Chem 321 Name Answer Key D. Miller 9.79 1. Using Activities, Calculate

Chem 321 Name Answer Key D. Miller

1. Using activities, calculate the pH of an aqueous solution formed by the addition of 20.0 mL of 0.100 M HCl to 200. mL of 0.016 M K CO . Assume that the ionic strength of the solution is fixed at 2 3 2- 0.10 M and that the activity coefficient for CO3 is 0.37. (12) 2- + - CO3 + H3O 6 HCO3 + H2Ofor μ = 0.10 M γCO32- = 0.37 init mol 0.0032 0.0020 γHCO3- = 0.775 mol after rxn 0.0012 0 0.0020

- 2- - a HCO3 /CO3 buffer is formed with [HCO3 ]init = 0.0020 mol/0.220 L = 0.00909 M

2- [CO3 ]init = 0.0012 mol/0.220 L = 0.00545 M Since the 10-3 rules apply,

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2. For each of the following multiple-choice questions, circle the letter preceding the correct (best) answer. Only one answer will be accepted for each. (3 ea)

The ionic strength of 0.010 M Na3PO4 aqueous solution is:

A. 0.040 M.

B. 0.12 M.

C. 0.020 M.

D. 0.060 M.

Activity coefficients generally increase as: A. the ionic strength increases. B. the charge on the ion increases.

C. the hydrated ion size increases. D. all of the above.

If, in an aqueous solution, [Fe2+]=[Fe3+], one should expect that the activity of Fe2+ is: A. smaller than the activity of Fe3+.

B. larger than the activity of Fe3+. C. the same as the activity of Fe3+. Chem 321 Name Answer Key D. Miller

2. (continued)

At the half-way point in the titration of NH3 with HCl, the pH of the solution is about equal to:

A. pKb of NH3.

+ B. pKa of NH4 .

C. both A and B.

D. none of the above.

If K3PO4 is titrated with HCl, the pH of the solution at the second equivalence point is about equal to: A. 9.6 B. 7.2

C. 4.7 D. 2.2

3. Short answer

-4 If a sample of formic acid (HCO2H, Ka = 1.80 x 10 ) is titrated with 0.10 M NaOH, which of the acid-base indicators listed on the DATA SHEET would be the most appropriate one to use. Justify your choice of indicator. (5)

HCO2H pKa = 3.74 At the equivalence point you expect pH ~ (12.5 + 3.74)/2 ~ 8.1. A suitable indictor undergoes a color change in a pH range that brackets the equivalence point pH. Thus, thymol blue, with a pH transition range of 8.0 - 9.6, is the best choice. Why is a strong acid used instead of a weak acid to titrate a weak base? Be specific. (5) The reaction between a strong acid and a weak acid strongly favors the products (K is very large). The reaction between a weak acid and a weak base does not favor the products as much (it may even favor the reactants). Thus, such a reaction is not as quantitative as is needed for a titration.

+ In solutions of weak acids the contribution to the H3O concentration from the autoionization of water is usually insignificant. Indicate one situation where this contribution is likely to be important. (3) + -7 The water contribution to the [H3O ] is about 10 M. This becomes important when (1) the concentration of acid is very small (e.g., 10-9 M) and (2) the acid hydrolyzes to a very -11 limited extent (e.g., Ka = 10 ). Chem 321 Name Answer Key D. Miller

3. (continued) If you wished to buffer a solution at pH=7.0, which of the following weak acids would you choose to make your buffer? Justify your choice. (4) H2SO3 HOBr HC7H5O2

pKa1 = 1.91 pKa = 8.64 pKa = 4.20 pKa2 = 7.18

For an effective buffer, pH ~ pKa ± 1. Thus H2SO3 is the best choice. Briefly describe how you would prepare a 0.10 M solution of this buffer at the desired pH. (3)

Measure out the necessary amount of H2SO3 to produce a 0,10 M solution and dissolve in deionized water (use ~ 3/4 of the desired final volume), Monitor the pH of this solution and add NaOH until the pH = 7.00. Dilute to the final volume with deionized water. Indicate whether an aqueous solution of each of the following is acidic, basic or neutral. Explain your choice in each case. (3 ea) - NaHSO3 - HSO3 is amphiprotic ACIDIC

- KClO4 - NEUTRAL, because ClO4 is the conj. base of a strong acid and is thus a neutral ion as is K+

When 15 mL of 0.10 M NaOH is mixed with 20 mL of 0.050 M H2CO3, what type of solution (weak acid, weak base, buffer, etc.) results? Explain your answer. (3) - - reactions: H2CO3 + OH 6 HCO3 + H2O

- - 2- HCO3 + OH 6 CO3 + H2O

- Since there are 0.0015 mol of OH and only 0.0010 mol of H2CO3, all of the H2CO3 will be - - 2- converted to HCO3 and then some of the HCO3 will be converted to CO3 forming a - 2- HCO3 / CO3 buffer.

Briefly explain why the solubility of AgSCN in water increases when KNO3 is added to the solution. (4)

Adding KNO3 to the solution increases the ionic strength. This results in better shielding of the Ag+ and SCN- ions, making it more difficult for them to recombine and allowing for more AgSCN to dissolve. Chem 321 Name Answer Key D. Miller

4. Using activities, calculate the pH of a 0.010 M Ba(OH)2 aqueous solution. (10) - 0.010 M Ba(OH)2 ! [OH ] = 0.020 M

μ = ½[(0.010 M)(+2)2 + (0.020 M)(-1)2] = 0.030 M

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