Monday, March 7, 2011

pH calculation of a very small concentration of a strong acid.

I have always felt that I didn't really "get" acid-base chemistry. Usually, when I'm presented with a chemistry problem, my mind just refuses to grasp what's being asked. I struggle to figure out which equations to use, and how to plug the numbers from the problem into the variables of the equation. Well, my experience this past week reassures me that I'm not the only one.

In my Biochemistry class, we got this problem:
Calculate the pH of a solution of the strong acid HCl at a concentration of 5x10-8 M.
This is a little bit of a tricky question, because the concentration of the acid is so low. But before getting into it, let's simplify the discussion a bit by assuming that we're talking about one liter of water. Then, the amount of acid that we're adding is 5x10-8 moles. To me, changing from concentrations into definite amounts makes it easier to understand.

A strong acid will dissociate more-or-less completely when added to water, so this is like adding a 5x10-8 moles of H+ ions into the water. Note that the number of H+ ions in one liter of pure water is 10-7 moles, so this is one-half of that. To use nice, small, round numbers, it's as we had a bottle of water (a very small bottle) that has 10 H+ ions, and we're adding an extra 5.

My first answer was based on this simple addition. I assumed that the number of H+ ions after the acid was added would be 1.5x10-7. Then the pH would be the negative logarithm of that:


This would be like assuming that if the bottle of water initially had 10 H+ ions, and you add 5, then in the end it would have 15.

The problem set was presented online, and after submitting our answers, we could go back and review them. The answer key gave the correct answer to this problem as "7.3". ... Wait, what? Neutral water has a pH of 7, and anything above 7 is considered alkaline. So how do you get an alkaline solution after adding a strong acid to water?

It's pretty easy to see where the "7.3" answer comes from. The usual way to solve these problems is to assume that the H+ concentration, after you add the strong acid, is equal to the concentration of the acid. See, for example, the first sentence of the Calculation of pH for weak and strong acids section of the pH article on Wikipedia: "In the case of a strong acid, there is complete dissociation, so the pH is simply equal to minus the logarithm of the acid concentration." So following that strategy:


But the problem is that this completely neglects the baseline concentration of the H+ ions in pure water. This is like assuming that if the bottle of water has 10 ions, and you add 5, then in the end it will have 5. It doesn't make sense.

In fact both of these answers are wrong. What happens, after you add the new H+ ions, is that some of them will react with the OH- ions to form neutral water molecules. The solution will be at equilibrium when the equilibrium equation for water is again satisfied:


To continue with the analogy I've been using, it's as if you have a bottle of water that has 10 H+ ions and 10 OH- ions. The equilibrium equation for water says that H+ times OH- must always equal 100, once equilibrium has been established. Now, you add 5 extra H+ ions. At this point the bottle has 15 H+ ions and 10 OH- ions. But that product is 150, so the solution is not at equilibrium. What happens? Well, some of the H+ ions will react with the OH- ions to form water. Each time that happens, the number of H+ ions and the number of OH- ions both decrease by one. So first there will be 14 H+ ions and 9 OH- ions, then 13 and 8. Now, 13 X 8 = 104, so in this simple-minded example, that's as close to equilibrium as we can get.

At each step, the difference in the number of H+ ions and OH- ions is always a constant, and is equal to the number of H+ ions that were added:


When we apply this reasoning to the original problem, we see that the difference in the concentrations of H+ and OH- is a constant, which is the amount of acid that was originally added:


We want to get rid of [OH-], so let's solve for that:


And we also know that the equilibrium equation for water must be satisified. Here it is again:


So let's plug the value for [OH-] into that:


And then put it into the form of a quadratic equation:


Then using the quadratic formula:



Clearly, the value must be positive, so we should pick "+", not "-":


Finally, we can calculate the pH:


At first, I wasn't very confident in this solution, so I dug around on the Internet. All of the examples I could find used very high concentrations of strong acid, and used the assumption that you could take the acid concentration as the H+ concentration. For example, here ("Find the pH of a 0.0025 M HCl solution.") and here ("How do you calculate pH of a 0.012M solution of HCL?").

Finally I found a good discussion in this online textbook, starting at section 3.1, "Strong acids and bases". Their procedure maps onto what I did above, almost exactly. Note that they refer to the hydronium ion H3O+ where I referred simply to the hydrogen ion H+.

Equation 17 from the book is:


which is the same as my quadratic equation above -- Ca is the concentration of the acid that was added, and Kw is the equilibrium constant of water.

At one point, the authors say:
As the acid concentration falls below about 10-6M, however, ... [H3O+] approaches √Kw, or 10-7M. The hydronium ion concentration can of course never fall below this value; no amount of dilution can make the solution
alkaline!
Which is another way of saying that you can't get a pH above 7, no matter how low the acid concentration is.

UPDATE:
You might notice that I linked to an old version of the Wikipedia article above. That's because when I was looking for resources for this post, I discovered that that section on Wikipedia not only had a lot of errors, but didn't even make much sense at all. So I rewrote it from scratch -- everything under "Calculations of pH", including "Strong acids and bases" and "Weak acids and bases".

5 comments:

  1. Hi! Well, this is all beyond me... but figuring that your interest in chemistry most likely began long ago, while a young boy, I have a question for you. My son (2nd grade) LOVES chemistry. He has a closet in our house that we've turned into his 'lab' and he loves to play with his chemistry set and microscope, etc. His new school is having a science fair in may (not judged, just for kids who want to have fun with science) and I am trying to help him pick out something that he can present. Unfortunately we're having a hard time finding something of 'quality' through the internet... either it's too dangerous or slightly boring having been done at all science fairs across the US. Any thoughts or ideas would be greatly appreciated!

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  2. Well, in keeping with the theme of this post, how about something to do with measuring pH? You can buy a whole bottle of 90 pH measuring strips from Amazon for $10. He could go out and measure the pH of soil samples and ponds and streams, and try to correlate his findings with acid rain. That's just a broad idea -- a lot of details would have to be filled in. Good luck!!

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  3. Yikes! I probably would have given up moving towards the quadratic equations and went off to start making serial dilutions because, we also Know that the pH of water is between 6-8 depending on where you got it from to begin with. Unless we are starting with some reliably deionized water.:)

    Good catch on that calculation though. I don't think I would have thought of it.

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  4. Everything seems alright.

    Just remember one thing, you can not use [H+] - [OH-] or as such.

    Try to write the equilibrium in an Acid–Base Reaction.

    Use pKa instead of just pH.

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  5. Hey! Your explanation about how to calculate the pH of a very low concentration of strong acid was extremely helpful. I had an essentially identical problem to what you had (except with a different concentration), and your step-by-step instructions really helped me understand that the dissociation of water into H+ and OH- affects the pH at low concentrations, so thanks again!

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