Acidification with Strong Acids to pH 3
Problem
Given is a groundwater sample with pH 6.9. How much nitric acid (HNO3) or how much sulfuric acid (H2SO4) is needed to obtain pH 3?
Procedure. For this task there is a special tool in aqion PRO available (i.e. dosage of chemicals to attain a target pH)1. Nonetheless, it can also be solved without the PRO version. This is done here indirectly using charge-balance adjustment (provided that the initial solution is charge-balanced).
Acidification with HNO3 (Monoprotic Acid)
Click on Open and select the groundwater sample “gw.sol”. The only thing we have to do is to change the pH parameter from 6.9 to 3.0. Then, click on Start.
Immediately you will get the message that the charge balance error is very large: 81.9. It’s no surprise because our change of pH leads to an immense excess of H+ ions. To re-adjust charge balance, we select the parameter NO3, and click on next.2
A schema pops up (which can be expanded by clicking on Details) as shown in the picture on the right. The adjusted amount of NO3- is just the required amount of
| HNO3 = 6.95 mM |
Thus, we need almost 7 mM of nitric acid to acidify the groundwater from pH 6.9 to 3.
Discussion. The result would be absolutely correct if the input water were exactly charge-balanced (i.e. CBE = 0). However, this is not quite the case; our input water has a very small but finite charge-balance error of 0.33 (with 0.04 mM excess of cations).
This means that the calculated value of 6.95 mM also contains a very small portion (namely 0.04 mM), which serves to correct the CBE of the input water. The exact value for the required dosage would therefore be “only” 6.91 mM.
Therefore: All calculations should always be carried out with charge-balanced waters. But be careful: What is the right parameter to adjust charge balance; is it pH or a cation or an anion? Depending on which parameter you choose, you get (slightly) different results. You are welcome to play it through with the program (which gives you a feeling for the uncertainties of the measured parameters).
Acidification with H2SO4 (Diprotic Acid)
Repeat the above procedure, but now select the parameter SO4 to adjust charge balance. The result is:
| H2SO4 = 3.55 mM |
Interesting result. Naively we would expect that it should be the half of the HNO3 amount, namely 3.47 mM, because the (strong) sulfuric acid is able to release two H+ ions.
The reason for the small deviation from 3.47 mM is that H2SO4 does not dissolve completely.3 A small residue of undissolved acid (in form of hydrogen sulfate, HSO4-) remains in the solution. And this small portion must also be considered when we add the sulfuric acid.
Further Reading
- the difference between strong and weak acids – see here
- analytical equations for N-protic acids – see here
Remarks & Footnotes