) to return equilibrium. (If *Q* = *K* the system is *at* equilibrium).* K* values cover a wide range and p*K*, analogous to pH, is more commonly used: p*K* = -log(*K*) and *K* = 10^{-pK}.

Consider Equation 8 at equilibrium (i.e. *Q *= *K*) and ask what happens if we add some acid H_{3}O^{+} to the system. By adding H_{3}O^{+} we increase the products, thus increasing the numerator and therefore increasing *Q* so that now *Q* > *K*. The reaction system therefore favours the left side (reactants) and H_{2}CO_{3} is produced.

So far this sounds like Le Chatelier, but now we consider what happens with equation 8 and equation 9 occurring simultaneously.

For Eq. 8, in seawater at 10 ^{o}C, p*K* = 6.00. At the preindustrial ocean pH of 8.25, the equilibrium ratio of the left side, carbonic acid, to the right side, bicarbonate, is about 1:170 (depending on salinity). The calculation is 10^{-6.00 }/ 10^{-8.25}. For Eq. 9, in seawater at 10 ^{o}C, p*K* = 9.19. At the preindustrial ocean pH, the equilibrium ratio of the left side, bicarbonate, to the right side, carbonate, is about 9:1 (the calculation is 10^{-9.19}/10^{-8.25}).

If we couple these reactions, we see that at typical ocean pH equation 8 produces a ratio of carbonic acid to bicarbonate (and carbonic acid to H_{3}O^{+}) of 1:170. If we do add a little acid (H_{3}O^{+}), then 170 parts of the acid stay as acid (H_{3}O^{+}) and only 1 part reacts to form carbonic acid and water. But equation 9 tells us that if acid is added then 1 part of the acid remains as H_{3}O^{+} but 9 parts of the acid, **and** therefore some of the carbonate, are consumed to produce bicarbonate. At typical seawater pH, the response of the system to the addition of acid is dominated by the consumption of carbonate shown in Equation 9.

Now we can see that adding acid to the system does indeed push equation 9 to the left. Though a naive application of Le Chatelier's Principal to seawater chemistry gets the 'right' answer, this is only because of a fortuitous combination of the values for the equilibrium constants. (Seawater is in fact even more complex and there are many other competing reactions going on at the same time).

**Thus, adding acid to the seawater-CO _{2} system massively changes the proportion of carbonate since it is consumed in the reaction. However, it hardly changes the proportion of bicarbonate since most of the carbon is already in the form of bicarbonate.**

Calculations based on these concepts can be used to produce a speciation plot like **Figure 3** which expresses the concentration of each species (expressed as a fraction of the whole) as a function of pH. In this figure the total sum of species containing carbon atoms is constant. What changes is the relative proportion of each species. We hasten to add that by adding CO_{2} to the ocean this is no longer true and the calculations become more difficult.

**Figure 3.** A speciation diagram for the carbonic acid system in seawater as a function of pH. The y-axis gives the fraction of each species present. A vertical line drawn at any pH value gives the relative proportion of each species. This plot is simplified to illustrate the concept; in real seawater several other factors like salinity, temperature and pressure are important.

We can run through some example calculations. Concentrations are given as moles per kg of seawater. For prior to the industrial revolution we have taken a pH of 8.25. Other input parameters are a total dissolved inorganic carbon concentration of 2100 x10^{-6} mol kg^{-1} (total dissolved inorganic carbon is the sum of the species described in post 5, temperature = 15 deg C, and the salinity = 35. Under these conditions, which are typical for the ocean, the concentration of H_{2}CO_{3}, HCO_{3}^{–} and CO_{3}^{2–} would have been 10 x10^{-6} mol kg^{-1}, 1830 x10^{-6} mol kg^{-1} and 260 x10^{-6} mol kg^{-1} of seawater, respectively (Roy constants used - see below).

The calculated values add to give the total dissolved inorganic carbon of 2100 x10^{-6} mol kg^{-1} and HCO_{3}^{–} and CO_{3}^{2–} are roughly in a 7:1 ratio (the carbonic acid does not really contribute, because, as expected it is 170 times less than the HCO_{3}^{–}). Today, a after a decrease in pH to 8.14 – and an increase of 29% in the concentration of H_{3}O^{+} – typical calculated concentrations of HCO_{3}^{–} and CO_{3}^{2–} are about 1880 x10^{-6} mol kg^{-1} and 210 ×10^{-6} mol kg^{-1}, respectively. You can see that HCO_{3}^{–} and CO_{3}^{2–} are now roughly in 9:1 ratio. The concentration of carbonate CO_{3}^{2–} has changed by –25% ((260-210)/210 × 100), but the concentration of bicarbonate HCO_{3}^{–} has only changed by 3% ((1880-1830)/1830) × 100.

We have made several simplifications here as the calculations are complex and beyond the scope of a blog post. Indeed, the full calculations are not encountered before postgraduate study. However, if you would like to try it yourself, you can download a program, written by one of us (KH) called SWCO2. It is available at the University of Otago. You will need to take some care – actual realistic values for the various entry parameters fall within a relatively narrow range.

If you do have a go yourself you will discover yet another layer of complexity: There are several sets of equilibrium constant, *K*, values to choose from. The reasons are complex and rest on the way that these values are determined **experimentally** and, though each set of values is internally consistent, each set uses a slightly different set of initial assumptions.

Now, just to further complicate things, we will introduce another equation:

*K* for this reaction is about 10^{-3}. That is, the ratio of left to right is about 1,000:1. This means that, to a first approximation, seawater (dominated by HCO_{3}^{–}), has only a little bit of CO_{2} and CO_{3}^{2-}. More importantly, it also shows that if we add CO_{2} to seawater, CO_{2} will spontaneously react with CO_{3}^{2–} to form 2HCO_{3}^{–} because *K* for the reverse reaction is 10^{3}. In a later post we put some numbers to this concept.

Ocean acidification is caused by absorbance of atmospheric CO_{2} by seawater. We now see that this acidification of surface seawater is causing the removal of CO_{3}^{2–}:

But why should this be a problem? We know from Equation 1 that calcification uses bicarbonate HCO_{3}^{–}, not carbonate CO_{3}^{2-}. Why does it matter if some carbonate from the oceans is converted to bicarbonate? Won't that actually help the shellfish? We explain why not, and just where the acid is coming from, over the next few posts.

Written by Doug Mackie, Christina McGraw , and Keith Hunter . This post is number 8 in a series about ocean acidification. Other posts: Introduction , 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, Summary 1 of 2, Summary 2 of 2.

Bernat 10:33 AM on 21 July, 2011Doug Mackieat 10:43 AM on 21 July, 2011Jeff Tat 21:24 PM on 21 July, 2011Christina McGrawat 09:52 AM on 22 July, 2011by themselvestell us anything. Our point is that we need to consider all these equationstogether. In several of the later posts we show how we can use theKandQvalues that we have introduced here to know the extent to which each equation occurs and how to work out what the overall outcome is. Read what we say again: That is if youreversethe reaction then it still favours, with a 1000:1 ratio the side with HCO_{3}^{-}over the side with CO_{2}. We are unclear how you draw your conclusion about figure 3.Sarahat 10:25 AM on 23 July, 2011_{2}to lower the pH then thefractionsbehave just like the figure shows, even as the total amount of carbon increases. You can alsoincreasethe pH by adding any kind of base, for example sodium hydroxide. And you can increase the pH by adding carbonate (CO3_{3}^{2-}). In that case the total amount of carbon also increases while the fractional distribution shifts to the right on the graph. Removing CO_{2}decreases the pH (by eq 8 it removes H_{3}O^{+}) and so the fractional distribution again shift right, even though this time the total amount of carbondecreases.. Figure 3 isvery powerfulprecisely because it separates the total amount of carbon from the fraction in each form.Doug Mackieat 11:05 AM on 23 July, 2011Jeff Tat 13:32 PM on 23 July, 2011Doug Mackieat 14:44 PM on 23 July, 2011_{2}system can be described by anytwoof the composition parameters. e.g. CT and pH are sufficient to determine AT, pH and CO_{2}. Your inability to understand the figure does not appear to be widely shared. By all means prepare a figure yourself and post it here.Sarahat 15:02 PM on 23 July, 2011^{-6}mol kg^{-6}. Adding more carbon to the system would make all three lines increase together. So, if the new concentration is 10% higher (it isn't in the real world!) all three lines would be 10% higher on that scale but their proportions would stay the same. Another way to think about this is to consider the CO_{3}^{-2}line at around pH 8.1. If you add a significant amount of CO_{2}then all three lines go up when the Y-axis is in concentration units. But, by eqn 12, adding CO_{2}causes some of the CO_{3}^{-2}to convert to HCO_{3}^{-}and the new distribution defines a new more acidic pH. Actually, the 2% increase in ocean carbon would only shift the concentration lines up (together) by a tiny amount (2%). But because the system at pH 8.1-8.2 is far down on the slippery slope of low carbonate, it makes a big change in percent CO_{3}^{-2}. Your suggestion that the ratio is more important than the concentration for defining the pH is correct. The complication is that in this system carbon is both the main thing controlling the pH (by the ratio of bicarbonate/carbonate)andthe thing making the pH shift (because CO_{2}) in water make carbonic acid, which reacts with carbonate to change the ratio, eqn 12). Neither the figure nor the post are "using only three equations to define four parameters". They are using the equations to understand the behavior of the system as a function of pH (H_{3}O^{+}). If you want to calculate the pH for a system containing these carbon species, then the equation you need is: 2H_{2}O --> H_{3}O^{+}+ OH^{-}. pKa=14.Jeff Tat 22:30 PM on 23 July, 2011JosHagelaarsat 02:11 AM on 24 July, 2011Doug Mackieat 10:33 AM on 24 July, 2011K1,K2 and [H_{3}O^{+}]. Try googling dissociation of a diprotic acid. If there is huge interest we may consider an extra post to go through the equations.Sarahat 13:25 PM on 25 July, 2011_{3}to the oceans over millions of years, which dissolves to the base CO_{3}^{2-}. Alkalinity (AT) is a measure of how much "excess" base is in the system, for example, how much CO_{3}^{2-}came in with Ca^{2+}instead of with H^{+}as the counter ion (to balance the charges, you can't add negative charge without adding + charges too). The charge balance in the system provides another constraining equation. The sum of charges of all the species present must add up to zero. The pH is determined by how much of the charge ends up being H_{3}O^{+}after all the equilibrium equations are satisfied.Doug Mackieat 13:44 PM on 25 July, 2011_{3}^{-}(Eq. 4) We discuss alkalinity in post 12. Seawater alkalinity is a more complex expression than that for freshwater and also includes species like B(OH)_{4}^{-}, SiO(OH)_{3}^{-}, and HF.Sarahat 22:23 PM on 26 July, 2011Doug Mackieat 14:26 PM on 25 August, 2011## second summary post