Chapter 11 Solutions

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 11

Chapter 11 11-1

The overall dissociation constant for H2S is [ H 3O  ]2 [S2- ]  K1 K 2 [ H 2S]

In a solution saturated with the gas, [H2S] is constant, therefore [S2- ] 

11-2

[ H 2S] K1 K 2 [ H 3 O  ]2

or

[S2- ] 

K [ H 3 O  ]2

The simplifications in equilibrium calculations involve assuming that the concentrations of one or more species can be approximated as 0.00 M. Adding or subtracting a concentration that can be approximated as 0.00 leads to a meaningful result. In contrast, multiplying or dividing by 0.00 in the equilibrium constant expression causes the constant to become equal to zero or infinity. Thus, the expression is meaningless.

11-3

A charge-balance equation is derived by relating the concentration of cations and anions such that no. mol/L positive charge = no. mol/L negative charge For a doubly charged ion, such as Ba2+, the concentration electrons for each mole is twice the molar concentration of the Ba2+. That is, mol/L positive charge = 2[Ba2+] Thus, the molar concentration of all multiply charged species is always multiplied by the charge in a charge-balance equation.

11-4

(a)

0.20

=

[H3AsO4] + [H2AsO4-] + [HAsO42-] + [AsO43-]

(b)

0.10

=

[H3AsO4] + [H2AsO4-] + [HAsO42-] + [AsO43-]

2(0.10) =

[Na+] =

0.20

Fundamentals of Analytical Chemistry: 8th ed. (c)

11-6

0.0500 + 0.100

=

[ClO-] + [HClO]

0.100

=

[Na+]

[F-] + [HF]

=

0.25 + 2[Ca2+]

[Na+]

=

0.25

(e)

0.100 =

[Na+] =

(f)

[Ba2+] =

[C2O42-] + [HC2O4-] + [H2C2O4]

(g)

[Ca2+] =

½([F-] + [HF])

(a)

[H3O+] =

[OH-] + [H2AsO4-] + 2[HAsO42-] + 3[AsO43-]

(b)

[Na+] + [H3O+]

=

[OH-] + [H2AsO4-] + 2[HAsO42-] + 3[AsO43-]

(c)

[Na+] + [H3O+]

=

[OH-] + [ClO-]

(d)

[Na+] + [H3O+]+ 2[Ca2+]

=

[F-] + [OH-]

(e)

2[Zn2+] + [Na+] + [H3O+]

=

[OH-] + 2[Zn(OH)42-]

(f)

2[Ba2+] + [H3O+]

=

[OH-] + 2[C2O42-] + [HC2O4-]

(g)

2[Ca2+] + [H3O+]

=

[OH-] + [F-]

(d)

11-5

Chapter 11

[OH-] + 2[Zn(OH)42-]

Step 1 Ag2C2O4  2Ag+ + C2O42H2C2O4 + H2O  H3O+ + HC2O4HC2O4- + H2O  H3O+ + C2O42Step 2 S = solubility = [Ag+]/2 Step 3 [Ag+]2[C2O42-] = Ksp = 3.510-11

(1)

[ H 3O  ][ HC 2 O -4 ]  K1  5.60  10 2 H 2C 2O 4 

(2)

[ H 3O  ][C 2 O 24- ]  K 2  5.42  10 5 [ HC 2 O 4 ]

(3)

Step 4 [Ag+] = 2([C2O42-] + [HC2O4-] + [H2C2O4])

(4)


Chapter 11

[H3O+] = 1.0  10-6

(5)

Step 5 No charge-balance equation because a buffer of unknown composition is present. Step 6 Unknowns: [Ag+], [C2O42-], [HC2O4-], [H2C2O4] Equations: (1), (2), (3) and (4) Step 7 No approximations needed. Step 8 Substituting (5) into (3) and rearranging gives



1.0  10 6 C 2 O 24[HC2O4 ] = 5.42  10 5 -



= 0.01845[C2O42-]

Substituting this relationship and (5) into (2) and rearranging gives [H2C2O4] =



1.0  10 6  0.01845 C 2 O 245.60  10 2



= 3.2910-7[C2O42-]

Substituting these two relationships into (4) gives [Ag+] = 2[C2O42-] + 2(0.01845[C2O42-]) + 2(3.2910-7[C2O42-]) [Ag+] = 2.037[C2O42-]

or

[C2O42-] = 0.4909[Ag+] Substituting this relationship into (1) and rearranging gives [Ag+] = (3.510-11/0.4909)1/3 = 4.1510-4 M Finally S = 4.1510-4/2 = 2.110-4 Substituting other values for [H3O+] yields the following solubility data:


11-7

Chapter 11

[H3O+], M

Solubility, M

(a)

1.010-6

2.110-4

(b)

1.010-7

2.110-4

(c)

1.010-9

2.110-4

(d)

1.010-11

2.110-4

Proceeding as in Problem 11-6, we write BaSO4  Ba2+ + SO42-

Ksp = 1.110-10

HSO4- + H2O  H3O+ + SO42-

K2 = 1.0210-2

S = [Ba2+] [Ba2+][SO42-] = 1.110-10

(1)

[ H 3O  ][SO 24 ]  1.0210-2 [ HSO 4 ]

(2)

Mass balance requires that [Ba2+] = [SO42-] + [HSO4-]

(3)

[H3O+] = 2.5

(4)

Substituting (4) into (2) gives upon rearranging [HSO4-] = (2.5)[SO42-]/(1.0210-2) = 245.1[SO42-]

(5)

Substituting (5) into (3) gives [Ba2+] = [SO42-] + 245.1[SO42-] = 246.1[SO42-] Substituting (6) into (1) gives 246.1[SO42-]2 = 1.110-10 [SO42-] = 6.710-7 M From (6)

[Ba2+] = 1.610-4 M = S

(6)


Chapter 11

Proceeding in a similar manner for the other [H3O+] gives

11-8

[H3O+]

S, mol/L

(a)

2.5

1.6510-4

(b)

1.5

1.2810-4

(c)

0.060

2.7510-5

(d)

0.200

4.7610-5

The following derivation applies to this and the following two problems. MS(s)  M2+ + S2H2S + H2O  H3O+ + HS-

K1 = 9.610-8

HS- + H2O-  H3O+ + S2-

K2 = 1.310-14

H2S + 2H2O  2H3O+ + S2-

K1K2 = 9.610-8  1.310-14 = 1.2510-21

S = solubility = [M2+] = [S2-] + [HS-] + [H2S] [M2+][S2-] = Ksp

(1)

[ H 3O  ][ HS  ]  K2 = 1.310-14 H 2S

(2)

[ H 3O  ]2 [S 2 ]  K1K2 = 1.2510-21 H 2S

(3)

From mass-balance consideration [M2+] = [S2-] + [HS-] + [H2S]

(4)

Substituting (2) and (3) into (4) gives [M2+] = [S2-] +

 [ H O  ] [ H 3 O  ]2  [ H 3O  ][S2 ] [ H 3O  ]2 [S2 ]    [S2 ]1  3  K2 K1 K 2 K2 K1 K 2  

Substituting (1) into (5)

(5)


(a)

[M2+] =

K sp  [ H 3O  ] [ H 3O  ]2  1    [ M 2 ]  K2 K1 K 2 

[M2+] =

 [ H O  ] [ H 3 O  ]2   K sp 1  3  K K K 2 1 2  

[M2+] =

 [ H 3O  ] [ H 3O  ]2   K sp 1   14 1.25  10 21   1.3  10

Substituting Ksp = 810-37 and [H3O+] = 2.010-1 into (6) gives [M2+] = solubility =

 0.202  0.20 8 10 37 1   14 1.25 10 21   1.3 10

= 5.1  10 9 M (b)

Substituting Ksp = 810-37 and [H3O+] = 2.010-4 into (6) gives solubility = 5.1  10 12 M

11-9

(a)

Substituting Ksp = 110-27 and [H3O+] = 2.010-1 into (6) gives solubility = 2  10 4 M

(b)

Substituting Ksp = 110-27 and [H3O+] = 2.010-4 into (6) gives solubility = 210-7 M

11-10 (a)

Substituting Ksp = 310-14 and [H3O+] = 2.010-5 into (6) gives solubility = 0.1 M

(b)

Substituting Ksp = 310-14 and [H3O+] = 2.010-7 into (6) gives solubility = 110-3 M

Chapter 11

(6)

Fundamentals of Analytical Chemistry: 8th ed. PbCO3  Pb2+ + CO32-

11-11

Chapter 11 Ksp = [Pb2+][CO32-] = 7.410-14

H2CO3 + H2O  H3O+ + HCO3- K1 =

[ H 3O  ][ HCO -3 ]  4.4510-7 [ H 2 CO 3 ]

HCO3- + H2O  H3O+ + CO32-

[ H 3O  ][CO 32- ]  4.6910-11  [ HCO 3 ]

K2 =

[Pb2+] = [CO32-] + [HCO3-] + [H2CO3] [H3O+] = 110-7 Proceeding as in problem 11-8 [M2+] =

2+

[Pb ] =

 [ H O  ] [ H 3 O  ]2   K sp 1  3  K2 K1 K 2  

7.4 10

14

7   1.0 10 7 2 1  1.0 10    4.69 10 11 4.45 10 7  4.69 10 11   

solubility = [Pb2+] = 1.410-5 M 11-12 Ag2SO3  2Ag+ + SO32-

Ksp = [Ag+]2[SO32-] = 1.510-14

H2SO3 + H2O  H3O+ + HSO3- K1 =

-

+

HSO3 + H2O  H3O + SO3

2-

[ H 3O  ][ HSO -3 ]  1.2310-2 [ H 2SO 3 ]

[ H 3O  ][SO 32- ]  6.610-8 K2 =  [ HSO3 ]

H2SO3 + 2H2O  2H3O+ + SO32-

K1K2 =

[ H 3O  ]2 [SO 32- ]  8.110-10 [ H 2SO 3 ]

(1) (2)

(3)

(4)

The mass balances are ½ [Ag+] = [SO32-] + [HSO3-] + [H2SO3]

(5)

[H3O+] = 110-8

(6)


Chapter 11

Substituting (2), (4) and (6) into (5) gives +

2-

½ [Ag ] = [SO3 ] +

1  10  [SO 8

6.6  10

23

]

-8

1  10  [SO 8 2

+

8.1  10

2   1  10 8 1  10 8     ½ [Ag ] = [SO3 ] 1   6.6  10 -8 8.1  10 10   

+

2-

Substituting (1) into this expression gives ½ [Ag+] =

8 8 2  K sp  1  1  10  1  10   2 -8 8.1  10 10  Ag   6.6  10





 1  108 2  1  10 8  [Ag+]3 = 21.510-14 1   6.6  10 -8 8.1  10 10    [Ag+] = 4.210-5 Solubility = 2.110-5 M 11-13 [Cu2+][OH-]2 = 4.810-20

[Mn2+][OH-]2 = 210-13

(a)

Cu(OH)2 precipitates first

(b)

Cu2+ begins to precipitate when [OH-] =

(c)

4.8 10 20 / 0.050 = 9.810-10 M

Mn2+ begins to precipitate when [OH-] =

2 10 13 / 0.040 = 2.210-6

[Cu2+] = 4.810-20/(2.210-6)2 = 9.610-9 M 11-14 Ba(IO3)2  Ba2+ + 2IO3BaSO4  Ba2+ + SO42To initiate precipitation of Ba(IO3)2

Ksp = 1.5710-9 Ksp = 1.310-10

-10

23

]


Chapter 11

[Ba2+] = 1.5710-9/(0.050)2 = 6.310-7 To initiate precipitation of BaSO4 [Ba2+] = 1.310-10/(0.040) = 3.210-9 M (a)

BaSO4 precipitates first

(b)

[Ba2+] = 3.210-9 M

(c)

When [Ba2+] = 6.310-7 M [SO42-] = 1.110-10/(6.310-7) = 1.710-4 M

11-15 (a)

[Ag+] = Ksp/[I-] = 8.310-17/(1.010-6) = 8.310-11 M

(b)

[Ag+] = Ksp/[SCN-] = 1.110-12/(0.070) = 1.610-11 M

(c)

[I-] when [Ag+] = 1.610-11 M [I-] = 8.310-17/(1.610-11) = 5.210-6 M [SCN-]/[I-] = 0.070/(5.210-6) = 1.3104

(d)

[I-] = 8.310-17/(1.010-3) = 8.310-14 M [SCN-] = 1.110-12/(1.010-3) = 1.110-9 M [SCN-]/[I-] = 1.110-9/(8.310-14) = 1.3104 Note that this ratio is independent of [Ag+] as long as some AgSCN(s) is present.

11-16 (a)

[Ba2+][SO42-] = 1.110-10

[Sr2+][SO42-] = 3.210-7

BaSO4 precipitation is complete when [SO42-] = 1.110-10/(1.010-6) = 1.110-4 M SrSO4 begins to precipitate when [SO42-] = 3.210-7/(0.050) = 6.410-6 M SrSO4 begins to precipitate before the Ba2+ concentration is reduced to 1.010-6 M.

Fundamentals of Analytical Chemistry: 8th ed. (b)

[Ba2+][SO42-] = 1.110-10

Chapter 11

[Ag+]2[SO42-] = 1.610-5

BaSO4 precipitation is complete when [SO42-] = 1.110-4 M Ag2SO4 begins to precipitate when [SO42-] = 1.610-5/(0.020)2 = 0.040 M Ag2SO4 does not precipitate before the Ba2+ concentration is reduced to 1.010-6 M. (c) Be2+ precipitates when [OH-] = (7.010-22/(0.020))1/2 = 1.910-10 M Hf4+ precipitates when [OH-] = (4.010-26/(0.010))1/4 = 1.410-6 M Be precipitation complete when [OH-] = (7.010-22/1.010-6)1/2 = 2.610-8 M Hf(OH)4 does not precipitate before the Be2+ concentration is reduced to 1.010-6 M. (d)

In3+ precipitates when [IO3-] = (3.310-11/(0.20))1/3 = 5.510-4 M Tl+ precipitates when [IO3-] = 3.110-6/(0.090) = 3.410-5 M Tl+ precipitation complete when [IO3-] = 3.110-6/1.010-6 = 3.1 M

In(IO3)3 begins to precipitate before the Tl+ concentration is reduced to 1.010-6 M. 11-17 AgBr  Ag+ + BrAg+ + 2CN-  Ag(CN)2-

5.010-13 = [Ag+][Br-] 1.31021 =

[Ag (CN) 2 ] [Ag  ][CN  ]2

(1) (2)

It is readily shown that the reaction CN- + H2O  HCN + OHproceeds to such a small extent that it can be neglected in formulating a solution to this problem. That is, [HCN] << [CN-], and only the equilibria shown need to be taken into account. Solubility = [Br-]


Chapter 11

Mass balance requires that [Br-] = [Ag+] + [Ag(CN)2-]

(3)

0.100 = [CN-] + 2[Ag(CN)2-]

(4)

We now have 4 equations and 4 unknowns. Based upon the large size of 2 let us assume that [CN-] << 2[Ag(CN)2-]

and

[Ag+] << [Ag(CN)2-]

(4) becomes

[Ag(CN)2-] = 0.100/2 = 0.0500

and (3) becomes

[Br-] = [Ag(CN)2-] = 0.0500

To check the assumptions, we calculate [Ag+] by substituting into (1) [Ag+] = 5.010-13/0.0500  110-11

( 110-11 << 0.0500)

To obtain [CN-] we substitute into (2) and rearrange [CN-] =

0.0500 1  10 1.3  10 21



11





= 2.010-6

( 210-6 << 0.100)

Thus, the two assumptions are valid and Solubility = [Br-] = 0.0500 M mass AgBr/200 mL

= 0.0500

mmol 0.1877 g AgBr  200 mL  mL mmol AgBr

= 1.877 g 11-18 CuCl(s)  Cu+ + ClCu+ + 2Cl-  CuCl2-

Ksp = [Cu+][Cl-] = 1.910-7 2 =

[CuCl-2 ] = 7.9104  - 2 [Cu ][Cl ]

(1) (2)

It is convenient to multiply (1) by (2) to give

[CuCl -2 ] = 1.910-77.9104 = 1.510-2 [Cl ]

(3)


Chapter 11

From a charge balance consideration, we can write (if we assume [H3O+] = [OH-]) [Cu+] + [Na+] = [Cl-] + [CuCl2-]

(4)

By rearranging (1) and (3) and substituting into (4) we obtain [Cl-] = [Na+] +

1.9  107 - 1.510-2[Cl-] [Cl ]

which rearranges to the quadratic 0 = 1.015[Cl-]2 - [Na+][Cl-] - 1.910-7

(5)

By using [Na+] = the NaCl analytical concentration, (5) can be solved to give the following [Cl-] (a) 2.0 M

(c) 0.020 M

(b) 0.20 M

(d) 0.0021 M

(e) 5.410-4 M

Note that the equilibrium [Cl-] concentration is larger than the NaCl analytical concentration for parts (d) and (e). The reason for this apparent anomaly is that the dissolution of CuCl to give Cu+ and Cl- contributes significantly to the equilibrium [Cl-] at the lower NaCl analytical concentrations. The solubility of CuCl can be obtained from the calculated [Cl-] and the expression

1.9  107 S = [Cu ] + [CuCl2 ] = + 1.510-2[Cl-]  [Cl ] +

-

Solution of this equation for each of the [Cl-] gives (a) 0.030 M

(c) 3.110-4 M

(b) 3.010-3 M

(d) 1.210-4 M

(e) 3.610-4 M

Fundamentals of Analytical Chemistry: 8th ed. 11-19 (a) CaSO4(s)  Ca2+ + SO42CaSO4(aq)  Ca2+ + SO42-

Chapter 11

Ksp = [Ca2+][SO42-] = 2.610-5

(1)

[Ca 2 ][SO 24- ] = 5.210-3 [CaSO 4 ( aq)]

(2)

Kd =

CaSO4(s)  CaSO4(aq)

(3)

The mass balance gives [Ca2+] = [SO42-]

(4)

We have 3 equations and 3 unknowns ([Ca2+], [SO42-], and [CaSO4(aq)] To solve we divide (1) by (2) to give [CaSO4(aq)] = Ksp/Kd = (2.610-5)/(5.210-3) = 5.010-3 Note that this is the equilibrium constant expression for (3) and indicates that the concentration of un-ionized CaSO4 is always the same in a saturated solution of CaSO4. Substituting (4) into (1) gives [Ca2+] = (2.610-5)1/2 = 5.110-3 and since S = [CaSO4(aq)] + [Ca2+] we obtain S = 5.010-3 + 5.110-3 = 0.0101 M %CaSO4(aq) = (5.010-3/1.0110-2)100% = 49% (b) (Note: In the first printing of the text, the answer in the back of the book was in error.) Here [CaSO4(aq)] is again equal to 5.010-3 and the mass balance gives [SO42-] = 0.0100 + [Ca2+]

(5)

Substituting (1) into (5) and rearranging gives 0 = [SO42-]2 - 0.0100[SO42-] - Ksp which may be solved using the quadratic equation to give [SO42-] = 0.0121

and

[Ca2+] = 2.1410-3

Fundamentals of Analytical Chemistry: 8th ed. S = 5.010-3 + 2.1410-3 = 7.1410-3 %CaSO4(aq) = (5.010-3/7.1410-3)100% = 70%

Chapter 11

Chapter 11 Solutions

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