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100

What the oxidation numbers for the following compounds?

(a) N2O

(b) NO3-

(c) NH4Cl

(a) +1, -2

(b) +5, -2

(c) -3, +1, -1

100

What is the rate constant of the reaction, given the following set of experiments? Be sure to give the correct units.

A + B → C + D

Exp     [A]      [B]    Initial Rate (M/s)

1      0.040  0.040     2.2 x 10-4

2      0.040  0.080     4.4 x 10-4

3      0.080  0.080     8.8 x 10-4

rate= k[A]1[B]1

k= rate / ([A]1[B]1)

k= (2.2 x 10-4 M s-1) / (0.040 M x 0.040 M)

k= 0.14 M-1s-1

100

Determine if the solution is acidic, basic, neutral, or cannot be determined from the information given. (Unless otherwise indicated, all solutions are 25ºC). 


1. pH = 3.8

2. pH > 4.00

3. [H+] = 3.6 x 10-10

4. 2.00 x 10-9 M HCl

5. At a temperature other than 25ºC, a solution has a higher [OH-] than [H+] and a pH of 7

6. pH = 6.90 at 37ºC (human body temperature; Kw = 2.5 x 10-14)

1. Acidic

2. Not enough information

3. Basic

4. Acidic, almost neutral

5. Basic

6. Basic, take the pKw/2 to determine the neutral pH

200

What are the oxidation numbers for the following compounds?

(a) Mg(OH)2

(b) Ca(ClO4)2

(c) Cr2O7-2

(a) +2, -2, +1

(b) +2, +7, -2

(c) +6, -2

200

The initial concentration of a reactant is 0.520 M and the rate constant is 0.00250 M-1s-1. Calculate the final concentration of the reactant after 10.0 min for a second-order reaction.

1/[A] = kt + 1/[A]0

1/[A] = (.0025 M-1s-1)(600 s) + (1/.52 M)

1/[A] = 3.423 M

[A]= 0.292 M

200

Calculate the pH of a mixture of 40.0 mL of 0.100 M HCl and 60.0 mL of 0.600 M HF. (Ka = 6.5 x 10-9)

Strong acid + weak acid:

HCl (4.00 mmol) + HF (36.0 mmol)

[H+] from HCl= 4.00 mmol/100.0 mL= 0.0400 M, pH contributed from only HCl= 1.40

R      HF     ⇌      H+     +      F-

I   0.360 M  0.0400 M           0

C    -x              +x             +x

E  0.360-x    0.0400 +x        x 

[H+] prior to dissociation is due to H+ from hydrochloric acid. The presence of this H+ in solution partially suppresses the dissociation of HF.

6.5 x 10-9 = (0.0400 + x)(x) / (0.360 -x) ignore first and third x.

x= 5.85 x 10-8 (very insignificant compared to the pH from HCl)

[H+]solution = 0.0400 M + 5.85 x 10-8 M = 0.0400 M

pH= 1.40

300

Balance Cl2 → Cl- + ClO3- in acidic solution using the half-reaction method.

Cl2 → Cl-                      Cl2 → ClO3-

Cl2 → 2 Cl-             Cl2 → 2 ClO3-

Cl2 → 2 Cl-         6 H2O + Cl2 → 2 ClO3- + 12 H+

2 e- + Cl2 → 2 Cl-       6 H2O + Cl2 → 2 ClO3- + 12 H+ + 10 e-

Lowest common denominator of 10 and 2= 10

(2 e- + Cl2 → 2 Cl-) x 5

(6 H2O + Cl2 → 2 ClO3- + 12 H+ + 10 e-) x 1

6 Cl2 + 6 H2O → 10 Cl- + 2 ClO3- + 12 H+

300

Sulfur dioxide reacts with oxygen gas to produce sulfur trioxide. At equilibrium, the partial pressures of each gas were found to be 0.100 atm for SO2, 0.300 atm for O2, and 0.450 atm for SO3. Calculate the equilibrium constant Kp.

2 SO2 + O2 ⇄ 2 SO3

Kp= [SO3]2/ ([SO2]2[O2]) = [0.450]2/ ([0.100]2[0.300]) = 67.5

300

If 6.0 mL of 1.0 M HCl is added to 100.0 mL of a buffer solution (0.24 M NH3 and 0.20 M NH4+), what is the resulting pH? Ka = 5.6 x 10-10.

HCl is SA, base in buffer will counteract:

R   NH3   +   HCl    -->   NH4+    +     Cl- 

I  24.0 mmol 6.0 mmol   20.0 mmol N/A

C   -6.0       -6.0          +6.0          +6.0 

E  18.0          0             26.0           w.c. 

pH= -log(5.6 x 10-10) + log (18.0/26.0)= 9.09

400

What is the cell potential of the reaction shown below at 298 K?

2 Al (s) + 3 Cu+2 (aq) → 2 Al+3 (aq) + 3 Cu (s)

E°cell= 2.00 V, [Al+3]= 0.100 M, [Cu+2]= 2.50 M

E= E° - (0.0591/n) log Q

Q (reaction quotient) = [Al+3]2/[Cu+2]3 = (0.100)2/(2.5)3

Q= 6.4 x 10-4, n= 6 (6 electrons transferred)

2 Al (s) → 2 Al+3 + 6 e-

E= 2.00 - (0.0591/6) log (6.4 x 10-4) = +2.03 V

400

I2(g) + Br2 (g) ⇌ 2 IBr (g)

2.00 M of I2 and 2.00 M of Br2 are initially present in a reaction mixture. Calculate the equilibrium concentration of IBr at 425 K. Kc= 100.0

*HINT: do NOT neglect x.

R   I2(g) + Br2 (g)  ⇌   2 IBr (g)

I    2.00      2.00              0      

C   -x           -x               +2x   

E   2-x         2-x              +2x    

Kc= 100= (2x)2/ ((2-x)(2-x)) Take the square root of both sides of the equation:

10= 2x / (2-x) ; x= 1.67 M

[IBr] at equilibrium= 2(1.67)= 3.34 M

400

The Ksp for Ca3(PO4)2 is 7.1 x 10-33. Calculate the molar solubility of Ca3(PO4)2.

R  Ca3(PO4)2   ⇄ 3 Ca+2 + 2 PO4-3

I                          0             0     

C                        +3x        +2x   

E                         3x            2x    

Ksp= [Ca+2]3[PO4-3]2= 7.1 x 10-33 = (3x)3(2x)2= 108x5,

(6.57 x 10-35)(1/5)= x = 1.46 x 10-7

500

What is the pH of a 0.820 M aqueous NH3 solution? Kb (NH3)= 1.8 x 10-5

R   NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)

I    0.820          N/A          0                0         

C     -x                           +x                +x      

E   0.820-x                     +x                +x       

1.8 x 10-5 = x2/0.820, x= +/- 3.84 x 10-3

[OH-]= +3.84 x 10-3, pOH= 2.42

pH= 14-2.42= 11.58

500

An equilibrium mixture of I2 (g), Cl2 (g), and ICl (g) at 298 K has partial pressures of 0.0100 atm, 0.0100 atm, and 0.0900 atm, respectively. What is the ΔG° for the reaction? (R= 8.314 J/mol K)

I2 (g) + Cl2 (g) ⇌ 2 ICl (g)

ΔG°= -RTlnK

K= (0.0900)2/ ((0.0100)1(0.0100)1)= 81.0

ΔG°= -(8.314 J/mol K)(298 K)ln(81.0)= -1.09 x 104 J/mol (-10.9 kJ/mol)