Chapter 16: Tricks and Tips

1. Do we really have to memorize all these weak acids?
   • No, although it doesn't hurt to become familiar with them. But in fact, all you really need to do is know the strong acids, since any other acid is, by definition, weak!

     Strong acids completely dissociate in water. The ones listed in your book are:

     • HClO₄
     • HI
     • HBr
     • HCl
     • H₂SO₄ (first ionization only)
     • HNO₃

     Anything else is weak. What do you look for in a weak acid?

     • The COOH group, indicating a carboxylic acid.
     • Any other H written at the beginning or end of a formula.
     • Conjugate acids of weak bases (example: NH₄⁺).
     • HF
     • HNO₂

2. What about weak and strong bases?
   • Again, just remember the strong bases are ionic hydroxides (or species that generate these upon reaction with water). Specifically, the Group 1 and 2 hydroxides:
     • Group 1: LiOH, NaOH, KOH, RbOH, CsOH
     • Group 2: Mg(OH)₂, Ca(OH)₂, S(OH)₂, Ba(OH)₂

     Most anything else is a weak base! What are some examples of weak bases we should watch for?

     • NH₃
     • Anything with a lone pair on nitrogen (including ammonia!)
     • Conjugate bases of weak acids.

3. What are some common math errors?
   • Here's a big one: (x)(x) = 2x instead of (x)(x) = x².

     I think the reason people mess this up is that they test the first statement by plugging in a value of 1 or 2 for x and decide the statement is true for ALL numbers. It's not.

     If you try substituting 6 for x you'd recognize that (6)(6) is NOT equal to 12! But it is equal to 6² = 36.

   • Another common one: (2x)² = 2x² instead of (2x)² = 4x². Easy to mess up in a rush!

4. Salt hydrolysis: Help!!!!
   • This is a lot easier than many people realize. Let's first review the procedure for a simple case such as NaNO₃:
     1. Separate (dissociate) the salt into an anion and cation. The cation is usually listed first. In our case, we have Na⁺(aq) and NO₃^-(aq).

     2. React each ion separately with water and write a balanced equation. You can take either H⁺ or HO^- from water. Make sure your charges balance on each side!:

        Na⁺(aq) + H₂O NaOH(aq) + H⁺(aq)

        NO₃^-(aq) + H₂O HNO₃(aq) + OH^-(aq)

     3. Look at the equations your wrote and ask "can these products exist in solution?"
        • For the anion, the reaction would produce HNO₃(aq). We recognize that HNO₃ is a strong acid, which means that it dissociates in water. In other words, this reaction can't happen as written (it lies completely to the left). Therefore, there is no reaction of NO₃^-(aq) with water and so there is no effect on the pH.

        • For the cation, the reaction would produce NaOH(aq), a strong base. Because NaOH is a strong base, it can't exist in water and this reaction lies 100% to the left. Therefore, no effect on pH this time, either.

     4. If they can exist, did we make H⁺(aq) or OH^-(aq)?

        Let's look at the formate anion as an example:

        HCOO^-(aq) + H₂O HCOOH(aq) + OH^-(aq)

        In step c with this particular reaction, we'd say that HCOOH is a weak acid (since it is not one of our 6 strong ones). Therefore, this equilibrium can occur and when it does, it will produce some amount of OH^-(aq). Therefore, this will make the reaction more basic.

        Notice that makes sense. We added a base (the conjugate base of the weak acid, HCOOH) and so the solution should become more basic.

        Do not be thrown off by the presence of the undissociated acid, HCOOH, on the right side of the equation. We are looking only for H⁺(aq) or OH^-(aq).

     There is, of course, the possibility that one reaction could form acid and the other could form base at the same time. In that case, we simply examine the k values for each of these reactions. Whichever one has the largest k will predominate!

5. Yes, but HOW do I know what my stuff dissociates into in water?
   • Let's use C₂H₅NH₃NO₃ as an example.

     There are a couple of clues here. First, cations are supposed to be written first and anions last (sometimes we get sloppy about that, so use caution). Second, your formula has an N with some H's on it -- a big clue.

     We could rewrite the formula C₂H₅NH₂HNO₃ instead. That's legal since it is all in the same order, but now look at it. On the far right we have HNO₃ and the rest is C₂H₅NH₂. In other words, taking away HNO₃ (one of our strong acids) leaves us with C₂H₅NH₂, ethylamine (a base).

     Notice that ethylamine has a formula like ammonia (a weak base) except that one of the H's of ammonia has been replaced by an ethyl (C₂H₅) group. So we know this is the salt formed by the acid-base reaction of ethylamine and HBr and that our two ions must therefore be ethylammonium, C₂H₅NH₃⁺, and nitrate, NO₃^-. Of course, we then have to determine how each affects pH from there.

     Running through this in a different way, if I start at the right side of the given formula and can recognize and remove the anion (NO₃^-), then whatever is left must be a cation, C₂H₅NH₃⁺. This cation looks exactly like the ammonium ion, NH₄⁺, except one H has been replaced by ethyl again. Therefore, this is the conjugate acid of ethylamine, ethylammonium.

     NOTE: there are a few complex anions that contain multiple halides (such as perchlorate, OCl₄^-), so use the latter approach with caution! Make sure your cation makes sense (i.e. don't make funky species like HO⁺ or RNH^- (where R = C's and H's etc.)). If all else fails, draw a Lewis structure if possible.

     In none of these salt hydrolysis problems will you generate a neutral molecule such as Cl₂, NH₃ etc. So, if you see something ending in Cl ₂, that must mean two Cl^- anions, not chlorine gas. In Chapter 18 we'll start to consider how neutral molecules can attach themselves (coordinate) to cations, but don't worry about that yet.