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Chapter 17: Tricks and Tips

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    How can I tell if a system is a buffer system?

    First, let any stoichiometric reactions or neutralizations occur and then decide what species are present. We can then decide if we have the two components necessary for a buffer system, namely: Let's try an example or two:

    1. A weak acid or weak base.
    2. The conjugate of that weak acid or weak base.
    3. Is a solution consisting of equal volumes of 0.1 M HCl and 0.2 M NH3 a buffer solution?

      • First we can write a chemical reaction for this system

        \[HCl + NH_3 \rightarrow NH_4Cl\]

        I've written this as irreversible because HCl won't exist in solution. The question now becomes a limiting reagent question. We can treat molarity and moles the same here because the volumes are equal. So, which is the limiting reagent?

        HCl is the limiting reageant. 0.1 moles of it will react with 0.1 moles of NH3 to form 0.1 moles of our salt. In other words, we can rewrite our equation as:

        \[0.1 HCl + 0.2 NH_3 \rightarrow 0.1 NH_4Cl + 0.1 NH_3\]

        Since we have a weak base (NH3) and its conjugate (NH4Cl) in nearly equal amounts after our initial stoichiometric reaction, this is a buffer system.

    4. Is a solution consisting of equal volumes of 0.2 M HCl and 0.1 mol NH3 a buffer system?

      • Using the same logic as above, only now with different coefficients, we now find that NH3 is the limiting reagent, not HCl.

        \[0.2 HCl + 0.1 NH_3 \rightarrow 0.1 NH_4Cl + 0.1 HCl\]

        In other words, after our initial reaction we have an equimolar mixture of a conjugate acid (NH4Cl ) and a strong acid (HCl). This does not meet our criteria for a buffer system!

    How do I know what equation to use in my buffer calculations?

    Remember that a buffer is a weak acid or base and its conjugate. These must exist in equilibria for the system to work, so there is only one equation. Put the acid or base on one side and the conjugate on the other.

    For example, if we made a buffer by mixing 0.1 moles of NaOH and 0.1 moles of CH3COOH we'd follow the process outlined in #1 above, namely do the stoichiometric reaction:

    \[0.1 NaOH + 0.1 CH_3COOH \rightarrow 0.1 CH_3COO^- + H_2O\]

    We know that since acetate ion is a weak base (it's the conjugate base of acetic acid, a weak acid), that it will undergo hydrolysis in water. We can write the hydrolysis reaction by putting the acid on one side of our equilibrium arrow and the conjugate on the other. Fill in water and balance and we're done!

    \[CH3COO^- + H_2O \rightleftharpoons CH_3COOH + HO^-\]

    It doesn't matter how we write this reaction. I could have just as easily written

    \[CH3COOH^+ H_2O \rightleftharpoons CH_3COO^- + H_3O+\]

    The key here is to recognize that if I use the first equation, I need to use Kb and if I use the second one I need to use Ka.

    Watch this distinction! You'll find many instances on the problem set and exam where you are given Ka when you really need Kb to solve the problem (or viuce-versa). Remember that you can always interconvert the two using the relationship

    \[K_aK_b = K_w = 10^{-14}\]

    or

    \[(pK_a)(pK_b) = pK_w = 14\]

    FINAL NOTE: You generally only need to write the one balanced reaction equation for Ka or Kb to solve these problems. You don't have to find both. But do note that the reaction neutralizes acids in one direction and neutralizes bases in the other. If it didn't do this, then it would not be a buffer!

    How do I handle addition of acid or base to a buffer?

    This is nothing we haven't already seen or done. Simply: We follow the same procedure when we deal with titrations. Titrations and buffers are closely related. In either case, we do an inital reaction and then solve for equilibrium values. Put another way, we do the initial reaction and then it's just Chapter 16 all over again (which is really just Chapter 15 all over again).

    1. Perform stoichiometric reactions first just as we did in question #1 above.
    2. Use the equilibrium concentrations that this gives you. For small amounts of added acid or base we have a salt left in the solution. We can find the pH etc. using our usual salt hydrolysis procedure or the Henderson-Hasselbalch equation (we'd get the same answer no matter which we did.

    Should I use the Ka or Kb to answer this question?

    • This is one of the most common errors students make in this Chapter! A problem may give you a Ka when you really need Kb or vice-versa. Don't be fooled into using the wrong one! I've already discussed this in # 3 above, but I'll repeat it here to drive the point home.

      Write out the equation for your reaction. For example, if we are asked to determine the pH of a NaOAc solution, which K do we need to use?

      We answer this by doing what we always do first, writing a balanced reaction for what happens when we dissolve the substance in water:

      \[AcO^- + H_2O \rightleftharpoons HOAc + HO^-\]

      Notice that the reaction is balanced. And also notice that when we dissolve our substance, it makes HO-. Therefore, this reaction is a base dissociation and we need to use the Kb expression, \[K_b = \dfrac{[HOAc][HO^-]}{[AcO^-]}\].

      People sometimes get confused because this reaction makes HOAc and ask "isn't this an acid?" Yes, it is, but is not dissociated so it doesn't affect the pH. In these kinds of reactions you are simply looking to see if the hydrolysis forms HO- or H3O+ when it reacts with water.

      The key to success is to write a balanced equation and then use the right K. Remember the relationship KaKb = Kw = 10-14, and you'll never miss a problem of this nature!


    Chapter 17: Tricks and Tips is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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