DOC PREVIEW
UT CH 302 - Practice Exam 2
Type Miscellaneous
Pages 7

This preview shows page 1-2 out of 7 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 7 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 7 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 7 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Spring 2009 CH302 Practice Exam 2 Answer Key 1. What would be the pH of a solution prepared by dissolving 120.1 g of CH3COOH and 82 g of NaCH3COO in 1 L of water? Acetic acid has a Ka of 1.8 x 10-5. 1. 5.05 2. 4.78 3. 4.12 4. 4.44 Correct Explanation: 120.1 g CH3COOH x (1 mol / 60.05 g) = 2 mol CH3COOH 82 g of NaCH3COO x (1 mol / 82.03 g) = 1 mol NaCH3COO For a buffer composed of a weak acid and its conjugate base, [H+] = Ka(Ca/Cb) = 1.8 x 10-5(2/1) = 3.6 x 10-5 pH = 4.44 Note, it is adequate to use the moles of HA and A- in place of their final concentrations. 2. Which of the following pairs of solutions would not result in a buffer upon mixing? 1. 100 mL of 10 mM NaOH & 80 mL of 20 mM NH4Cl 2. 20 mL of 0.3 M NaF & 12 mL of 0.4 M HCl 3. 0.4 L of 10 mM HClO3 & 0.5 L of 8 mM C6H5NH2 Correct 4. 2 L of 1.35 M Ba(OH)2 & 3 L of 2 M CHOOH Explanation: A buffer prepared by a neutralization reaction requires a weak acid mixed with less strong base or a weak base mixed with less strong acid. The only pair of solutions which fails to satisfy this constraint is 0.4 L of 10 mM HClO3 and 0.5 L of 8 mM C6H5NH2. 3 Consider the following acids and their provided pKas. Rank them in terms of increasing strength of their conjugate bases. CH3COOH pKa = 4.75 CH3CHOHCOOH pKa = 3.85 CHOOH pKa = 3.74 CH3CH2COOH pKa = 4.88 1. CHOOH < CH3CHOHCOOH < CH3COOH < CH3CH2COOH Correct 2. CH3CH2COOH < CHOOH < CH3CHOHCOOH < CH3COOH 3. CH3COOH < CH3CH2COOH < CHOOH < CH3CHOHCOOH 4. CH3CHOHCOOH < CH3COOH < CH3CH2COOH < CHOOH Explanation: The strength of an acid's conjugate base is directly proportional to the pKa of the acid. 4 Which of the following buffers could absorb the greatest amount of strong base before being exhausted? 1. 45 mL of 2 mM N2H5Cl, 4 mM N2H4 2. 3.2 L of 0.4 M HClO, 0.5 NaClO Correct 3. 2 L of 9 mM HF, 7 mM NaF 4. 0.3 L of 0.4 M NH4Cl, 0.6 M NH3 5. 20 mL of 5 M CHOOH, 4 M NaCHOO Explanation: Any strong base added to a buffer will react with and be neutralized by the weak acid species that is present. The buffer with the greatest amount of weak acid is 3.2 L of 0.4 M HClO, 0.5 NaClO. 5. If one added 200 mL of 6 M HCl to 1 L of a buffer composed 4.2 M CH3COOH and 6.6 M NaCH3COO, what would be the resulting pH? The Ka of CH3COOH is 1.8 x 10-5. 1. 5.32. 4.9 3. 5.1 4. 4.7 Correct Explanation: The general reaction that takes place is A- + H+  HA. Initial amounts of each reactant are 6.6, 1.2 and 4.2 moles respectively. After the reaction goes to completion, the equilibrium concentrations are 5.4, 0 and 5.4 respectively. There is no need to calculate the final concentrations. For a buffer composed of a weak acid and its conjugate base, [H+] = Ka(Ca/Cb) = 1.8 x 10-5(5.4/5.4) = 1.8 x 10-5 pH = 4.7 6. How many buffer regions and equivalence points would be visible on the titration curve of a weak tetraprotic acid? 1. 3, 1 2. 3, 4 3. 1, 4 4. 4, 1 5. 4, 4 Correct Explanation: Each ionizable proton will produce one buffer region and one equivalence point. 7. A 100 mL sample of 0.1 M H3PO4 is titrated with 0.2 M NaOH. What is the pH of the solution after 100 mL of NaOH has been added? Phosphoric acid has Ka1 = 7.5 x 10-3, Ka2 = 6.2 x 10-8 and Ka3 = 2.1 x 10-13. 1. 4.10 2. 8.51 3. 4.67 4. 7.40 5. 9.94 Correct Explanation: 0.1 M H3PO4 x 100 mL = 0.01 moles H3PO4 0.2 M NaOH x 100 mL = 0.02 moles OH- Two equivalents of OH- have been added, and the solution will be at the second equivalence point, composed primarily of the amphoteric species HPO42-. [H+] = (Kax·Kay)1/2 = (6.2 x 10-8·2.1 x 10-13)1/2 = 1.14 x 10-10 pH = 9.94 8. What will be the pH at the first equivalence point of a titration of 0.2 M H2SO4 with 0.2 M NaOH? The Ka for HSO4- is 2 x 10-2. 1. 1.45 Correct 2. 1.35 3. 7.00 4. not enough information Explanation: At its first equivalence point, the predominant species of this titration will be HSO4-. Because the analyte and titrant are equimolar, the [HSO4-] will be half of Ca, or 0.1 M. This is not an amphoteric solution because HSO4- cannot function as a base. Because the Ka of HSO4- is so large, a full solution is required. R HSO4-  H+ SO42- I 0.1 0 0 C - x + x + x E 0.1 - x x x Ka = 2 x 10-2 = (x)(x)/(0.1 -x) x2 + 2 x 10-2x - 2 x 10-3 = 0x = [H+] = 0.0358 pH = 1.45 9. All of the salts below have the same approximate molar solubility except for one. Which is it? 1. TlBr Ksp = 4.00 x 10-6 2. PbI2 Ksp = 7.47 x 10-9 3. AgSCN Ksp = 1.16 x 10-12 Correct 4. CsIO4 Ksp = 5.16 x 10-6 Explanation: Silver thiocyanate is a salt composed of one cation and one anion and thus its molar solubility is approximately equal to the square root of its Ksp. The best approximation is therefore (10-12)1/2 = 10-6. 10. The Ksp of MgNH4PO4 at 25 °C is 2.5 x 10-13. What is its molar solubility at this temperature? (Hint: do the RICE diagram for this one.) 1. 3.2 x 10-4 2. 4.0 x 10-5 3. 6.3 x 10-5 Correct 4. 1.2 x 10-3 Explanation: MgNH4PO4 will dissolve into Mg2+ NH4+ and PO43-. Each will be produced in equal proportions and so Ksp = x3 = 2.5 x 10-13 11.2 What would be the molar solubility of Sn(OH)2 (Ksp = 10-26) in pH 13 NaOH solution? 1. 1 x 10-24 Correct 2. 4 x 10-24 3. 1 x 10-28 4. 4 x 10-28 5. not enough information Explanation: [OH-] = 0.1 M molar solubility = (Ksp/[OH-]2) = [10-26/(10-1)2] = 10-24 12. Consider the table below. Which anion would be the best for separating Pb2+ from Ca2+? Which would be the worst? Ksp values C2O4- CO32- SO42- IO3- Pb2+ 2.74 x 10-11 3.3 x 10-14 1.6 x 10-8 1.2 x 10-13 Ca2+ 2.57 x 10-9 8.7 x 10-9 4.93 x 10-5 6.44 x 10-7 1. C2O4- & SO42- 2. IO3- & SO42- 3. CO32- & IO3- 4. IO3- & C2O4- Correct 5. CO32- & C2O4- Explanation: The Ksp values for IO3- are farthest apart and the values for C2O4- are closest together. 13. A student used the equation [H+] = (Ka·Ca)1/2 to calculate [H+] and got a value of 0.4 M. The actual value was determined experimentally to …


View Full Document

UT CH 302 - Practice Exam 2

Documents in this Course
Exam 2

Exam 2

6 pages

Exam 3

Exam 3

8 pages

Acids

Acids

21 pages

Exam 3

Exam 3

7 pages

SYLLABUS

SYLLABUS

15 pages

ex1s08

ex1s08

11 pages

Load more
Download Practice Exam 2
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Practice Exam 2 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Practice Exam 2 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?