CHAPTER 16 CHEMICAL EQUILIBRIUM - Chemeketa ...faculty.chemeketa.edu/lemme/CH 121/solutions/Hein9thCh16.pdf13. If pure HI is placed in a vessel at 700 K, some of it will decompose

CHAPTER 16

CHEMICAL EQUILIBRIUM

SOLUTIONS TO REVIEW QUESTIONS

1. At 25°C both tubes would appear the same and contain more molecules in the gaseousstate than the tube at 0°C, and less molecules in the gaseous state than the tube at 80°C.

2. The reaction is endothermic because the increased temperature increases theconcentration of product present at equilibrium.

3. In an endothermic process heat is absorbed or used by the system so it should be placedon the reactant side of a chemical equation. In an exothermic process heat is given off bythe system so it belongs on the product side of a chemical equation.

4. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.

5. Free protons do not exist in water because they are hydrated forming

6. The sum of the pH and the pOH is 14. A solution whose pH is would have a pOHof 15.

7. Acids stronger than acetic acid are: benzoic, cyanic, formic, hydrofluoric, and nitrousacids (all equilibrium constants are greater than the equilibrium constant for acetic acid).Acids weaker than acetic acid are: carbolic, hydrocyanic, and hypochlorous acids (allhave equilibrium constants smaller than the equilibrium constant for acetic acid). All haveone ionizable hydrogen atom.

8. The order of solubility will correspond to the order of the values of the solubility productconstants of the salts being compared. This occurs because each salt in the comparisonproduces the same number of ions (two in this case) for each formula unit of salt thatdissolves. This type of comparison would not necessarily be valid if the salts beingcompared gave different numbers of ions per formula unit of salt dissolving. The order is:

9. (a)

Each salt gives 3 ions per formula unit of salt dissolving. Therefore, the salt withthe largest (in this case ) is more soluble.

(b) has a greatermolar solubility than even though its is smaller, because the produces more ions per formula unit of salt dissolving than BaCrO4.

Ag2CrO4KspBaCrO4,Ag2CrO4Ksp Ag2CrO4 = 1.9 * 10-12.Ksp BaCrO4 = 8.5 * 10-11;

Ag2CrO4Ksp

Ksp Ag2CrO4 = 1.9 * 10-12.Ksp Mn(OH)2 = 2.0 * 10-13;

AgC2H3O2, PbSO4, BaSO4, AgCl, BaCrO4, AgBr, AgI, PbS.

-1

H3O+.(H+)

(NO2)

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HEINS16-229-255v4.qxd 12/30/06 2:42 PM Page 229

Let

Let

has the greater solubility.

10.

Concentration After Initial Concentrations Added Equilibrium Shifts

_ _ _ _ _ 0.11 M

0.010 mol

_ _ _ _ _ 0.09 M

The initial concentration of in the buffer solution is very low because of the large excess of acetate ions. 0.010 mol of HCl is added to one liter of thebuffer solution. This will supply The added creates a stress on the rightside of the equation. The equilibrium shifts to the left, using up almost all the added reducing the acetate ion by approximately 0.010 M, and increasing the acetic acid byapproximately 0.010 M. The concentration of will not increase significantly and thepH is maintained relatively constant.

11. In a saturated sodium chloride solution, the equilibrium is

Bubbling in HCl gas increases the concentration of creating a stress, which willcause the equilibrium to shift to the right, precipitating solid NaCl.

12. The rate of a reaction increases when the concentration of one of the reactants increases.The increase in concentration causes the number of collisions between the reactants toincrease. The rate of a reaction, being proportional to the frequency of such collisions, asa result, will increase.

Cl-,

Na+(aq) + Cl-(aq) ÷ NaCl(s)

H+

H+,H+0.010 M H+.

11.8 * 10-5 M2H+

0.10 MC2H3O2

-1.9 * 10-5 M1.8 * 10-5MH+

0.10 MHC2H3O2

HC2H3O2 ÷ H+ + C2H3O2-

Ag2CrO4

y = A3

1.9 * 10-12

4 = 7.8 * 10-5 mol Ag2CrO4>L

Ksp = [2y]2[y] = 1.9 * 10-12

y = molar solubility of Ag2CrO4

Ksp = [Ag+]2[CrO4

2-]Ag2CrO4(s) ÷ 2 Ag+ + CrO4

2-

y = 28.5 * 10-11 = 9.2 * 10-6 mol BaCrO4>LKsp = [y][y] = 8.5 * 10-11

y = molar solubility of BaCrO4

Ksp = [Ba2+][CrO4

2-]BaCrO4(s) ÷ Ba2+ + CrO4

2-

- 230 -

- Chapter 16 -


13. If pure HI is placed in a vessel at 700 K, some of it will decompose. Since the reaction isreversible HI molecules will react to produce and

14. An increase in temperature causes the rate of reaction to increase, because it increasesthe velocity of the molecules. Faster moving molecules increase the number andeffectiveness of the collisions between molecules resulting in an increase in the rate ofthe reaction.

15. A catalyst speeds up the rate of a reaction by lowering the activation energy. A catalyst isnot used up in the reaction.

16.

When A and B are initially mixed, the rate of the forward reaction to produce C and D is at its maximum. As the reaction proceeds, the rate of production of C and Ddecreases because the concentrations of A and B decrease. As C and D are produced,some of the collisions between C and D will result in the reverse reaction, formingA and B. Finally, an equilibrium is achieved in which the forward rate exactly equalsthe reverse rate.

17.

As water is added (diluting the solution from 1.0 M to 0.10 M), the equilibrium shifts tothe right, yielding a higher percent ionization.

18. The statement does not contradict Le Chatelier’s Principle. The previous question dealswith the case of dilution. If pure acetic acid is added to a dilute solution, the reaction willshift to the right, producing more ions in accordance with Le Chatelier’s Principle. But,the concentration of the un-ionized acetic acid will increase faster than the concentrationof the ions, thus yielding a smaller percent ionization.

19. At different temperatures, the degree of ionization of water varies, being higher at highertemperatures. Consequently, the pH of water can be different at different temperatures.

20. In pure water, and are produced in equal quantities by the ionization of the watermolecules, Since and they will always be identical for pure water. At 25°C, they each have the value of 7, but athigher temperatures, the degree of ionization is greater, so the pH and pOH would bothbe less than 7, but still equal.

21. In water the silver acetate dissociates until the equilibrium concentration of ions isreached. In nitric acid solution, the acetate ions will react with hydrogen ions to formacetic acid molecules. The removes acetate ions from the silver acetateequilibrium allowing more silver acetate to dissolve. If HCl is used, a precipitate of silver

HNO3

pOH = - log[OH-],pH = - log[H+],H2O ÷ H+ + OH-.OH-H+

HC2H3O2 + H2O ÷ H3O+ + C2H3O2-

A + B ÷ C + D

I2.H2(H2 + I2 ÷ 2 HI)

- Chapter 16 -

- 231 -


chloride would be formed, since silver chloride is less soluble than silver acetate. Thus,more silver acetate would dissolve in HCl than in pure water.

22. When the salt, sodium acetate, is dissolved in water, the solution becomes basic. Thedissolving reaction is

The acetate ion reacts with water. The reaction does not go to completion, but some ions are produced and at equilibrium the solution is basic.

23. A buffer solution contains a weak acid or base plus a salt of that weak acid or base, suchas dilute acetic acid and sodium acetate.

When a small amount of a strong acid is added to this buffer solution, the reacts with the acetate ions to form un-ionized acetic acid, thus neutralizing the addedacid. When a strong base, is added it reacts with un-ionized acetic acid to neutralizethe added base. As a result, in both cases, the approximate pH of the solution is maintained.

24. All four K expressions are equilibrium constants. They describe the ratio between theconcentrations of products and reactants for different types of reactions when atequilibrium. is the equilibrium constant expression for the ionization of a weak acid,

is the equilibrium constant expression for the ionization of a weak base, is theequilibrium constant expression for the ionization of water and is the equilibriumconstant expression for a slightly soluble salt.

Ksp

KwKb

Ka

OH-,

H+(H+)

NaC2H3O2(aq) ¡ Na+(aq) + C2H3O2-(aq)

HC2H3O2(aq) ÷ H+(aq) + C2H3O2-(aq)

C2H3O2

-(aq) + H2O(l) ÷ OH-(aq) + HC2H3O2(aq)

OH-

NaC2H3O2(s) H2O99:

Na+(aq) + C2H3O2

-(aq)

AgC2H3O2(s) ÷ Ag+(aq) + C2H3O2-(aq)

- 232 -

- Chapter 16 -


CHAPTER 16

SOLUTIONS TO EXERCISES

1. Reversible systems

(a)

(b)

2. Reversible systems

(a)

(b)

3. Equilibrium system

(a) The reaction is exothermic with heat being evolved.

(b) The addition of will shift the reaction to the right until equilibrium isreestablished. The concentration of and will be increased. Theconcentration of the will be decreased.

4. Equilibrium system

(a) The addition of will shift the reaction to the left until equilibrium isreestablished. The concentration of and will be increased. Theconcentration of will be decreased.

(b) The addition of heat will shift the reaction to the left.

5.

Change or stress imposed Direction of reaction, Changes in numberon the system at left or right, to of moles

equilibrium reestablish equilibrium

(a) Add right I D I

(b) Remove left I D D

(c) Decrease volume ofreaction vessel right D D I

(d) Increase temperature left I I D

information to determine? = insufficientN = No Change;D = Decrease;I = Increase;

H2

N2

NH3H2N2

N2(g) + 3 H2(g) ÷ 2 NH3(g) + 92.5 kJ

H2ON2NH3, O2

N2

4 NH3(g) + 3 O2(g) ÷ 2 N2(g) + 6 H2O(g) + 1531 kJ

NH3

O2N2, H2O,O2

4 NH3(g) + 3 O2(g) ÷ 2 N2(g) + 6 H2O(g) + 1531 kJ

SO2(l) ÷ SO2(g)

H2O(l) 100°C

CRD

H2O(g)

Na2SO4(s) ÷ 2 Na+(aq) + SO4

2-(aq)

H2O(s) 0°C

CRD

H2O(l)

- 233 -


6.

Change or stress imposed Direction of reaction, Changes in numberon the system at left or right, to of moles

equilibrium reestablish equilibrium

(a) Add left I I I

(b) Increase volume ofreaction vessel left I I D

(c) Add a catalyst no change N N N

(d) Add and ? ? I I

information to determine

7. Direction of shift in equilibrium:

Increased Increased Pressure AddReaction Temperature (Volume Decreases) Catalyst

(a) right right no change

(b) left no change no change

(c) left right no change

8. Direction of shift in equilibrium:

Increased Increased Pressure AddReaction Temperature (Volume Decreases) Catalyst

(a) right left no change

(b) left left no change

(c) left left no change

9. Equilibrium shifts

(a) right

(b) left

(c) none

10. Equilibrium shifts

(a) left

(b) right

(c) right

? = insufficientN = No Change;D = Decrease;I = Increase;

NH3H2

NH3

NH3H2N2

N2(g) + 3 H2(g) ÷ 2 NH3(g) + 92.5 kJ

- 234 -

- Chapter 16 -


11. (a) (c)

(b)

12. (a) (c)

(b)

13. (a) (c)

(b) (d)

14. (a) (c)

(b) (d)

15. If the ion concentration is decreased:

(a) pH is increased

(b) pOH is decreased

(c) is increased

(d) remains the same. is a constant at a given temperature.

16. If the ion concentration is increased:

(a) pH is decreased (pH of 1 is more acidic than that of 4)

(b) pOH is increased

(c) is decreased

(d) remains unchanged. is a constant at a given temperature.

17. The basis for deciding if a salt dissolved in water produces an acidic, a basic, or a neutralsolution, is whether or not the salt reacts with water. Salts that contain an ion derivedfrom a weak acid or base will produce an acidic or a basic solution.

(a) CaBr2 neutral (c) basic

(b) acidic (d) basicK3PO4NH4NO3

NaCN

KwKw

[OH-]

H+

KwKw

[OH-]

H+

Ksp = [Pb2+]3 [AsO43-]2Ksp = [Ca2+][C2O4

2-]

Ksp = [Tl3+][OH-]3Ksp = [Mg2+][CO32-]

Ksp = [Ca2+]3 [PO43-]2Ksp = [Pb2+][CrO4

2-]

Ksp = [Zn2+][OH-]2Ksp = [Ag+][Cl-]

Keq =[CH4][H2S]

2

[CS2][H2]4

Keq =[N2O5]2

[NO2]4[O2]Keq =

[H+][H2PO4-]

[H3PO4]

Keq =[H+][HCO3

-]

[H2CO3]

Keq =[CO2][CF4]

[COF2]2Keq =[NH3]2[H2O]4

[NO2]2[H2]7

- Chapter 16 -

- 235 -


18 The basis for deciding if a salt dissolved in water produces an acidic, a basic, or a neutralsolution, is whether or not the salt reacts with water. Salts that contain an ion derivedfrom a weak acid or base will produce an acidic or a basic solution.

(a) acidic (c) acidic

(b) basic (d) Kl neutral

19. (a)

(b)

20. (a)

(b)

21. (a)

(b)

22. (a)

(b)

23. When excess acid gets into the blood stream it reacts with to form un-ionizedthus neutralizing the acid and maintaining the approximate pH of the blood.

24. When excess base gets into the blood stream it reacts with to form water. Thenionizes to replace thus maintaining the approximate pH of the blood.

25. (a)

Let

(b)

(c) Percent ionization

[H+]

[HC2H3O2] 11002 = ¢2.1 * 10-3

0.25≤11002 = 0.84%

pH = - log [H+] = - log12.1 * 10-32 = 2.68

x = 210.25211.8 * 10-52 = 2.1 * 10-3 M = [H+]

x2 = 10.25211.8 * 10-52 Ka =

[H+][C2H3O2-]

[HC2H3O2]=

x2

0.25= 1.8 * 10-5

[HC2H3O2] = 0.25 - x = 0.25 (since x is small)

[H+] = [C2H3O2-] = x

x = molarity of H+

HC2H3O2(aq) ÷ H+(aq) + C2H3O2

-(aq)

H+,H2CO3

H+

H2CO3,HCO3

-(H+)

ClO2

-(aq) + H2O(l) ÷ OH-(aq) + HClO2(aq)

OCl-(aq) + H2O(l) ÷ OH-(aq) + HOCl(aq)

NH4+(aq) + H2O(l) ÷ H3O+(aq) + NH3(aq)

HCO3

-(aq) + H2O(l) ÷ OH-(aq) + H2CO3(aq)

SO32-(aq) + H2O(l) ÷ OH-(aq) + HSO3

-(aq)

NH4+(aq) + H2O(l) ÷ H3O+(aq) + NH3(aq)

C2H3O2-(aq) + H2O(l) ÷ OH-(aq) + HC2H3O2(aq)

NO2

-(aq) + H2O(l) ÷ OH-(aq) + HNO2(aq)

NaC2H3O2

CuSO4NH4Cl

- 236 -

- Chapter 16 -


26. (a)

Let

(b)

(c) Percent ionization

27.

28.

29.

Let

[HC2H3O2] = initial concentration - x = initial concentration

[H+] = [C2H3O2-] = x

x = molarity of H+

Ka =[H+][C2H3O2

-]

[HC2H3O2]= 1.8 * 10-5

HC2H3O2 ÷ H+ + C2H3O2-

Ka =[H+][A-]

[HA]=13.4 * 10-322

0.497= 2.3 * 10-5

[HA] = 0.500 M - 0.0034 M = 0.497 M

[H+] = [A-] = (0.500 M)(0.0068) = 3.4 * 10-3 M

HA ÷ H+ + A-

Ka = Keq =[H+][HCO3

-]

[H2CO3]= 4.8 * 10-7

[HA] = 0.025 M - 0.00011 M = 0.025 M

[H+] = [A-] = (0.025 M)10.00452 = 1.1 * 10-4 M

HA ÷ H+ + A-

[H+]

[HC6H5O] 11002 = ¢5.7 * 10-6

0.25≤11002 = 2.3 * 10-3 %

pH = - log [H+] = - log15.7 * 10-62 = 5.24

x = 210.25211.3 * 10-102 = 5.7 * 10-6 = [H+]

x2 = 10.25211.3 * 10-102 Ka =

[H+][C6H5O-]

[HC6H5O]=

x2

0.25= 1.3 * 10-10

[HC6H5O] = 0.25 - x = 0.25 (since x is small)

[H+] = [C6H5O-] = x

x = molarity of H+

HC6H5O(aq) ÷ H+(aq) + C6H5O-(aq)

- Chapter 16 -

- 237 -


Since is small, the degree of ionization is small. Therefore, the approximation,concentration, is valid.

(a)

(b)

(c)

30.

Let

[H+] = [ClO-] = x

x = molarity of H+

Ka =[H+][ClO-]

[HClO]= 3.5 * 10-8

HClO ÷ H+ + ClO-

pH = - log14.2 * 10-42 = 3.38

¢4.2 * 10-4 M

0.010 M≤11002 = 4.2% ionized

x = 21.8 * 10-7 = 4.2 * 10-4 M

x2 = 10.010211.8 * 10-52 = 1.8 * 10-7

1x21x20.010

= 1.8 * 10-5

[HC2H3O2] = 0.010 M

pH = - log11.3 * 10-32 = 2.89

¢1.3 * 10-3 M

0.10 M≤11002 = 1.3% ionized

x = 21.8 * 10-6 = 1.3 * 10-3 M

x2 = 10.10211.8 * 10-52 = 1.8 * 10-6

1x21x20.10

= 1.8 * 10-5

[HC2H3O2] = 0.10 M

pH = - log14.2 * 10-32 = 2.38

a4.2 * 10-3 M

1.0 Mb11002 = 0.42% ionized

x = 21.8 * 10-5 = 4.2 * 10-3 M

x2 = 11.0211.8 * 10-52

1x21x21.0

= 1.8 * 10-5

[H+] = [C2H3O2-] = x [HC2H3O2] = 1.0 M

initial concentration - x = initialKa

- 238 -

- Chapter 16 -


Since is small, the degree of ionization is small. Therefore, the approximation,concentration, is valid.

(a)

(b)

(c)

31.

First, find the This is calculated from the pH expression,

Ka =[H+][A-]

[HA]=12 * 10-4212 * 10-42

0.37= 1 * 10-7

[H+] = [A-] = 2 * 10-4 [HA] = 0.37

[H+] = 2 * 10-4pH = - log[H+] = 3.7.[H+].

Ka =[H+][A-]

[HA]HA ÷ H+ + A-

pH = - log11.9 * 10-52 = 4.72

¢1.9 * 10-5 M

0.010 M≤11002 = 0.19% ionized

x = 23.5 * 10-10 = 1.9 * 10-5 M

1x21x20.010

= 3.5 * 10-8 x2 = 10.010213.5 * 10-82[HClO] = 0.010 M

pH = - log15.9 * 10-52 = 4.23

a5.9 * 10-5 M

0.10 Mb11002 = 0.059% ionized

x = 23.5 * 10-9 = 5.9 * 10-5 M

1x21x20.10

= 3.5 * 10-8 x2 = 10.10213.5 * 10-82[HClO] = 0.10 M

pH = - log11.9 * 10-42 = 3.72

a1.9 * 10-4 M

1.0 Mb11002 = 1.9 * 10-2% ionized

x = 23.5 * 10-8 = 1.9 * 10-4 M

1x21x2

1.0= 3.5 * 10-8 x2 = 11.0213.5 * 10-82

[H+] = [ClO-] = x [HClO] = 1.0 M

initial concentration - x = initialKa

[HClO] = initial concentration - x = initial concentration

- Chapter 16 -

- 239 -


32. See problem 51 for a discussion of calculating from pH.

33. 1.0 M NaOH yields (100% ionized)

34. 3.0 M HNO3 yields (100% ionized)

35.

(a)

(b)

(c)

(x) (x)

0.895= 1.8 * 10-5

Ka =[H+][C2H3O2

-]

[HC2H3O2]= 1.8 * 10-5

xx0.895 M

HC2H3O2 ∆ H+ + C2H3O2-

pOH = - log1OH-2 = - log (0.333) = 0.478 pH = 14.0 - 0.478 = 13.5

0.333 M KOH yields [OH-] = 0.333 M(100% ionized)

pH = - log[H+] = - log (0.250) = 0.602 pOH = 14.0 - 0.602 = 13.4

0.250 M HBr yields [H+] = 0.250 M (100% ionized)

pH + pOH = 14.0 pOH = 14.0 - pH

[OH-] =Kw

[H+]=

1 * 10-14

3.0= 3.3 * 10-15

pOH = 14 - pH = 14 - (–0.47) = 14.47

pH = - log 3.0 = -0.47

[H+] = 3.0 M

[H+] =Kw

[OH-]=

1.0 * 10-14

1.0= 1 * 10-14

pH = 14 - pOH = 14.00

pOH = - log 1.0 = 0.00

[OH-] = 1.0 M

Ka =[H+][A-]

[HA]=11.3 * 10-3211.3 * 10-32

0.23= 7.3 * 10-6

[H+] = 1.3 * 10-3 = [A-] [HA] = 0.23

- log [H+] = 2.89 [H+] = 1.3 * 10-3

HA ÷ H+ + A- Ka =[H+][A-]

[HA] pH = 2.89

[H+]

- 240 -

- Chapter 16 -


36.

(a)

(b)

(c)

37. Calculate the

(a)

(b)

38. Calculate the

(a) [OH-] =1.0 * 10-14

4.0 * 10-9 = 2.5 * 10-6[H+] = 4.0 * 10-9

[OH-] =Kw

[H+][OH-].

[OH-] =1.0 * 10-14

2.8 * 10-6 = 3.6 * 10-9[H+] = 2.8 * 10-6

[OH-] =1.0 * 10-14

1.0 * 10-4 = 1.0 * 10-10[H+] = 1.0 * 10-4

[OH-] =Kw

[H+][OH-].

pOH = 14.0 - 5.74 = 8.3

pH = - log (1.8 * 10-6) = 5.74

x = 1.8 * 10-6 = [H+]

x2 = (0.0250)(1.3 * 10-10) x = 23.3 * 10-12

(x) (x)

0.0250= 1.3 * 10-10

Ka =[H+][C6H5O-]

[HC6H5O]= 1.3 * 10-10

xx0.0250 M

HC6H5O ∆ H+ + C6H5O-

pH = - log [H+] = - log10.1252 = 0.903 pOH = 14.0 - 0.903 = 13.1

(b) 0.125 M HCl yields [H+] = 0.125 M (100% ionized)

pOH = - log [OH-] = - log10.00102 = 3.00 pH = 14.0 - 3.00 = 11.0

0.0010 M NaOH yields [OH-] = 0.0010 M (100% ionized)

pH + pOH = 14.0 pOH = 14.0 - pH

pOH = 14.0 - 2.40 = 11.6

pH = - log(4.0 x 10-3) = 2.40

x = 4.0 * 10-3 = [H+]

x2 = (0.895)(1.8 * 10-5) x = 21.6 * 10-5

- Chapter 16 -

- 241 -


(b)

39. Calculate the

(a)

(b)

40. Calculate the

(a)

(b)

41. The molar solubilities of the salts and their ions are indicated below the formulas in theequilibrium equations.

(a)

(b)

(c) First change

(d) First change

a0.0019 g AgCl

Lb a 1 mol

143.4 gb = 1.3 * 10-5 M AgCl

g>L ¡ mol>LKsp = [Ca2+][SO4

2-] = 14.9 * 10-322 = 2.4 * 10-5

4.9 * 10-34.9 * 10-3

CaSO4(s) ÷ Ca2+ + SO4 2-

¢0.67 g CaSO4

L≤ a 1 mol

136.1 gb = 4.9 * 10-3 M CaSO4

g>L ¡ mol>LKsp = [Ag+]2[CrO4

2-] = 115.6 * 10-52217.8 * 10-52 = 1.9 * 10-12

7.8 * 10-5217.8 * 10-52Ag2CrO4(s) ÷ 2 Ag+ + CrO4

2-

Ksp = [Ba2+][SO4

2-] = 13.9 * 10-522 = 1.5 * 10-9

3.9 * 10-53.9 * 10-5

BaSO4(s) ÷ Ba2+ + SO4

2-

[H+] =1 * 10-14

7.3 * 10-4 = 1.4 * 10-11[OH-] = 7.3 * 10-4

[H+] =1.0 * 10-14

4.5 * 10-6 = 2.2 * 10-9[OH-] = 4.5 * 10-6

[H+] =Kw

[OH-][H+].

[H+] =1 * 10-14

1 * 10-8 = 1 * 10-6[OH-] = 1 * 10-8

[H+] =1.0 * 10-14

6.0 * 10-7 = 1.7 * 10-8[OH-] = 6.0 * 10-7

[H+] =Kw

[OH-][H+].

[OH-] =1.0 * 10-14

8.9 * 10-2 = 1.1 * 10-13[H+] = 8.9 * 10-2

- 242 -

- Chapter 16 -


42. The molar solubilities of the salts and their ions are indicated below the formulas in theequilibrium equations.

(a)

(b)

(c) First change

(d) First change

43. The molar solubilities of the salts and their ions will be represented in terms of x belowtheir formulas in the equilibrium equations.

(a)

x = A3

3.9 * 10-11

4 = 2.2 * 10-4 M

Ksp = [Ca2+][F-]2 = (x)(2x)2 = 4x3 = 3.9 * 10-11

2xx

CaF2 ÷ Ca2+ + 2 F-

Ksp = [Zn2+][OH-]2 = 12.34 * 10-6214.68 * 10-622 = 5.13 * 10-17

212.34 * 10-622.34 * 10-6

Zn(OH)2(s) ÷ Zn2+ + 2 OH-

¢2.33 * 10-4 g Zn(OH)2

L≤ a 1 mol

99.41 gb = 2.34 * 10-6 M Zn(OH)2

g>L ¡ mol>LKsp = [Ag+]3[PO4

3-] = 14.83 * 10-52311.61 * 10-52 = 1.81 * 10-18

1.61 * 10-5311.61 * 10-52Ag3PO4(s) ÷ 3 Ag+ + PO4

3-

¢6.73 * 10-3 g Ag3PO4

L≤ a 1 mol

418.7 gb = 1.61 * 10-5 M Ag3PO4

g>L ¡ mol>LKsp = [Pb2+][IO3

-]2 = 14.0 * 10-5218.0 * 10-522 = 2.6 * 10-13

2(4.0 * 10-5)4.0 * 10-5

Pb(IO3)2(s) ÷ Pb2+ + 2 IO3

-

Ksp = [Zn2+][S2-] = 13.5 * 10-1222 = 1.2 * 10-23

3.5 * 10-123.5 * 10-12

ZnS(s) ÷ Zn2+ + S2-

Ksp = [Ag+][Cl-] = 11.3 * 10-522 = 1.7 * 10-10

1.3 * 10-51.3 * 10-5

AgCl(s) ÷ Ag+ + Cl-

- Chapter 16 -

- 243 -


(b)

44. The molar solubilities of the salts and their ions will be represented in terms of belowtheir formulas in the equilibrium equations.

(a)

(b)

45. (a)

(b)

46. (a)

(b)

47. The molar concentrations of ions, after mixing, are calculated and these concentrations aresubstituted into the equilibrium expression. The value obtained is compared to the ofthe salt. If the value is greater than the precipitation occurs. If the value is less than the

no precipitation occurs.

Volume after mixing = 200. mL

100. mL 0.001 M Pb(NO3)2 ¡ 100. mL 0.001 M Pb2+100. mL 0.010 M Na2SO4 ¡ 100. mL 0.010 M SO4

2-

Ksp,Ksp ,

Ksp

a9.2 * 10-6 mol BaCrO4

Lb(0.100 L)a253.3 g BaCrO4

molb = 2.3 * 10-4 g BaCrO4

¢1.1 * 10-4 PbSO4

L≤ (0.100 L)a303.3 g PbSO4

molb = 3.3 * 10-3 g PbSO4

a1.9 * 10-10 mol Fe(OH)3

Lb(0.100 L)a106.9 g Fe(OH)3

molb = 2.0 * 10-9 g Fe(OH)3

¢2.2 * 10-4 mol CaF2

L≤ (0.100 L)a78.08 g CaF2

molb = 1.7 * 10-3g CaF2

x = 28.5 * 10-11 = 9.2 * 10-6 M

Ksp = [Ba2+][CrO4

2-] = (x) (x) = x2 = 8.5 * 10-11xx

BaCrO4 ∆ Ba2+ + CrO4

2-x = 21.3 * 10-8 = 1.1 * 10-4 M

Ksp = [Pb2+][SO4

2] = (x) (x) = x2 = 1.3 * 10-8

xx

PbSO4 ÷ Pb2+ + SO4

2-

x

x = A4

3.4 * 10-38

27 = 1.9 * 10-10 M

Ksp = [Fe3+][OH-]3 = 1x213x23 = 27x4 = 3.4 * 10-38

x 3x

Fe(OH)3 ÷ Fe3+ + 3 OH-

- 244 -

- Chapter 16 -


Concentrations after mixing:

which is less than therefore, precipitation occurs.

48. The molar concentrations of ions, after mixing, are calculated and these concentrationsare substituted into the equilibrium expression. The value obtained is compared to the of the salt. If the value is greater than the precipitation occurs. If the value is lessthan the no precipitation occurs.



49. The concentration of in 1.0 L of 0.10 M NaBr. Substitute this concentration in the expression and solve for the in equilibrium with

50.

Substitute the concentration in the expression and solve for in equilibriumwith

¢2.6 * 10-12 mol Ag+

L≤ a1 mol AgBr

1 mol Ag+ b(1.0 L) = 2.6 * 10-12 mol AgBr will dissolve

[Ag+] =5.2 * 10-13

[Br-]=

5.2 * 10-13

0.20= 2.6 * 10-12 M

0.20 M Br-.[Ag+]KspBr -

¢0.10 mol MgBr2

L≤ ¢ 2 mol Br-

1 mol MgBr2≤ = ¢0.20 mol Br-

L≤ = 0.20 M Br- in solution.

¢5.2 * 10-12 mol Ag+

L≤ ¢1 mol AgBr

1 mol Ag+ ≤ (1.0 L) = 5.2 * 10-12 mol AgBr will dissolve

[Ag+] =5.2 * 10-13

[Br-]=

5.2 * 10-13

0.10= 5.2 * 10-12 M

Ksp = [Ag+][Br-] = 5.2 * 10-13

0.10 M Br-.[Ag+]Ksp

Br -Br- = 0.10 M

2.2 * 10-9,Ksp = 1.7 * 10-10[Ag+][Cl-] = 13.3 * 10-5216.7 * 10-52 = 2.2 * 10-9

11.0 * 10-4 M Cl-2a100. mL

150. mLb = 6.7 * 10-5 M Cl-

11.0 * 10-4 M Ag+2a50.0 mL

150. mLb = 3.3 * 10-5 M Ag+


100. mL 1.0 * 10-4 M NaCl ¡ 100. mL 1.0 * 10-4 M Cl-50.0 mL 1.0 * 10-4 M AgNO3 ¡ 50.0 mL 1.0 * 10-4 M Ag+

Ksp,Ksp ,

Ksp

3 * 10-6,Ksp = 1.3 * 10-8[Pb2+][SO4

2-] = 15.0 * 10-3215 * 10-42 = 3 * 10-6Pb2+ = 0.0005 MSO4

2- = 0.0050 M

- Chapter 16 -

- 245 -


51.

52.

53. Initially, the solution of NaCl is neutral.

Change in in the unbuffered solution

54. Initially,

Change in units in the buffered solution

55. The concentration of solid salt is not included in the equilibrium constant because theconcentration of solid does not change. It is constant and part of the Ksp .

Ksp

pH = 4.74 - 4.72 = 0.02

pH = - log11.9 * 10-52 = 4.72

Final [H+] = 1.9 * 10-5

pH = - log11.8 * 10-52 = 4.74

[H+] =Ka[HC2H3O2]

[C2H3O2-]

[H+] = 1.8 * 10-5

pH = 7.0 - 1.70 = 5.3 units

pH = - log12.0 * 10-22 = 1.70

Final H+ = 2.0 * 10-2 M

pH = - log11 * 10-72 = 7.0

[H+] = 1 * 10-7

pH = - log11.8 * 10-52 = 4.74

[H+] = 11.8 * 10-52a0.20

0.20b = 1.8 * 10-5 M

[H+] = Ka¢ [HC2H3O2]

[C2H3O2-]≤ [HC2H3O2] = 0.20 M [C2H3O2

-] = 0.20 M

Ka =[H+][C2H3O2

-]

[HC2H3O2]= 1.8 * 10-5

HC2H3O2 ÷ H+ + C2H3O2

-

pH = - log13.6 * 10-52 = 4.44

[H+] = 11.8 * 10-52a0.20

0.10b = 3.6 * 10-5 M

[H+] = Ka¢ [HC2H3O2]

[C2H3O2

-]≤ [HC2H3O2] = 0.20 M [C2H3O2

-] = 0.10 M

Ka =[H+][C2H3O2

-]

[HC2H3O2]= 1.8 * 10-5

HC2H3O2 ÷ H+ + C2H3O2-

- 246 -

- Chapter 16 -


56. The energy diagram represents an exothermic reaction because the energy of theproducts is lower than the energy of the reactants. This means that energy was given offduring the reaction.

57.

58.

The reaction is a 1 to 1 mole ratio of hydrogen to iodine. The data given indicates thathydrogen is the limiting reactant.

59.

(a) and will produce 4.00 mol HI assuming 100% yield.However, at 79% yield you get

(b) The addition of makes the iodine present in excess and the the limiting reactant. The yield increases to 85%.

There will be 15% unreacted and plus the extra added.

present; also

In addition to the 0.30 mol of unreacted will be the added.

present.

(c)

The formation of 3.16 mol HI required the reaction of and At equilibrium, the concentrations are:

1.58 mol H2.1.58 mol I2

K =[HI]2

[H2][I2]

0.27 mol + 0.30 mol = 0.57 mol I2

0.27 mol I2I2 ,

0.30 mol I2 .10.15212.0 mol H22 = 0.30 mol H2

I2I2H2

(2.00 mol H2)¢ 2 mol HI

1 mol H2≤10.852 = 3.4 mol HI

2.00 mol H20.27 mol I2

4.00 mol HI * 0.79 = 3.16 mol HI

2.00 mol I22.00 mol H2

H2 + I2 ÷ 2 HI

(2.10 mol H2)¢ 2 mol HI

1 mol H2≤ = 4.20 mol HI

H2 + I2 ÷ 2 HI

Pot

enti

al e

nerg

y

Progress of reaction

Activationenergy

Activationenergy withcatalyst

Energyabsorbed

- Chapter 16 -

- 247 -


In the calculation of the equilibrium constant, the actual number of moles ofreactants and products present at equilibrium can be used in the calculation in placeof molar concentrations. This occurs because the reaction is gaseous and the litersof HI produced equals the sum of the liters of and reacting. In the equilibriumexpression, the volumes will cancel.

60.

At equilibrium, moles present are:

0.500 mol HI;

61.

The concentrations are:

Keq =0.011

10.0050210.0752 = 29

Cl2 =1.50 mol

20. L= 0.075 M

PCl3 =0.10 mol

20. L= 0.0050 M

PCl5 =0.22 mol

20. L= 0.011 M

Keq =[PCl5]

[PCl3][Cl2]

PCl3(g) + Cl2(g) ÷ PCl5(g)

0.788 - 0.250 = 0.538 mol I2

2.98 - 0.250 = 2.73 mol H2

(200. g I2)a 1 mol

253.8 gb = 0.788 mol I2 initially present

(6.00 g H2)a 1 mol

2.016 gb = 2.98 mol H2 initially present

(0.500 mol HI)a1 mol H2

2 mol HIb = 0.250 mol H2 reacted

(0.500 mol HI)a 1 mol I2

2 mol HIb = 0.250 mol I2 reacted

(64.0 g HI)a 1 mol

127.9 gb = 0.500 mol HI present

H2 + I2 ÷ 2 HI

I2H2

Keq =13.1622

10.42210.422 = 57

3.16 mol HI; 2.00 - 1.58 = 0.42 mol H2 = 0.42 mol I2

- 248 -

- Chapter 16 -


62. temperature increase. This increase is equal to seven 10°Cincrements. The reaction rate will be increased by

63.

0.30 M y y

64. pH of an acetic acid-acetate buffer

65. Concentration of in solution

Determine the moles of and in solution

Excess in

Concentration of in solution (total )

Now using the calculate the remaining in solution.

[Ba2+] =8.5 * 10-11

2.5 * 10-2 = 3.4 * 10-9 mol>L[Ba2+][2.5 * 10-2] = 8.5 * 10-11[Ba2+] [CrO4

2- ] =Ba2+Ksp,

2.5 * 10-3 mol CrO42-

0.100 L= 2.5 * 10-2 M CrO4

2-

volume = 100 mLCrO42-

solution = 0.0025 mol 12.5 * 10-32 after 0.005 mol BaCrO4 precipitate.CrO42-

a0.15 mol

Lb (0.050 L) = 0.0075 mol CrO4

2-

a0.10 mol

Lb (0.050 L) = 0.0050 mol Ba2+

CrO42-Ba2+

BaCrO4 ∆ Ba2+ + CrO4

2- Ksp = 8.5 * 10-11

BaCl2(aq) + Na2CrO4(aq) ∆ BaCrO4(s) + NaCl(aq)

Ba2+

pH = - log12.7 * 10-52 = 4.57

H+ =11.8 * 10-5210.302

0.20= 2.7 * 10-5

Ka =[H+][C2H3O2

-]

[HC2H3O2]=

[H+][0.20]

0.30= 1.8 * 10-5

HC2H3O2 ∆ H+ + C2H3O2-

pH = - log11.3 * 10-52 = 4.89 (on acidic solution)

y = [H+] = 1.3 * 10-5

y2 = 15.6 * 10-10210.302 = 1.7 * 10-10

[NH3][H+]

[NH4+]

=[y][y]

[0.30]= 5.6 * 10-10

y = [H+] = [NH3]

NH4+ ∆ NH3 + H+ Keq = 5.6 * 10-10

27 = 128 times.100°C - 30°C = 70°C

- Chapter 16 -

- 249 -


66. Hypochlorous acid

Equilibrium concentrations:

Propanoic acid


Hydrocyanic acid


(neglecting )

67. Let dissolved

(a)

(b) ¢2.1 * 10-4 mol CaF2

L≤ (0.500 L)a78.08 g

molb = 8.2 * 10-3 g CaF2

¢2.1 * 10-4 mol CaF2

L≤ ¢ 2 mol F-

1 mol CaF2≤ = 4.2 * 10-4 M F

-

¢2.1 * 10-4 mol CaF2

L≤ ¢ 1 mol Ca2+

1 mol CaF2≤ = 2.1 * 10-4 M Ca2+

y = B3 3.9 * 10-11

4= 2.1 * 10-4 M (CaF2 dissolved)

Ksp = [Ca2+][F-]2 = 1y212y22 = 4y3 = 3.9 * 10-11

y = molar solubility

2yy

CaF2(s) ÷ Ca2+ + 2 F-y = M CaF2

Ka =[H+][CN-]

[HCN]=18.9 * 10-622

0.20= 4.0 * 10-10

8.9 * 10-6[HCN] = 0.20 - 8.9 * 10-6 = 0.20 M

[H+] = [CN-] = 8.9 * 10-6 M

HCN ÷ H+ + CN-

Ka =[H+][C3H5O2

-]

[HC3H5O2]=11.4 * 10-3211.4 * 10-32

0.15= 1.3 * 10-5

[HC3H5O2] = 0.15 - 1.4 * 10-3 = 0.15 M 1neglecting 1.4 * 10-32[H+] = [C3H5O2

-] = 1.4 * 10-3 M

HC3H5O2 ÷ H+ + C3H5O2

-

Ka =[H+][OCl-]

[HOCl]=15.9 * 10-5215.9 * 10-52

0.10= 3.5 * 10-8

[HOCl] = 0.1 - 5.9 * 10-5 = 0.10 M 1neglecting 5.9 * 10-52[H+] = [OCl-] = 5.9 * 10-5 M

HOCl ÷ H+ + OCl-

- 250 -

- Chapter 16 -


68. The molar concentrations of ions, after mixing, are calculated and these concentrationsare substituted into the equilibrium expression. The value obtained is compared to the of the salt. If the value is greater than the precipitation occurs. If the value is lessthan the no precipitation occurs.

(a)



(b)

Volume after mixing



(c) Convert to

Final volume

Concentration after mixing:

which is greater than therefore, no precipitation occurs.

6.0 * 10-7,Ksp = 1.3 * 10-6

[Ca2+][OH-]2 = 10.015210.006322 = 6.0 * 10-7

(0.01 M OH-)¢ 250 mL

4.0 * 102 mL≤ = 0.0063 M OH-

(0.041 M Ca2+)¢ 150 mL

4.0 * 102 mL≤ = 0.015 M Ca2+

= 4.0 * 102 mL

250 mL 0.01 M NaOH ¡ 250 mL 0.01 M OH-

¢1.0 g Ca(NO3)2

0.150 L≤ a 1 mol

164.1 gb ¢ 1 mol Ca2+

1 mol Ca(NO2)2≤ = 0.041 M Ca2+

g Ca2+g Ca(NO3)2

2.2 * 10-9,Ksp = 1.7 * 10-10

[Ag+][Cl-] = 13.3 * 10-5216.7 * 10-52 = 2.2 * 10-9

11.0 * 10-42a100. mL

150. mLb = 6.7 * 10-5 M Cl-

11.0 * 10-42a50.0 mL

150. mLb = 3.3 * 10-5 M Ag+

= 150. mL

100. mL 1.0 * 10-4 M NaCl ¡ 100. mL 1.0 * 10-4 M Cl-50.0 mL 1.0 * 10-4 M AgNO3 ¡ 50.0 mL 1.0 * 10-4 M Ag+

3 * 10-6,Ksp = 1.3 * 10-8

[Pb2+][SO4

2-] = 15.0 * 10-3215 * 10-42 = 3 * 10-6

Pb2+ = 0.0005 MSO4

2- = 0.0050 M


100. mL 0.001 M Pb(NO3)2 ¡ 100. mL 0.001 M Pb2+100. mL 0.010 M Na2SO4 ¡ 100. mL 0.010 M SO4

2-

Ksp,Ksp ,

Ksp

- Chapter 16 -

- 251 -


69. With a known concentration, the concentration can be calculated using thevalue.

(a)

(b) in solution

70. If exceeds the precipitation will occur.

is smaller than the value. Therefore, no precipitate of will form.

71.

Both cations are present in equal concentrations (0.10 M). Therefore, as is added,the of will be exceeded before that of precipitates first.

72.

73.

(molar concentration)

74.

Let y = [NH3] Keq =[NH3]2

[N2][H2]3 = 4.0

N2 + 3 H2 ÷ 2 NH3

Ksp = [Ba2+][F-]2 = 11.8 * 10-2213.6 * 10-222 = 2.3 * 10-5

211.8 * 10-221.8 * 10-2BaF2(s) ÷ Ba2+ + 2 F-

a2.7 * 10-4 mol

0.015 Lb = 1.8 * 10-2 M BaF2 dissolved

(0.048 g BaF2)a 1 mol

175.3 gb = 2.7 * 10-4 mol BaF2

Keq =[SO3]2

[SO2]2[O2]=

111.02214.202210.60 * 10-32 = 1.1 * 104

2 SO2(g) + O2(g) ÷ 2 SO3(g)

BaSO4SrSO4.BaSO4Ksp

SO4

2-[Ba2+][SO4

2-] = 1.5 * 10-9 [Sr2+][SO4

2-] = 3.5 * 10-7

PbCl2Ksp[Pb2+][Cl-]210.050210.01022 = 5.0 * 10-60.010 M NaCl ¡ 0.010 M Cl-0.050 M Pb(NO3)2 ¡ 0.050 M Pb2+Ksp = [Pb2+][Cl-]2 = 2.0 * 10-5

Ksp ,[Pb2+][Cl-]2

= 7.0 * 10-7 g BaSO4 remain in solution

¢3.0 * 10-8 mol BaSO4

L≤ (0.100 L)a 233.4 g

molb

M SO4

2- = M BaSO4

[SO4

2-] =Ksp

[Ba2+]=

1.5 * 10-9

0.050= 3.0 * 10-8 M SO4

2- in solution

Ksp = [Ba2+][SO4

2-] = 1.5 * 10-9 Ba2+ = 0.050 M

BaSO4(s) ÷ Ba2+ + SO4

2-Ksp

SO4

2-Ba2+

- 252 -

- Chapter 16 -


75. Total volume of (0.0400 L)

no precipitation should occur.

76. First change

(molar solubility)

77.

Three ways to increase ozone

(a) increase heat

(b) increase amount of

(c) increase pressure

(d) remove as it is made

78.

Conditions on the second day

(a) the temperature could have been cooler

(b) the humidity in the air could have been higher

(c) the air pressure could have been greater

79.

(c) is the correct answer

CO(g) + H2O(g) ÷ CO2(g) + H2(g)

H2O(l) ÷ H2O(g)

O3

O2

3 O2(g) + heat ÷ 2 O3(g)

Ksp = [Hg2

2+][I-]2 = 14.64 * 10-10219.28 * 10-1022 = 4.00 * 10-28

214.64 * 10-10 M24.64 * 10-10 M

Hg2I2 ÷ Hg2

2+ + 2 I-

¢3.04 * 10-7 g Hg2I2

L≤ a 1 mol

655.0 gb = 4.64 * 10-10 M Hg2I2

g Hg2I2 ¡ mol Hg2I2

4.7 * 10-7 6 7.6 * 10-7

[Sr2+][SO4

2-] = 16.3 * 10-4217.5 * 10-42 = 4.7 * 10-7

[SO4

2-] =12.0 * 10-3 M210.0150 L2

0.0400 L= 7.5 * 10-4 M

[Sr2+] =11.0 * 10-3 M2(0.0250 L)

0.0400 L= 6.3 * 10-4 M

Ksp = [Sr2+][SO4

2-] = 7.6 * 10-7

mixture = 40.0 mL

y = 8.0 M = [NH3]

y = 264y2 = 64 4.0 =y2

12.0212.023

- Chapter 16 -

- 253 -


With equal concentrations of products and reactants, the value will equal 1.

80. (a) (c)

(b) (d)

81. 2A B C

1.0 M 1.0 M 0 Initial conditions

0.30 Equilibrium concentrations

0.4 M 0.7 M 0.30 M

82. Since the second reaction is the reverse of the first, the value of the second reactionwill be the reciprocal of the value of the first reaction.

(first reaction)

83.

reacts with and equilibrium shifts to the right.

(a) After an initial increase, will be neutralized and equilibrium shifts to the right.

(b) will be reduced (reacts with ). Equilibrium shifts to the right.

(c) increases as equilibrium shifts to the right.

(d) decreases and equilibrium shifts to the right.

84.

x = 0.014 M CaSO4

x = 22.0 * 10-41x21x2 = 2.0 * 10-4

Let x = moles CaSO4 that dissolve per L = [Ca2+] = [SO4

2-]

Ksp = [Ca2+][SO4

2-] = 2.0 * 10-4

CaSO4(s) ÷ Ca2+(aq) + SO 42-(aq)

[HNO2]

[NO2

-]

OH-[H+]

[OH-]

H+OH-HNO2(aq) ÷ H+(aq) + NO2

-(aq)

Keq =1

2.2 * 10-3 = 450Keq =[ICl]2

[I2][Cl2]

Keq =[I2][Cl2]

[ICl]2 = 2.2 * 10-3

Keq

Keq

Keq =[C]

[A]2[B]=

0.30

10.42210.72 = 3

1.0 - 0.301.0 - 210.302÷+

Keq =[H+]6

[Bi3+]2[H2S]3Keq =[H2O(l)]

[H2O(g)]

Keq = [CO2 (g)]Keq =[O3]2

[O2]3

Keq

Keq =[CO2][H2]

[CO][H2O]= 1

- 254 -

- Chapter 16 -


85.

change

86. Treat this as an equilibrium where and

W 2 S K

144 0 0 amount before cracking

number of kernels after cracking

87.

0.50 M 0.50 M 0 0 Initial conditions

Equilibrium concentrations

Take the square root of both sides

[SO2] = [NO2] = 0.05 M

[SO3] = [NO] = 0.45 M

x = 0.45 Mx

0.50 - x= 9.0

Keq =[SO3][NO]

[SO2][NO2]=

x2

10.50 - x22 = 81

xx0.50 - x0.50 - x

NO(g)+SO3(g)÷NO2(g)+SO2(g)

Keq =12x221x2144 - x

=150221252

119= 5.3 * 102

x = 25 kernels; 50 shell halves; 119 whole nuts left

144 + 2x = 194; 2x = 50

144 - x + 2x + x = 194 total pieces

x =x2x144 - x

+÷K = kernelsW = whole nuts, S = shell halves,

Ksp = 11.0 * 10-3212.0 * 10-322 = 4.0 * 10-9

[Pb2+] = 1.0 * 10-3; [F-] = 211.0 * 10-32 = 2.0 * 10-3

Ksp = (Pb2+)(F-)2

¢0.098 g PbF2

0.400 L≤ a 1 mol

245.2 gb = 1.0 * 10-3 mol>L = 1.0 * 10-3 M PbF2

g PbF2 ¡ mol PbF2

PbF2(s) ÷ Pb2+ + 2 F-

¢0.014 mol CaSO4

L≤ (0.600 L)a136.2 g

molb = 1.1 g CaSO4

M ¡ moles ¡ grams

- Chapter 16 -

- 255 -


Documents

CHAPTER 16 CHEMICAL EQUILIBRIUM - Chemeketa ...faculty.chemeketa.edu/lemme/CH 121/solutions/Hein9thCh16.pdf13. If pure HI is placed in a vessel at 700 K, some of it will decompose