Solutions to Buffer Problems

Below are the solutions to AP Buffer problems #3 through #5. I know that I said that I would only supply the solutions to #3 and #4, but I liked a couple of the aspects of #5. One aspect of #5 is part b when the Kb value is to be calculated, but no concentrations are given to do this. This can be accomplished by knowing the weak acid that the base came from. Kw can be divided by Ka to determine the Kb. Tomorrow's quiz is another old AP Chemistry free-response question dealing with acid-base equilibrium with a focus on buffers. There will also be a question on determination of acid strength. Remember, this can be determined by knowing the Ka value or the structure of the acid. The structure, to be a "stronger" weak acid is to either have more oxygens or a more electronegative element in the acid structure. Come see me before class if you have any further questions. Remember, there is a study session tomorrow at 6:30 a.m.

Solutions to Acid / Base Equilibrium Problems 4 through 6

Below are the solutions to the acid / base equilibrium problems #4 through #6. The solutions to #4 are included even though they were presented in class. Tomorrow, you will have a quiz over the types of problems presented in the packet of old AP Chemistry exam questions dealing with acid / base equilibrium. In the bulleted text below, I have attempted to explain some of the more difficult parts of the questions.

• #5 part (c) caused some confusion in class. The acid (hypochlorous, HOCl) is reacted with the strong base sodium hydroxide (NaOH). Whenever a weak acid or base is reacted with a strong base or acid, the strong base or acid reacts completely. The reaction is not shown at equilibrium, but rather with only a right arrow showing a complete shift to the right. All of the hydroxide is consumed and converted into water and the conjugate base OCl^-1. Notice that the spectator ion Na^+1 was not included in the net ionic equation. This balanced equation will be needed for problem (d) (ii).


• #5 part (d) (ii) is a very involved problem. The moles of hydroxide ion will react completely with the weak acid HOCl. All of the moles (or millimoles) of hydroxide will be converted into water and the conjugate base OCl^-1. That is the reason that the B.C.E. table is shown. The moles (or millimoles) left of the HOCl and OCl^-1 after the reaction is completed will be used to determine the initial concentrations of each to re-establish equilibrium. Each ending mole value (or millimole value) will be divided by the total volume of 0.0400 liters (for moles) or 40.0 mL (for millimoles). The initial concentrations of each will then be used to determine the hydronium ion concentration when equilibrium is re-established.


• #6 (a) (iv) shows a B.C.E. table to illustrate that all of the moles of hydroxide ion (equal to the moles of NaOH found in question 6 (a) (iii)) will convert into moles of the conjugate base benzoate (C6H5COO^-1). The intent of the question is to determine if you know that the moles of the strong base will completely react and convert to the conjugate base.


• #6 (a) (v) is a different problem in that the initial concentration of benzoic acid (C6H5COOH) is not given. This can be treated just like any other unknown variable in an I.C.E. table. Once the equilibrium concentratio of benzoic acid is determined, the volume given in the problem can be used to determine the moles of benzoic acid present. The initial and equilibrium concentrations are considered to be the same since the change amount is miniscule compared to the initial value.