Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110

Chem 110, Dr. Paselk

Hour Exam I

 Name
Fall 2002

(100 pts)
 Lab

Answer Key

This examination is scheduled for fifty (50) minutes. At the end of this fifty minute period you will have five (5) minutes to place your examination at the front of the examination room.

1. (15 @ 3 ea) True/False. Indicate whether the following statements are True (T) or False (F).

A state function depends mostly on the initial and final conditions of a process, with only a slight contribution by the temperature. _______F________

Oxidation half-reactions occur at the anode. _______T________

The Nernst Equation is used to calculate the free energy of a galvanic cell. _______F________

Ammonia is a strong base. _______F________

For covalent molecules the electronic geometry is the same as the molecular shape _______F________

2. (20 @ 4 ea) Answer the following in the spaces provided: Significant Figures Count!

What is the solubility of BaCO3 if it’s Ksp=1.6 x 10-9? __4.0 x 10-5___

Which of the following materials is best to prepare a buffer with a pH of 5.5? _______c________

(a) H2S / NaSH Ka= 1.0 x 10-7

(b) HOCl / NaOCl Ka= 3.5 x 10-8

(c) HC3H5O2 / Na C3H5O2 Ka= 1.3 x 10-5

For which of the following reactions is D So expected to be negative? _______a & c________

(a) H2O(g) ´ H2O(l)

(b) Br2(g) + Cl2(g) ´ 2 BrCl(g)

(c) 2H2(g) + O2(g) ´ 2 H2O(l)

A particular chemical reaction has value of D H= -2.0 J and D S= -73 J/K @ 25°C. Is it spontaneous? _No, non-spontaneous__

What is the shape of IF4-1? _Square planar__

3. (10 @ 5 ea) Solve the following stoichiometry problems, stating answers to the proper number of significant figures. SHOW WORK FOR CREDIT!

a. What is the concentration of hydrogen ions in a solution created by adding 500.0 mL of 1.5 M NaOH to 500.0 mL of 1.0 M H2SO4 (Ka2= 1.2 x 10-2)

   H+  + HSO4-  + OH-  H20 + SO42-
Before rxn  0.5 mol   0.5 mol    0.75 mol   - 0
after rxn  0  
0.25 mol
   0   -  
0.25 mol

50:50 mix of acid and salt after reaction, therefore pH = pKa2 = 1.9, and

[H+] = 1.2 x 10-2

b. What is the concentration of ammonium ions in a solution created by adding 500.0 mL of 0.50 M HCl to 500.0 mL of 0.50 M ammonia?

Equal numbers of moles of strong acid (HCl) and weak base (NH3) are combined, so all of weak base is converted to salt.

Therefore. concentration of ammonium = (0.50M)(0.500L)/(1L) = 0.25M

4. ( 10) For the reaction

SF4(g) + F2(g) SF6(g)

the value for D G° is —374 J/K· mole. Find the value of D G°f for SF4(g) given that: for F2(g) S° = 203 J/mol, while for SF6(g) S° = 292 J/mol and D H°f = -1209 kJ/mol. SHOW WORK FOR CREDIT!

DSF6(g) = DH° - TD S° = -1209 kJ/mol - (298K)(292 J/K*mol) = -1.296 x 106J/mol

DF2(g) = DH° - TD S° = 0 - (298K)(203 J/K*mol) = -6.05 x 104J/mol

Drxn = Dprod - Dreactants = DSF6(g) - (60.5 kJ/mol + DSF4(g)) = -374 J/mol

DSF4(g) = -1296 kJ/mol + 605 kJ/mol + 0.374 kJ/mol = -1235 kJ/mol

 

5. (10) Calculate the pH of a solution of 0.95 M acetic acid and 1.25 M potassium acetate. The Ka of acetic acid (HC2H3O2) is 1.8 x 10-5. SHOW WORK FOR CREDIT!

pH = pKa - [HA]/[A-]; pKa = -logKa = - log(1.8 x 10-5) = 4.75

pH = 4.75 - log(0.95/1.25) = 4.75 + 0.119 = 4.87

6. (10) Calculate the [Cu2+] in a solution prepared by dissolving 0.10 mol of copper(II) nitrate in 1.00 L of 1.00 M ammonia (assume no volume change). For the copper ammonia complex ion Kdiss = 1.0 x 10-12; Kb = 1.8 x 10-5 for ammonia. SHOW WORK FOR CREDIT!

Cu(NH3)42+ Cu2+ + 4 NH3

Kdiss = [Cu2+][NH3]4/[Cu(NH3)42+]; [NH3] = 1.00M - 4(0.10 M) = 0.60 M

[Cu2+] = Kdiss([Cu(NH3)42+]/[NH3]4 = (1.0 x 10-12)(0.10M)/(0.60M)4

[Cu2+] = 7.7 x 10-13M

7. (6) Write the NET ionic equation for the reaction which occurs when the following are mixed. (If there is no reaction, indicate by "NR." )

a. Copper(II) nitrate is mixed with excess ammonia.

Cu2+ + 4 NH3 Cu(NH3)42+

b. Sodium sulfide is mixed with excess hydrochloric acid.

S- + 2 H+ H2S

c. Tin(IV) sulfide is mixed with 6 M hydrochloric acid.

SnS2(s) + 6 Cl- + 4 H+ SnCl62- + 2 H2S

8. (19) Consider the galvanic cell diagramed below:

Sn | Sn2+(0.10M) || Ag+(0.20M) | Ag

where E° for Sn2+/Sn = - 0.133V and E° for Ag+/Ag = +0.799V

a. Write out the balanced net ionic equation for the cell. SHOW WORK!

 Sn2+ + 2 e-   Sn  E° = - 0.133 V
 Ag+ + e-   Ag  E° = + 0.799 V
 Reversing the tin reaction and adding:  
 Sn   Sn2+ + 2 e-  E° = + 0.133 V
 2(Ag+ + e- Ag)
 E° = + 0.799 V
 2Ag+ + Sn   Sn2+ + 2Ag  E° = + 0.932 V

b. Calculate the cell potential, E, for this cell. SHOW WORK FOR CREDIT!

E = E° - (RT/nF)lnQ; Q = [Sn2+]/[Ag+]2

E = 0.932 V - {(8.314 J/K*mol)(298 K)/2(96,485 C/mol)}ln(0.10)/(020)2

E = 0.932 V - (+1.176 x 10-2 V) = 0.920 V (V = J/C)

c. How much free energy is available from this cell? SHOW WORK FOR CREDIT!

DG = -nFE = - 2 mol(96,485 C/mol)(0.920 V) = - 1.77 x 10-2


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Last modified 13 June 2006