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Chem 126 Common 2 March 7, 2008 Dr. Ellis

[ ]

( )

+5.6 1 / 0.200 0.219 . Because HI is a strong acid it exists as H and I 1 128

ions in solution. log log 0.219 0.66

gHI mol HIHI x L Mg HI

pH H

−

+

= =

= − = − =

( )5log log 8.86 10 4.05

14 9.95

pOH OH x

pH pOH

− − = − = − = = − =

NONE OF THE MULTIPLE CHOICE PROBLEMS REQUIRES EXTENSIVE OR TIME COMSUMING CALCULATIONS. IF YOUR METHOD REQUIRES EXTENSTIVE CALCULATIONS IT IS EITHER WRONG OR HARDER THAN WHAT IS REQUIRED. 1. What is the equilibrium constant for the reaction: 2 ( ) ( ) 3C(g)+D(s)A s B g+ ⇌ , given that in a 5.00 L

container A = 0.35 moles, B = 0.20 moles, C = 0.25 moles, and D = 0.21 moles at equilibrium? (a) 0.13 M2 (b) 0.027 M2 (c) 1.25 M2 (d) 0.39 M2 (e) 0.0031 M2

2. Some silicon dioxide, water, silane (SiH4), and oxygen are at equilibrium in a sealed container at a particular

temperature: ( ) ( ) ( ) ( )2 2 4 22 2 0SiO s H O g SiH g O g H+ + ∆ >⇌ In which of the following cases does the reaction NOT proceed to right to reestablish equilibrium?

(a) Some silicon dioxide is added. (b) Some water vapor is added. (c) Some oxygen is removed. (d) The temperature is increased. (e) The volume of the container is increased.

3. All of the following compounds are soluble in water and are strong electrolytes except

(a) NaI. (b) PbI2. (c) Ca(CH3COO)2. (d) NH4CH3COO. (e) MgCl2.

4. The concentration of an HCl solution is 0.25 M. What is the pH? ( )log log 0.25 0.60pH H + = − = − =

(a) -1.38 (b) -0.60 (c) 0 (d) 0.60 (e) 1.38

5. 5.60 g of HI is dissolved in enough water to make a 0.200 L solution. What is the pH?

(a) -0.75 (b) -1.45 (c) 1.52 (d) 0.75 (e) 0.66

6. Calculate the pH of a 4.43 x 10-5 M barium hydroxide, Ba(OH)2, solution.

(a) 4.05 (b) 9.95 (c) 4.35 (d) 9.65 (e) none of the above

[ ] [ ] [ ][ ]

( )3 320.0500.20 0.250.040 ; 0.050 ; 0.0031

5.0 5.0 0.040eq

C Mmol molB M C M K ML L B M

= = = = = = =

Chem 126 Common 2 March 7, 2008 Dr. Ellis

4.77 5log . , 10 10 1.70 10pOHpOH OH So OH x M− − − − − = − = = =

30.66 35 23.1

10.25 115 28.75

1

mol mLmmol HNO x mmolLmol mLmmol KOH x mmol

L

= =

= =

( ) ( ) ( ) ( )3 2 3

23.1 28.75 -23.1 -23.1

E 0 5.65

HNO aq KOH aq H O l KNO aqI mmol mmolC mmol mmol

mmol

+ → +

5.65log log 1.42; 14 12.5835 115

mmolpOH OH pH pOHmL

− = − = − = = − = +

( )log log 0.04 1.40 (strong acid + weak acid; ignore weak acid)pH H + = − = − =

( ) ( ) ( )

[ ]

4

4

6.03 7 4

3.6 10 0 0 3.6 10

10 10 9.33 10 ; 3.6 10 9.33 10pH

HB aq H aq B aq

I x MC x x xE x x x x

x H x M HB x x

+ −

−

−

+ − − − −

+

− + +

−

= = = = = −

⇌

[ ]( )

7 4

279

4

3.59 10

9.33 102.4 10

3.59 10a

x M

xH BK x M

HB x

− −

−+ −

−−

=

= = =

7. The pOH of a NaOH solution is 4.77. What is the concentration of the OH- ion?

(a) 0.00848 M (b) 5.89x10-10 M (c) 3.40 x 10-5 M (d) 0.00424 M (e) 1.70x10-5 M

8. What is the pH of a solution formed by mixing 35.0 mL of 0.66 M HNO3 and 115 mL of 0.250 M KOH?

(a) 1.42 (b) 12.69 (c) 12.58 (d) 13.21 (e) none of the above

9. A solution consists of 0.75 M phenol (C6H5OH, a weak acid, with Ka = 1.0 x10-10 M) and 0.040 M HNO3.

What is the pH?

(a) 1.40 (b) 0.125 (c) 0.10 (d) 1.30 (e) none of the above 10. A 3.6x10-4 M aqueous solution of a weak acid, HB, has a pH of 6.03. Calculate Ka, the acid-dissociation

constant, for HB. (HB is some unknown weak acid. Its actual composition is irrelevant for solving the problem.)

(a) 2.6x10-3 M

(b) 5.7x10-7 M

(c) 6.0x10-5 M

(d) 2.4x10-9 M

(e) none of the above

11. Which of the following solutions is neutral?

(a) CH3NH2(aq) (b) CsI(aq) (c) KCH3COO(aq) (d) HCOOH(aq) (e) FeCl3(aq)

Chem 126 Common 2 March 7, 2008 Dr. Ellis

12. For the reaction: ( ) ( ) ( ) ( )3 3 4 3aq aq aq + aqNH CH COOH NH CH COO+ −+ " , what is the conjugate base?

(a) NH3 (b) CH3COOH (c) NH4+ (d) CH3COO- (e) OH-

13. Which of the following solutions has the highest solubility for iron (III) hydroxide?

(a) pure water

(b) water buffered to a pH of 2 Le Chatelier’s Principle:

(c) water buffered to a pH of 5 It is most soluble in the solution with the lowest [OH-]

(d) water buffered to a pH of 10 because stress is too little [OH-]

(e) water buffered to a pH of 12

14. What is the concentration of a 5.00 L HBr solution if 30.0 mL of it is completely neutralized by 95.7 mL

0.25 M NaOH? (a) 0.798 M

(b) 0.754 M

(c) 0.250 M

(d) 0.190 M

(e) 0.0479 M

15. In which of the following reactions does the underlined compound act as a Lewis base?

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

4

4

3 2

2 3 3

3 33 3

.

.

. where the N is bonded to the Ga

I NH aq H O l NH aq OH aq

II NH aq H O l NH aq H O aq

III NH Ga OH NH Ga OH

+ −

+ +

+ +

+ +

+

⇌

⇌

⇌

(a) only I (b) only II (c) only III (d) II and III (e) I, II, and III

16. Calculate the pH of a solution that is 0.21 M in HF and 1.43 M KF.

(HF: Ka = 6.76 x 10-4 M) (a) 0.68 (b) 1.92 (c) 4.00 (d) 3.17 (e) 2.34

( )

[ ]

4log log 6.76 10 3.17

1.43log 3.17 log 4.000.21

a a

oa

o

pK K x

BpH pK

HB

−

−

= − = − =

= + = + =

0.25 95.7 1 23.931 1

23.93 0.7975 0.79830

mol NaOH mL NaOH mmol HBrx x mmol HBrL mmol NaOH

mmol HBr mol M HBrmL L

=

= =

( ) ( ) ( ) ( )2NaOH aq HBr aq H O l NaBr aq+ → +

Chem 126 Spring 2007 Common 2 March 9 Dr. Ellis

0.85 0 0 0.85

I MC x x xE x x x

− + +

−

[ ]

( )

22 2

2

2

nd

1.15 100.85

1.15 10 0.85 0.0989 (first iteration with x = 0)

0.0929 (2 iteration)

H ClO x xHClO x

x x x

+ −

−

−

= =−

= − =

=

= rd th

+

0.0933 (3 and 4 iteration )

H 0.093M =

[ ]2

0.0933100% 100% 11%0.85

o

x x xHClO

= =

PROBLEMS (12 points each; 36 total) I. The Ka for Chlorous acid, HClO2, is 1.15 x 10-2 M.

(a) (4 PT) What is the [H+] for a 0.85 M solution of this acid? Show work, including an ICE table. Use method of successive approximation for full credit. ( ) ( ) ( )2 2HClO aq H aq ClO aq+ −+⇌

(b) (4 PT) What is the percent dissociation of chlorous acid? Percent dissociation =

(c) (4 PT) What happens to the percent dissociation as the solution of chlorous acid is diluted by the addition of water? Why? (Circle correct choice and give reason.)

-It decrease because ______________________________________________________

-It stays the same because _____________________________________________________

-It increases because dilution stress—too few species in solution; reaction moves to make more species (using Le Chatelier’s principle)

Chem 126 Spring 2007 Common 2 March 9 Dr. Ellis

II. Separation of ions by precipitation. A solution contains 0.30 M AgNO3 and 0.50 M Ca(ClO4)2.

Sodium carbonate will be added to the solution. The Ksp for silver carbonate is 8.5 x 10-12 M3. The Ksp

for calcium carbonate is 5 x 10-9 M2.

(a) (2 PT) Write the equation for the dissolution of Ag2CO3, (Include (s) and (aq) and charge on ions)

( ) ( ) ( )22 3 32Ag CO s Ag aq CO aq+ −+⇌

(b) (2 PT) Write the solubility product expression for the dissolution of silver carbonate (Ksp = ?): 2 2 12 3

3 8.5 10spAg CO K x M+ − − = =

(c) (2 PT) At what carbonate concentration does silver carbonate start to precipitate?

( )

12 32 113 2 2

8.5 10 9.4 100.3

spK x MCO x MMAg

−− −

+ = = =

(d) (2 PT) At what carbonate concentration does calcium carbonate start to precipitate? Hint, first do

steps (a) and (b) above.

( ) ( ) ( )2 2 2 2 9 23 3 3 5 10spCaCO s Ca aq CO aq Ca CO K x M+ − + − − + = = ⇌

9 32 83 2

5 10 1.0 100.5

spK x MCO x MMCa

−− −

+ = = ==

(e) (2 PT) As the carbonate ion concentration increases, what compound is the first to precipitate? Why?

Silver carbonate because it starts to precipitate at a smaller carbonate conc. (9.4 x10-11 is smaller than

1x10-8).

(f) (2 PT) For the compound which precipitates first, what is the concentration of its remaining ions

when the second compound begins to precipitate? 0.5 0.512 32

82 23 3

8.5 10 0.0291 10

sp spK K x MAg Ag Mx MCO CO

−+ +

−− −

= = = =

Chem 126 Spring 2007 Common 2 March 9 Dr. Ellis

( ) ( ) ( ) ( )14 2

52 10

10 2.09 104.79 10

0.444 - - - 0 0 0.444

wb

a

K MCN aq H O l HCN aq OH aq K x MK x M

I MC x x xE x

−− − −

−+ + = = =

− + +

−

⇌

[ ]

( )

( )

25

5

2.09 100.444

2.09 10 0.444 0 0.00305

log 0.00305 2.52 14 2.52 11.48

x x

HCN OH x xxCN

x x

pOH pH

−

−

−

−

= =−

≈ − =

= − = = − =

( ) ( ) ( ) ( )2 3 37 53 4 42 3 2 1.4 10

0.35 0 3 2

spCu PO s Cu aq PO aq K x MI excess MC x x xE excess

+ − −+ =

− + +

⇌

( ) ( )

( )

3 2 3 22 3 374

37

373 2 37

0.35+3 2

Note that x is the solubility.

0.35+3 2 1.4 10

3 must be very close to 0 because 1.4 10 is so small. Thus,

1.4 100.35 4 1.4 10 ;

x x

Cu PO x x x

x x

xx x x

+ − −

−

−−

= =

= =( )

193 9.0 10 /

0.35 4x mol L−=

III (a) (6 PT) What is the pH of a 0.444 M potassium cyanide, KCN, solution? For HCN, the Ka is 4.79 x 10-10 M. Include (i) writing the reaction (the net ionic reaction, that is, what takes place when KCN is added to water), (ii) finding the value of its equilibrium constant (Kb), and (iii) setting up the ICE table. (b) (6 PT) What is the solubility of copper (II) phosphate in a 0.35 M solution of Cu(NO3)2. The Ksp is 1.4 x 10-37 M5. Include (i) writing the reaction for the dissolution of copper (II) phosphate, (ii) setting up an ICE table, and (iii) writing the solubility product expression (Ksp = ?).