1. You are faced with the job of titrating acid HX, a weak acid whose Ka = 2.0 X 10-6, with a standardized KOH
titrant whose concentration is 0.2000 M. Assuming that there is present 10.00 mmoles of HX in 50.00 mL of original solution, what is the pH at the equivalence point of this titration.

2. Calculate the pH for the following systems:
a) A solution consisting of 9.10 grams of methylammonium acetate, (CH3NH3+)(-OAc) (MW = 91.00) which has been dissolved in 50.00 mL of distilled water and then diluted to a final solution of 200.0 mL.
Ka for HOAc is 1.8 × 10-5 and Kb for CH3NH2 is 5.0 × 10-4 ANS: 7.72
b) The first equivalence point when 10.00 mmoles of phthalic acid, H2P
(MW = 166.0, K1 = 8.0 × 10-4, K2 = 4.0 × 10-6), in 100.0 mL of initial solution is titrated with 0.2000 M KOH (MW = 56.00). Assume that the 2 protons titrate separately.

3. You wish to prepare 750 mL of a buffer whose pH value is 9.10 by mixing together sodium bicarbonate, NaHCO3 (MW = 84.00) and sodium carbonate, Na2CO3 (MW = 106.00). If it is necessary to have the total carbonate concentration no more than 0.1500 M, how much (in grams) of each component is needed? (NOTE: for carbonic acid, K1 = 4.3 × 10-7, K2 = 4.8 × 10-11)

4. To save you time in your antacid analysis you decide to buy a 4-liter container of pre-standardized HCl. The label says the concentration is 0.1200 M but you decide to check for yourself. Titrating four 50.00 mL aliquots of the HCl yields molarities of 0.1176, 0.1182, 0.1185, and 0.1195 respectively. ( y = 0.4738, y2 = 0.0561235) At the 95% level of confidence, is the molarity on the label correct? Explain your reasoning.
ANS: C.I. = .0013 M; label is incorrect

5. In honor of the NCAA "Final Four" in basketball we present the following two "Universal truths"
TRUTH 1 -- "Foul the shooter with the lowest Free-Throw Average"
TRUTH 2 -- "A good team should have a Free-Throw Average of 75% or better"

Below are the Free Throw Averages for a mythical Final Four starting five.
61.7%, 78.6%, 72.2%, 77.9%, 73.8%
a) Using the data-handling principles of this course, decide on how (if at all) TRUTH 1 should be applied in this case. Does one player shoot fouls (significantly) poorer than the others? ANS: NO (by outlier test)
b) Again using the data-handling methods of this course decide whether or not TRUTH 2 applies to this team. Is this a "good team"? ANS: C.I = 8.4% , YES a good team

Clueless:

This is the standard titration problem. You need to add as many millimoles of base as there are millimoles of acid to produce only one species with acid/base properties, X-, the conjugate of HX. Then, using the Kb for X-, you calculate the pOH (and then the pH) of this solution.

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Hints:

A. You know you have 10.00 mmoles of HX. You therefore need to add 10.00 mmoles of KOH. You also know that the KOH is "packaged" to come 0.2000 mmoles of OH- per mL of KOH solution, so you can figure how much KOH solution needs to be added.

B. Once you've determined how much KOH solution had to have been added to add 10.00 mmoles of KOH (to just neutralize the 10.00 mmoles of HX) you know the final volume of solution, which now contains only one species with acid/base properties, X-. This allows you to calculate the concentration of X-.

C. Knowing the concentration of X- and determining its Kb from the Ka of its conjugate, HX, you can calculate [OH-], leading directly to pOH and thence to pH.

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Clueless:

Part a. You have only one formula here, (always not counting water, which only changes the concentration, never the #mmoles). There is not possible, therefore, a major reaction, so the number of mmoles of this species will not change. You need to decide whether this formula is for an acid (weak or strong), a base (weak or strong), or a salt. Those are the only possibilities here.

Part b. The phrase "first equivalence point" tells you you're dealing with a polyprotic system (even before you notice that TWO Ka's are given). You know how many mmoles of the polyprotic you have, so you'll know how many mmoles of the base KOH need to be added. You also know that when you titrate a polyprotic to its first equivalence point you get an INTERMEDIATE SALT.

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Hints:

Part a. Examine the formula to decide what it is. The decision as whether you have a strong acid or strong base should be trivial. You are responsible for knowing only 6 strong acids and 3 strong bases, and this formula is not any of them. The TWO K values presented to you should tip you to the fact that you've not only got a salt here, it is the salt of a weak acid and also a weak base. That should tell you how to solve this problem.

Part b. When you titrate any polyprotic to an intermediate salt, you should know from your notes how to calculate the pH for this system. It's an automatic.

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Clueless:

This is a buffer determination problem. You are told that it is a 2-component buffer, NaHCO3 and Na2CO3. You need to decide which is the acid and which is the conjugate so that you can use the Henderson-Hasselbalch equation to calculate the ratio of these two components which will fix the pH of the solution to the desired value.

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Hints:

A. Draw the pH line for this system (it will be the carbonic acid system, since you're given the K's for the carbonic acid system). Identify all formulas on the pH line and place the pK values on the line.
B. You should be in a position to know which pK value to use for the H-H equation.
C. Write the H-H eqn, specifying the formulas of the acid and conjugate.
D. Solve the H-H eqn to determine the ratios of the acid and conjugate. At the moment you have 2 "unknowns", the conc of acid, and the conc of conjugate.
E. From the information that the "total carbonate concentration" is to be no more than 0.1500 M, you can set one of your unknowns to the classic value X. The other unknown will then be (0.1500 - X). Substituting these into the value for the ratio of these two will permit you to calc the concentration of the acid and conjugate.
F. You now are faced with preparing a solution of, in this case, 750 mL for a formula whose concentration you've just determined in step 5 above. This is a straightforward chem 111 problem set calculation (P.S. 130)

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Clueless:

This is the standard statistics decision-making problem. The "supposed to be" value (what we've termed the Decision Limit) is 0.1200 M. You are trying to ascertain whether the acid really is 0.1200 M or whether there is some Real Effect present which has made the value of this acid really different from 0.1200 M. The way this was done here was to make four measurements and then decide, based on those measurements, whether this real effect is present. (It's further been agreed that this decision can only be made with 95% assurance of being right). You need to decide what you think, based on the 4 measurements, what the best estimate of the concentration for this solution is and also how much this estimate is affected by random error. You need both these two pieces before you can make your decision.

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Hints:

A. First point, always. Statistics works on notion that measurements are REPLICATES. If you see a "suspicious" looking value (high or low) it's always conceivable that it is not a replicate, but that a GROSS ERROR crept into that (and only that) measurement at the time it was taken, resulting in the anomalous value. Here the values look pretty close to each other, suggesting the non-likelihood of an outlier, but they (the highest and the lowest) can be always tested via a Q-test.
B. Determine what the Decision Limit is, the value to be tested against.
C. Determine what you think the value for the conc of HCl really is in the 4-liter container. Almost always the mean value is the preferred choice. Almost certainly it will be numerically different than the decision limit.
D. Now determine what the effect of random error (in 95% of the cases) will be on this mean value. Do this by the calculation of a 95% Confidence Interval. This will allow you to add and subtract the 95% C.I. to the mean getting you a range of values, centered by the mean. This range of values are your 95% Confidence Limits.
E. See where your Decision Limit lies, with respect to the 95% "Confidence Limit Box" you have built around your mean value. If the mean is "inside the box" you have not demonstrated a real difference between your mean and the D.L. If "outside the box" you declare (with only 95% assurance, however) that the 4-liter container does NOT have the concentration specified by the Decision Limit.

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Clueless:

Part a) If one of the players IS a poorer free-throw shooter than the others, statistically that is equivalent to his not being a replicate but an outlier. This is what you test for.

Part b) Here is another case of using statistics to compare measured values (free throw %) with a D.L. (75%) and decide if there is a real cause present (a poorer capability to shoot free throws).

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Hints:

Part a) The suspect value is, of course, 61.7%. Numerically it represents the player who is suspected of being the poorest performer. You test this value as a possible OUTLIER via a Q-test.

Part b) Compare best estimate of what you think the team shoots with the D.L. of 75% be seeing if 75% fits in the 95% Confidence Limits Box generated by the data. If the D.L. fits in the box you have demonstrated no real cause present (so this is a "good" team). Were the D.L. to lie "above the box" then there would be a real difference declared (with 95% assurance of being correct in this assessment) that this was NOT a good team because they shot at less than 75%.

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