Chapter 15 Solutions

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

Chapter 15 15-1

The HPO42- is such a weak acid (Ka3 = 4.510-13) that the change in pH in the vicinity of the third equivalence point is too small to be observable.

15-2

(a)

NH4+ + H2O  NH3 + H3O+

Ka = 5.7010-10

OAc- + H2O  HOAc + OH-

Kb =

K w 1.00  10 14   5.71  10 10 5 K a 1.75  10

Since the K’s are essentially identical, the solution should be approximately neutral. (b)

NO2- + H2O  HNO2 + OH-

Solution will be basic

(c)

Neither K+ nor NO3- reacts with H2O.

Solution will be neutral

(d)

HC2O4- + H2O  C2O42- + H3O+

Ka2 = 5.4210-5

HC2O4- + H2O  H2C2O4 + OH-

Kb2 =

1.00  10 14  1.79  10 13 2 5.60  10 Solution will be acidic

(e)

C2O42- + H2O  HC2O4- + OH-

Kb =

1.00  10 14  1.84  10 10 5 5.42  10 Solution will be basic

(f)

HAsO42- + H2O  AsO43- + H3O+

Ka3 = 3.210-12

HAsO42- + H2O  H2AsO4- + OH-

Kb2 =

1.00  10 14  9.1 10 8 7 1.1 10 Solution will be basic

(g)

H2AsO4- + H2O  HAsO42- + H3O+ Ka2 = 1.110-7 H2AsO4- + H2O  H3AsO4 + OH-

Kb3 =

1.00  10 14  1.7  10 12 3 5.8  10 Solution will be acidic

Fundamentals of Analytical Chemistry: 8th ed.

(h)

Chapter 15

AsO43- + H2O  HAsO42- + OH-

Kb1 =

1.00  10 14  3.1 10 3 12 3.2  10 Solution will basic

15-3

H3PO4 + H2O  H2PO4- + H3O+

Ka1 = 7.1110-3

H2PO4- + H2O  HPO42- + H3O+

Ka2 = 6.3210-8

Here both Ka2 and Kw/Ka1 are small and we may assume that [H2PO4-] >> [H3PO4] and [HPO42-] at the first equivalence point. If we further assume that Kw > 1 we may use equation 15-16 to give [H3O+] = (7.1110-36.3210-8)1/2 = 2.1210-5 pH = -log(2.1210-5) = 4.674 Bromocresol green would be satisfactory 15-4

(Note: In the first printing of the text, the answer in the back of the book was in error.) H2PO4- + H2O  HPO42- + H3O+

Ka2 = 6.3210-8

HPO42- + H2O  PO43- + H3O+

Ka3 = 4.510-13

Here again both Ka3 and Kw/Ka2 are small and we may assume that [HPO42-] >> [H2PO4-] and [PO43-] at the first equivalence point. However, the further approximation Kw > 1 is not valid. [H3O+] =

0.05  1.1  10 7  1.0  10 14 = 2.410-5 1  0.05 /(5.8  10 3 )

pH = -log(2.410-5) = 4.62 Bromocresol green (3.8 to 5.4) (b)

P2- + H2O  HP- + OH-

Kb1 =

[OH - ][HP - ] 1.00  10 14   2.56  10 9 [ P 2- ] 3.91  10 6

[OH-] = [HP-] and we assume [P2-] = 0.05 – [OH-]  0.05 [OH-] = (0.052.5610-9)1/2 = 1.1310-5 pH = 14.00 – (-log(1.1310-5)) = 9.05 Phenolphthalein (8.3 to 10) (c)

As in part (b) [OH-] = (0.051.0010-14/4.3110-5)1/2 = 3.4110-6 pH = 14.00 – (-log(3.4110-6)) = 8.53 Cresol purple (7.6 to9.2)

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

Chapter 15

Here we are able to use equation 15-16 [H3O+] = (1.4210-71.1810-10)1/2 = 4.0910-9 pH = -log(4.0910-9) = 8.39 Cresol purple (7.6 to 9.2)

(e)

NH3C2H4NH32+ + H2O  NH3C2H4NH2+ + H3O+

Ka1 = 1.4210-7

[H3O+] = (0.051.4210-7)1/2 = 8.4310-5 pH = -log(8.4310-5) = 4.07 Bromocresol green (3.8 to 5.4) (f)

Proceeding as in part (a) +

[H3O ] =

0.05  6.6  10 8  1.0  10 14 = 2.5510-5 2 1  0.05 /(1.23  10 )

pH = -log(2.5510-5) = 4.59 Bromocresol green (3.8 to 5.4) (g)

Proceeding as in part (b) we obtain pH = 9.94 Phenolphthalein (8.5 to 10.0)

15-7

(a)

H3AsO4 + H2O  H2AsO4- + H3O+

Ka1 = 5.810-3

[H 3 O  ][H 2 AsO -4 ] [H 3 O  ] 2   5.8  10 3 [H 3 AsO 4 ] 0.0400  [H 3 O  ]

0 = [H3O+]2 + 5.810-3[H3O+] – 5.810-30.0400 Solving using the quadratic formula gives [H3O+] = 1.2610-2 and pH = 1.90 Proceeding in the same way we obtain (b)

2.20

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

(c)

1.64

(d)

1.77

(e)

4.21

(f)

NH2C2H4NH2 + H2O  NH3C2H4NH2+ + OH-

Kb1 =

1.00  10 14  8.47  10 5 10 1.18  10

[OH - ]2  8.4710-5 0.0400 [OH-] = (0.04008.4710-5)1/2 = 1.8410-3 pH = 14.00 – (-log(1.8410-3)) = 11.26

15-8

(a)

[H3O+] =

0.0400  1.1  10 7  1.0  10 14 = 2.3610-5 3 1  0.0400 /(5.8  10 )

pH = -log(2.3610-5) = 4.63 Proceeding as in part (a) we obtain

15-9

(b)

pH = 2.95

(c)

pH = 4.28

(d)

pH = 4.60

(e)

pH = 9.80

(f)

pH = 8.39

(a)

AsO43- + H2O  HAsO42- + OH-

Kb1 =

[OH - ][HAsO 24- ] [OH - ] 2   3.12  10 3 3[AsO 4 ] 0.0400  [OH ]

0 = [OH-]2 + 3.1210-3[OH-] – 3.1210-30.0400 Solving using the quadratic formula gives

1.00  10 14  3.12  10 3 3.2  10 12

Fundamentals of Analytical Chemistry: 8th ed. [OH-] = 9.7210-3 (b)

and

Chapter 15

pH = 14.00 – (-log(9.7210-3)) = 11.99

C2O42- + H2O  HC2O4- + OH-

Kb1 =

[OH-] = (0.04001.8410-10)1/2 = 2.7210-6 and

1.00  10 14  1.84  10 10 5 5.42  10 pH = 8.43

Proceeding as in part (b) we obtain (c)

pH = 9.70

(d)

pH = 9.89

(e)

Proceeding as in part (a) gives pH = 12.58

(f)

NH3C2H4NH32+ + H2O  NH3C2H4NH2+ + H3O+

Ka1 = 1.4210-7

[H3O+] = (0.04001.4210-7)1/2 = 7.5410-5 and

pH = 4.12

15-10 (a)

H3PO4 + H2O  H2PO4- + H3O+

Ka1 = 7.1110-3

[H 3 O  ][H 2 PO-4 ] (0.0200  [H 3 O  ])[H 3 O  ]  7.1110 = [H 3 PO4 ] 0.0500  [H 3 O  ] -3

Rearranging gives 0 = [H3O+]2 +(0.0200+7.1110-3)[H3O+] – (7.1110-3)(0.0500) and solving the quadratic gives (b)

[H3O+] = 9.6710-3 and pH = 2.01

H2AsO4- + H2O  HAsO42- + H3O+

Ka2 = 1.1110-7

[H 3 O  ][HAsO 24- ] (0.0500  [H 3 O  ])[H 3 O  ] (0.0500)[H 3 O  ]   1.1110 = 0.0300 [H 2 AsO -4 ] 0.0300  [H 3 O  ] -7

[H3O+] = 6.6610-8 (c)

and

pH = 7.18

HCO3- + H2O  CO32- + H3O+ Proceeding as in part (b) we obtain

(d)

Ka2 = 4.6910-11 [H3O+] = 2.3410-11 and pH = 10.63

H3PO4 + HPO42-  2H2PO4For each milliliter of solution, 0.0200 mmol Na2HPO4 reacts with 0.0200 mmol

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

H3PO4 to give 0.0400 mmol NaH2PO4 and to leave 0.0200 mmol H3PO4. Thus, we have a buffer that is 0.0200 M in H3PO4 and 0.0400 M in NaH2PO4. Proceeding as in part (a) we obtain (e)

[H3O+] = 2.8210-3 and pH = 2.55

HSO4- + H2O  SO42- + H3O+

Ka2 = 1.0210-2 [H3O+] = 8.6610-3

and

pH = 2.06

15-11 (a)

Proceeding as in 15-10(a) we obtain [H3O+] = 3.4810-3

and

pH = 2.46

(b)

Proceeding as in 15-10(b) we obtain [H3O+] = 3.1010-8

and

pH = 7.51

(c)

HOC2H4NH3+ + H2O  HOC2H4NH2 + H3O+

Proceeding as in part (a) we obtain

Ka = 3.1810-10

Proceeding as in 15-10(b) we obtain [H3O+] = 3.7310-10 and (d)

pH = 9.43

H2C2O4 + C2O42-  2HC2O4For each milliliter of solution, 0.0240 mmol H2HPO4 reacts with 0.0240 mmol C2O42- to give 0.0480 mmol HC2O4- and to leave 0.0120 mmol C2O42-. Thus, we have a buffer that is 0.0480 M in HC2O4- and 0.0120 M in C2O42-. Proceeding as in 15-10(a) we obtain [H3O+] = 2.1710-4 and pH = 3.66

(e) 15-12 (a)

Proceeding as in 15-10(b) we obtain [H3O+] = 2.1710-4 and pH = 3.66 (NO2)3C6H2OH + H2O  (NO2)3C6H2O- + H3O+

Ka = 0.43

[H 3 O  ][(NO 2 ) 3 C 6 H 2 O - ] (0.0100  x) x  0.43 = [(NO 2 ) 3 C 6 H 2 OH] 0.0200  x

Rearranging gives 0 = x2 +(0.0100+0.43)x – (0.43)(0.0200) and solving the quadratic gives

x = 1.8710-2

the total [H3O+] = 0.0100 + x = 0.0287 (b)

Proceeding as in part (a) we obtain

and pH = 1.54

[H3O+] = 1.01310-2 and pH = 1.99

Fundamentals of Analytical Chemistry: 8th ed.

(c)

Chapter 15

CO32- + H2O  HCO3- + OH-

2.1310-4 =

Kb1 =

1.00  10 14  2.13  10 4 11 4.69  10

[OH - ][HCO 3- ] (0.0100  x) x  0.100  x [CO 32- ]

Rearranging gives 0 = x2 +(0.0100+2.1310-4)x – (2.1310-4)(0.100) and solving the quadratic gives

x = 1.7810-3

the total [OH-] = 0.0100 + x = 0.0118 (d) 15-13 (a) (b)

Proceeding as in part (c) we obtain

and pH = 12.07

[OH-] = 1.01710-2 and pH = 12.01

Proceeding as in 15-12(a) we obtain [H3O+] = 1.28710-2 and pH = 1.89 Recognizing that the first proton of H2SO4 completely dissociates we obtain HSO4- + H2O  SO42- + H3O+

Ka2 = 1.0210-2

[H 3 O  ][SO 24- ] (0.0100  0.0150  x) x  1.0210 = 0.0150  x [HSO -4 ] -2

Rearranging gives 0 = x2 +(0.0250+1.0210-2)x – (1.0210-2)(0.0150) and solving the quadratic gives

x = 3.9110-3

the total [H3O+] = 0.0250 + x = 0.0289

and pH = 1.54

(c)

Proceeding as in 15-12(c) we obtain [OH-] = 0.0382 and pH = 12.58

(d)

CH3COO- + H2O  CH3COOH + OH-

Kb1 =

1.00  10 14  5.7  10 10 1.75  10 5

CH3COO- is such a weak base that it makes no significant contribution to [OH-]. Therefore, 15-14 (a)

[OH-] = 0.010 and pH = 12.00

Let us compare the ratio [H2SO3]/[HSO3-] with that of [SO32-]/[HSO3-] at [H3O+] = 1.0010-6. The larger ratio will contain the predominant acid/base pair.

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

[H 2SO 3 ] [H 3 O  ] 1.00  10 6    8.1  10 5 2 K a1 [HSO 3 ] 1.23  10 [SO 32- ] K a2 6.6  10 8    0.066 [HSO 3- ] [H 3 O  ] 1.00  10 6

Clearly the predominant pair is SO32-/HSO3- and its acid/base ratio is 1/0.066 = 15.2 (b)

Substituting [H3O+] = 1.0010-6 into the expressions for K1, K2 and K3 yields

[H 2 Cit - ] = 745 [H 3 Cit]

[HCit 2- ] = 17.3 [H 2 Cit - ]

[Cit 3- ] = 0.40 [HCit 2- ]

The large size of the first two ratios and the small size of the third indicate the HCit2- is a predominant species in this solution. To compare [Cit3-] and [H2Cit-] we invert the second ratio. Then [H2Cit-]/[HCit2-] = 1/17.3 = 0.058 Thus, the predominant acid/base system involves [Cit3-] and [HCit2-] and their acid/base ratio is [HCit2-]/[Cit3-] = 1/0.40 = 2.5 (c)

Proceeding as in part (a) we obtain [HM-]/[M2-] = 0.498

(d) (Note: In the first printing of the text, the answer in the back of the book was in error.) Proceeding as in part (a) we obtain [HT-]/[T2-] = 0.0232 15-15 (a)

Proceeding as in Problem 15-14(a) with [H3O+] = 1.0010-9 we obtain [H2S]/[HS-] = 0.010

(b)

Formulating the three species as BH22+, BH+ and B, where B is the symbol for NH2C2H5NH2.

Fundamentals of Analytical Chemistry: 8th ed. [H 3 O  ][BH  ] = K1 = 1.4210-7 and 2 [BH 2 ]

[BH22+]/[BH+] =

[B]/[BH+] =

Chapter 15

[H 3 O  ][B] = K2 = 1.1810-10  [BH ]

1.00  10 9 = 0.0070 1.42  10 7

1.18  10 10 = 0.118 1.00  10 9

[BH22+] is clearly < [B] and [BH+]/[B] = 1.00/0.118 = 8.5 (c)

Proceeding as in Problem 15-14(b) we find [H2AsO4-]/[HAsO42-] = 9.110-3

(d)

Proceeding as in Problem 15-14(a) we find [HCO3-]/[CO32-] = 21

15-16 pH = 7.30

[H3O+] = antilog (-7.30) = 5.01210-8

[H3O+][HPO42-]/[H2PO4-] = 6.3210-8 [HPO42-]/[H2PO4-] = 6.3210-8 / (5.01210-8) = 1.261 HPO42- + H3PO4  2H2PO4no. mmol H3PO4 present = 400  0.200 = 80.0 no. mmol H2PO4- in the buffer = 2  80.0 = 160.0 no. mmol HPO42- needed for the buffer = 1.261  160.0 = 201.8 Thus, we need 80.0 mmol of Na2HPO4 to react with the H3PO4 and an additional 201.8 mmol to provide the needed concentration of HPO42- or 281.8 mmol. mass Na2HPO42H2O = 281.8 mmol  0.17799 g/mmol = 50.2 g 15-17 pH = 5.75

[H3O+] = antilog (-5.75) = 1.77810-6

[H3O+][P2-]/[HP-] = 3.9110-6

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

[P2-]/[HP-] = 3.9110-6 / (1.77810-6) = 2.199 P2- + H2P  2HPno. mmol H2P present = 750  0.0500 = 37.5 no. mmol HP- in the buffer = 2  37.5 = 75.0 no. mmol P2- needed in the buffer = 2.199  75.0 = 164.9 Thus, we need 37.5 + 164.9 = 202.4 mmol of K2P. mass K2P = 202.4 mmol  0.24232 g K2P /mmol = 49.0 g K2P 15-18 no. mmol NaH2PO4 = 50.0  0.200 = 10.0 (a)

no. mmol H3PO4 formed = no. mmol HCl added = 50.0  0.120 = 6.00

cH3PO4 = 6.00/100 = 0.0600 M cNaH2PO4 = (10.0 – 6.00)/100 = 0.0400 Proceeding as in Problem 15-10(a) we obtain pH = 2.11 (b)

cNa2HPO4 = 6.00/100 = 0.0600 M cNaH2PO4 = (10.0 – 6.00)/100 = 0.0400

Proceeding as in Problem 15-10(b), using K2 gives pH = 7.38 15-19 no. mmol KHP = 100  0.150 = 15.0 (a)

no. mmol P2- = 100  0.0800 = 8.00 no. mmol KHP = 15.0 – 8.00 = 7.00

cHP- = 7.00/200 = 0.0350 M

cP2- = 8.00/200 = 0.0400

Proceeding as in Problem 15-10(b) we obtain pH = 5.47

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

cH2P = 8.00/200 = 0.0400 M

Chapter 15

cHP- = (15.00 - 8.00)/200 = 0.0350

Proceeding as in Problem 15-10(a) we obtain pH = 2.92 15-20 pH = 9.60

[H3O+] = antilog (-9.60) = 2.51210-10

[H3O+][CO32-]/[HCO3-] = 4.6910-11 [CO32-]/[HCO3-] = 4.6910-11/2.51210-10 = 0.1867 Let VHCl = mL 0.200 M HCl and VNa2CO3 = mL 0.300 M Na2CO3 Since the solutions are dilute, the volumes will be additive VHCl + VNa2CO3 = 1000 mL Assume [CO32-] ≈ cNa2CO3 = ( VNa2CO3  0.300 - VHCl  0.200)/1000 [HCO3-] ≈ cHCO- = VHCl  0.200/1000 3

Substituting these relationships into the ratio [CO32-]/[HCO3-] gives 0.300 VNa 2CO3  0.200VHCl 0.200VHCl

= 0.1867

0.300 VNa2CO3 - 0.200VHCl = 0.023734 VHCl 0.300(1000 – VHCl) = 0.23734 VHCl VHCl = 300/0.5373 = 558 mL

VNa2CO3 = 1000 – 558 = 442 mL Thus mix 442 mL of 0.300 M Na2CO3 with (1000 – 442) = 558 mL of 0.200 M HCl. 15-21 [H3O+][HPO42-]/[HPO4-] = 6.3210-8 [HPO24- ] 6.32  10 8  = 0.632 [H 2 PO-4 ] 1.00  10 7

(1)

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

Let VH3PO4 and VNaOH be the volume in milliliters of the two reagents. Then

VH3PO4 + VNaOH = 1000 mL

(2)

From mass-balance considerations we may write that in the 1000 mL no. mmol NaH2PO4 + no. mmol Na2HPO4 = 0.200  VH3PO4

(3)

no. mmol NaH2PO4 + 2  no. mmol Na2HPO4 = 0.160  VNaOH

(4)

Equation (1) can be written as

no. mmol Na 2 HPO4 /1000 no. mmol Na 2 HPO4 = 0.632  no. mmol NaH 2 PO4 /1000 no. mmol NaH 2 PO4

(5)

Thus, we have four equations, (2), (3), (4) and (5), and four unknowns: VH3PO4 , VNaOH, no. mmol NaH2PO4 and no. mmol Na2HPO4. Subtracting Equation (3) from (4) yields no. mmol Na2HPO4 = 0.160 VNaOH – 0.200 VH3PO4

(6)

Substituting Equation (6) into (3) gives no. mmo NaH2PO4 + 0.160 VNaOH – 0.200 VH3PO4 = 0.200 VH3PO4 no. mmo NaH2PO4 = -0.160 VNaOH + 0.400 VH3PO4 Substituting Equations (6) and (7) into (5) gives

0.160VNaOH  0.200VH3P O4 0.400VH3P O4  0.160VNaOH

= 0.632

This equation rearranges to 0.2611 VNaOH = 0.4528 VH3PO4 Substituting Equation (2) gives 0.2611 (1000 - VH3PO4 ) = 0.4528 VH3PO4

(7)

Fundamentals of Analytical Chemistry: 8th ed.

VH3PO4 = 261.1/0.7139 = 366 mL and

Chapter 15 VNaOH = 1000 – 366 = 634 mL

Thus, mix 366 mL H3PO4 with 634 mL NaOH 15-22 [H3O+][HAsO42-]/[H2AsO4-] = 1.110-7 [HAsO42-]/[H2AsO4-] = 1.110-7 / (1.0010-6) = 0.11

(1)

As in Problem 15-21 we now develop four independent equations that allow calculation of VHCl, VNa3AsO4 , no. mmol of HAsO42- and no. mmol of H2AsO4-.

VNa3AsO4 + VHCl = 1000 mL

(2)

no. mmol NaH2AsO4 + no. mmol Na2HAsO4 = 0.500 VNa3AsO4

(3)

2  no. mmol NaH2AsO4 + no. mmol Na2HAsO4 = 0.400 VHCl

(4)

no. mmol Na2HAsO4 / no. mmol NaH2AsO4 = 0.11

(5)

Proceeding as in Problem 15-21 we solve equations (2), (3), (4) and (5) and conclude that 704 mL of the HCl should be mixed with 296 mL of the Na3AsO4. 15-23 (a) For Na2M, K1 = 1.310-2 and K2 = 5.910-7. This initial pH is high and two equally spaced end points will be encountered. Thus, curve C. (b) Curve A. (c) Curve C for the reasons given in part (a). 15-24 (a) Titrations with NaOH of a solution containing a mixture of two weak acids HA1 and HA2. HA1 is present in a greater concentration and has a dissociation constant that is larger by a factor of about 104. (b) Titration of a typical monoprotic weak acid.

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

(c) Titration of a mixture of a weak base, such as Na2CO3, and an acid salt, such as NaHCO3. 15-25 For the titration of a mixture of H3PO4 and H2PO4- the volume to the first end point would have to be smaller than one half the total volume to the second point because in the titration from the first to second end points both analytes are titrated, whereas to the first end point only the H3PO4 is titrated. For Problem 15-26 we set up spreadsheets that will solve a quadratic equation to determine [H3O+] or [OH-], as needed (See approach taken for Problems 14-41 through 43). While approximate solutions are appropriate for many of the calculations, the approach taken represents a more general solution and is somewhat easier to incorporate in a spreadsheet. As an example consider the titration of a weak diprotic acid with a strong base. Before the 1st equivalence point:

[H2A] =

c

Vi H 2 A  ci NaOHVNaOH

i H 2A

V

i H2A

[HA-] =

and

 VNaOH

ci NaOHVNaOH 

V

i H2A

 VNaOH







- [H3O+]

+ [H3O+]

Substituting these expressions into the equilibrium expression for H2A and rearranging gives



 c  K a1 ci H 2 AVi H 2 A  ci NaOHVNaOH  V 0 = [H3O+]2 +  i NaOH NaOH  K a1  [H3O+]  Vi H A  VNaOH  Vi H 2 A  VNaOH 2  





From which [H3O+] is directly determined. At the 1st equivalence point we use Equation 15-13 [H3O+] =

K a2 ciH 2 AViH 2A /(ViH 2A  VNaOH )  K w 1  ciH 2AViH 2 A /( K a1 (ViH 2 A  VNaOH ))

After the 1st and before the 2nd equivalence point:







Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

[HA-] =

2  c

Vi H 2 A  ci NaOHVNaOH

i H2A

V

 VNaOH

i H2A

c

[A2-] =

and

V

i NaOH NaOH

V

i H2A



 ciH 2 AViH 2 A

 VNaOH



 - [H O+] 3

 + [H O+] 3

Substituting these expressions into the equilibrium expression for HA- and rearranging gives







 ci NaOHVNaOH  ciH 2 AViH 2 A  K a2 2  ci H 2AVi H 2A  ci NaOHVNaOH  K a2  [H3O+] 0 = [H3O+]2 +    Vi H 2 A  VNaOH Vi H 2 A  VNaOH  











From which [H3O+] is directly determined. At and after the 2nd equivalence point:

c

[A2-] =

V

i H2A

[OH-] =

and

c

VH 2 A

i H 2A



 VNaOH V

i NaOH NaOH

V

i H2A



- [HA-]

 2  ci H 2AVi H 2A  VNaOH





+ [HA-]

Substituting these expressions into the equilibrium expression for A2- and rearranging gives









 ci NaOHVNaOH  2  ci H 2 AVi H 2 A K w ci H 2 AViH 2A K   w  [HA-] 0 = [HA-]2 +   Vi H 2 A  VNaOH K a2  K a2 Vi H 2A  VNaOH 









From which [HA-] can be determined and [OH-] and [H3O+] subsequently calculated. A similar approach is taken for the titration of a weak base with a strong acid.

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

15-26 A

B

C

D

E

F

1 Part (a)

Vi, H2SO3

50.00

Kw, H2O

1.00E-14

2

ci, H2SO3

0.1000

c, NaOH

0.2000

3

Ka1, H2SO3

1.23E-02

4

Ka2, H2SO3

6.60E-08

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Vol. NaOH, mL 0.00 12.50 20.00 24.00 25.00 26.00 37.50 45.00 49.00 50.00 51.00 60.00

b 1.2300E-02 5.2300E-02 6.9443E-02 7.7165E-02

c -1.2300E-03 -4.9200E-04 -1.7571E-04 -3.3243E-05

2.6316E-03 2.8571E-02 4.2105E-02 4.8485E-02 1.5152E-07 1.9803E-03 1.8182E-02

-4.1684E-09 -1.8857E-09 -6.9474E-10 -1.3333E-10 -7.5758E-09 -7.5008E-09 -6.8871E-09

Veq. pt. -

[OH ]

8.6963E-05 1.9840E-03 1.8182E-02

25.00 +

[H3O ] 2.9456E-02 8.1403E-03 2.4443E-03 4.2843E-04 2.6179E-05 1.5830E-06 6.6000E-08 1.6500E-08 2.7500E-09 1.1499E-10 5.0404E-12 5.4999E-13

pH 1.53 2.09 2.61 3.37 4.58 5.80 7.18 7.78 8.56 9.94 11.30 12.26

Spreadsheet Documentation F3 = C1*C2/F2 B6 = $F$2*A6/($C$1+A6)+$C$3 C6 = -$C$3*($C$2*$C$1-$F$2*A6)/($C$1+A6) E6 = (-B6+SQRT(B6^2-4*C6))/2 F6 = -LOG(E6) E10 = SQRT(($C$4*$C$1*$C$2/($C$1+A10)+$F$1)/(1+$C$1*$C$2/($C$1+A10)/$C$3)) B11 = ($F$2*A11-$C$1*$C$2)/($C$1+A11)+$C$4 C11 = -$C$4*(2*$C$2*$C$1-$F$2*A11)/($C$1+A11) B15 = ($F$2*A15-2*$C$1*$C$2)/($C$1+A15)+$F$1/$C$4 C15 = -($F$1/$C$4)*($C$2*$C$1)/($C$1+A15) D15 = (-B15+SQRT(B15^2-4*C15))/2+($F$2*A15-2*$C$1*$C$2)/($C$1+A15) E15 = $F$1/D15

Fundamentals of Analytical Chemistry: 8th ed. A

B

Chapter 15

C

D

E

F

1 Part (b)

Vi, H2NC2H4NH2

50.00

Kw, H2O

1.00E-14

2

ci, H2NC2H4NH2

0.1000

c, NaOH

0.2000

3

2+ Ka1, H3NC2H4NH3 + Ka2, H2NC2H4NH3

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Vol. HCl, mL 0.00 12.50 20.00 24.00 25.00 26.00 37.50 45.00 49.00 50.00 51.00 60.00

Veq. pt.

1.42E-07

25.00

1.18E-10 -

b 8.4746E-05 4.0085E-02 5.7228E-02 6.4950E-02

c -8.4746E-06 -3.3898E-06 -1.2107E-06 -2.2904E-07

[OH ] 2.8690E-03 8.4389E-05 2.1147E-05 3.5263E-06

2.6316E-03 2.8571E-02 4.2105E-02 4.8485E-02 1.4200E-07 1.9803E-03 1.8182E-02

-4.4477E-09 -2.0121E-09 -7.4129E-10 -1.4227E-10 -7.1000E-09 -7.0297E-09 -6.4545E-09

1.6890E-06 7.0422E-08 1.7606E-08 2.9343E-09

+

[H3O ] 3.4855E-12 1.1850E-10 4.7287E-10 2.8359E-09 4.0960E-09 5.9206E-09 1.4200E-07 5.6800E-07 3.4080E-06 8.4191E-05 1.9837E-03 1.8182E-02

pH 11.46 9.93 9.33 8.55 8.39 8.23 6.85 6.25 5.47 4.07 2.70 1.74

Spreadsheet Documentation F3 = C1*C2/F2 B6 = $F$2*A6/($C$1+A6)+$F$1/$C$4 C6 = -$F$1/$C$4*($C$2*$C$1-$F$2*A6)/($C$1+A6) D6 = (-B6+SQRT(B6^2-4*C6))/2 E6 = $F$1/D6 F6 = -LOG(E6) E10 = SQRT(($C$4*$C$1*$C$2/($C$1+A10)+$F$1)/(1+$C$1*$C$2/($C$1+A10)/$C$3)) B11 = ($F$2*A11-$C$1*$C$2)/($C$1+A11)+$F$1/$C$3 C11 = -($F$1/$C$3)*(2*$C$2*$C$1-$F$2*A11)/($C$1+A11) B15 = ($F$2*A15-2*$C$2*$C$1)/($C$1+A15)+$C$3 C15 = -($C$3)*($C$1*$C$2/($C$1+A15)) E15 = (-B15+SQRT(B15^2-4*C15))/2+(A15*$F$2-2*$C$1*$C$2)/(A15+$C$1)

Fundamentals of Analytical Chemistry: 8th ed. A

B

Chapter 15

C

D

E

F

1 Part (c)

Vi, H2SO4

50.00

Kw, H2O

1.00E-14

2

ci, H2SO4

0.1000

c, NaOH

0.2000

3

Ka1, H2SO4

4

Ka2, H2SO4

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Vol. NaOH, mL 0.00 12.50 20.00 24.00 25.00 26.00 37.50 45.00 49.00 50.00 51.00 60.00

b 1.1020E-01 5.0200E-02 2.4486E-02 1.2903E-02 1.0200E-02 1.2832E-02 3.8771E-02 5.2305E-02 5.8685E-02 9.8039E-13 1.9802E-03 1.8182E-02

Veq. pt.

25.00

1.02E-02 c -1.0200E-03 -8.1600E-04 -7.2857E-04 -6.8919E-04 -6.8000E-04 -6.4421E-04 -2.9143E-04 -1.0737E-04 -2.0606E-05 -4.9020E-14 -4.8534E-14 -4.4563E-14

-

[OH ]

2.2140E-07 1.9802E-03 1.8182E-02

+

[H3O ] 1.0859E-01 5.2926E-02 3.1682E-02 2.3285E-02 2.1471E-02 1.9764E-02 6.4452E-03 1.9779E-03 3.4905E-04 4.5166E-08 5.0500E-12 5.5000E-13

Spreadsheet Documentation F3 = C1*C2/F2 B6 = ($C$1*$C$2-$F$2*A6)/($C$1+A6)+$C$4 C6 = -$C$4*($C$2*$C$1)/($C$1+A6) E6 = (-B6+SQRT(B6^2-4*C6))/2+($C$1*$C$2-$F$2*A6)/($C$1+A6) F6 = -LOG(E6) B10 = ($F$2*A10-$C$1*$C$2)/($C$1+A10)+$C$4 C10 = -$C$4*(2*$C$2*$C$1-$F$2*A10)/($C$1+A10) E10 = (-B10+SQRT(B10^2-4*C10))/2 B15 = ($F$2*A15-2*$C$1*$C$2)/($C$1+A15)+$F$1/$C$4 C15 = -($F$1/$C$4)*($C$2*$C$1)/($C$1+A15) D15 = (-B15+SQRT(B15^2-4*C15))/2+($F$2*A15-2*$C$1*$C$2)/($C$1+A15) E15 = $F$1/D15

pH 0.96 1.28 1.50 1.63 1.67 1.70 2.19 2.70 3.46 7.35 11.30 12.26

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

For Problems 15-27 and 15-28 we set up spreadsheets that will solve a quadratic equation to determine [H3O+] or [OH-], as needed (See approach taken for Problems 14-41 through 43). While approximate solutions are appropriate for many of the calculations, the approach taken represents a more general solution and is somewhat easier to incorporate in a spreadsheet. As an example consider the titration of a volume of a mixture (ViMix) of a strong (SA) and weak (HA) acid with a strong base (B).

ci HAViMix  - [A-] Vi Mix  VB 

Before the 1st equivalence point:

[HA] =

and

[H3O+] =

ViMixcSA  ciBVB  ViMix  VB 

+ [A-]

Substituting these expressions into the equilibrium expression for HA and rearranging gives

 V c  ciBVB    K a ci HAViMix   K a  [A-] + 0 = [A-]2 +  iMix SA Vi Mix  VB   ViMix  VB   From which [A-] is determined and [H3O+] may be calculated. From the 1st equivalence to before the 2nd equivalence point: [HA] =

and

[A-] =

ci HAVi Mix  (ciBVB  ViMix ciSA ) Vi Mix  VB 

- [H3O+]

ci BVB  ciSAViMix  + [H O+] 3 Vi Mix  VB 

Substituting these expressions into the equilibrium expression for HA- and rearranging gives

 c V  ciSA ViMix    K a ci HAVi Mix  (ciBVB  ViMix ciSA )   K a  [H3O+] + 0 = [H3O+]2 +  i B B Vi Mix  VB   Vi Mix  VB   From which [H3O+] is directly determined.

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

ci HAViMix  - [HA] Vi Mix  VB 

At and after the 2nd equivalence point:

[A-] =

and

[OH-] =

ci BVB  ci HAViMix  ciSAViMix  Vi Mix  VB 

+ [HA]

Substituting these expressions into the equilibrium expression for A- and rearranging gives

 c V  ci HAViMix  ciSA ViMix  K w   K w ci HAViMix   [HA] +  0 = [HA]2 +  i B B Vi Mix  VB  Ka  K a Vi Mix  VB   From which [HA] can be determined and [OH-] and [H3O+] subsequently calculated. A similar approach is taken for the titration of a mixture of a strong and weak base with a strong acid.

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

15-27 A

B

C

D

1

Vi, Mixture

2

ci, NaOH

3

ci, H2NNH2

8.00E-02

4

+ Ka, H2NNH3

1.05E-08

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

Vol. HClO4, mL 0.00 10.00 20.00 24.00 25.00 26.00 35.00 44.00 45.00 46.00 50.00

b 1.0000E-01 5.0001E-02 1.4287E-02 2.7037E-03 9.5238E-07 2.6325E-03 2.3530E-02 4.0426E-02 1.0500E-08 2.0833E-03 1.0000E-02

E

50.00

Kw, H2O

0.1000

c, HClO4

c -7.6190E-08 -6.3492E-08 -5.4422E-08 -5.1480E-08 -5.0794E-08 -4.7619E-08 -2.2409E-08 -2.0263E-09 -4.4211E-10 -4.3750E-10 -4.2000E-10

-

[OH ] 1.0000E-01 5.0001E-02 1.4290E-02 2.7216E-03 2.2490E-04 1.7966E-05 9.5230E-07 5.0124E-08

F 1.00E-14 0.2000

+

[H3O ] 9.9999E-14 1.9999E-13 6.9981E-13 3.6743E-12 4.4464E-11 5.5660E-10 1.0501E-08 1.9950E-07 2.1021E-05 2.0835E-03 1.0000E-02

pH 13.00 12.70 12.15 11.43 10.35 9.26 7.98 6.70 4.68 2.68 2.00

Spreadsheet Documentation B6 = ($C$1*$C$2-$F$2*A6)/($C$1+A6)+$F$1/$C$4 C6 = -($F$1/$C$4)*($C$3*$C$1)/($C$1+A6) D6 = (-B6+SQRT(B6^2-4*C6))/2+($C$1*$C$2-$F$2*A6)/($C$1+A6) E6 = $F$1/D6 F6 = -LOG(E6) B10 = ($F$2*A10-$C$1*$C$2)/($C$1+A10)+($F$1/$C$4) C10 = -($F$1/$C$4)*($C$3*$C$1-($F$2*A10-$C$1*$C$2))/($C$1+A10) D10 = (-B10+SQRT(B10^2-4*C10))/2 C14 = ($F$2*A14-$C$1*$C$2-$C$1*$C$3)/($C$1+A14)+$C$4 D14 = -($C$4)*($C$3*$C$1)/($C$1+A14) E14 = (-B14+SQRT(B14^2-4*C14))/2+($F$2*A14-$C$1*$C$2-$C$1*$C$3)/($C$1+A14)

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

15-28 A

B

C

D

1

Vi, Mixture

2

ci, HClO4

3

ci, HCOOH

8.00E-02

4

Ka, HCOOH

1.80E-04

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

Vol. KOH, mL 0.00 10.00 20.00 24.00 25.00 26.00 35.00 44.00 45.00 46.00 50.00

b 1.0018E-01 5.0180E-02 1.4466E-02 2.8827E-03 1.8000E-04 2.8116E-03 2.3709E-02 4.0606E-02 5.5556E-11 2.0833E-03 1.0000E-02

E

50.00

Kw, H2O

0.1000

c, KOH

c -1.4400E-05 -1.2000E-05 -1.0286E-05 -9.7297E-06 -9.6000E-06 -9.0000E-06 -4.2353E-06 -3.8298E-07 -2.3392E-12 -2.3148E-12 -2.2222E-12

-

[OH ]

1.5294E-06 2.0833E-03 1.0000E-02

F 1.00E-14 0.2000

+

[H3O ] 1.0014E-01 5.0238E-02 1.4965E-02 4.6975E-03 3.0097E-03 1.9073E-03 1.7731E-04 9.4295E-06 6.5385E-09 4.8000E-12 1.0000E-12

pH 1.00 1.30 1.83 2.33 2.52 2.72 3.75 5.03 8.19 11.32 12.00

Spreadsheet Documentation B6 = ($C$1*$C$2-$F$2*A6)/($C$1+A6)+$C$4 C6 = -$C$4*($C$3*$C$1)/($C$1+A6) E6 = (-B6+SQRT(B6^2-4*C6))/2+($C$1*$C$2-$F$2*A6)/($C$1+A6) F6 = -LOG(E6) B10 = ($F$2*A10-$C$1*$C$2)/($C$1+A10)+$C$4 C10 = -$C$4*($C$3*$C$1-($F$2*A10-$C$1*$C$2))/($C$1+A10) E10 = (-B10+SQRT(B10^2-4*C10))/2 B14 = ($F$2*A14-$C$1*$C$2-$C$1*$C$3)/($C$1+A14)+$F$1/$C$4 C14 = -($F$1/$C$4)*($C$3*$C$1)/($C$1+A14) D14 = (-B14+SQRT(B14^2-4*C14))/2+($F$2*A14-$C$1*$C$2-$C$1*$C$3)/($C$1+A14) E14 = $F$1/D14

Fundamentals of Analytical Chemistry: 8th ed. 15-29 (a)

Chapter 15

2H2AsO4-  H3AsO4 + HAsO42[H 3 O  ][H 2 AsO -4 ] K1 = = 5.810-3 [H 3 AsO 4 ]

(1)

K2 =

[H 3 O  ][HAsO 24- ] = 1.110-7 [H 2 AsO 4 ]

(2)

K3 =

[H 3 O  ][AsO 34- ] = 3.210-12 [HAsO 24- ]

(3)

Dividing Equation (2) by Equation (1) leads to K 2 [H 3 AsO 4 ][HAsO 24- ]  = 1.910-5 - 2 K1 [H 2 AsO 4 ]

which is the desired equilibrium constant expression. (b)

2HAsO42-  AsO43- + H2AsO4-

Here we divide Equation (3) by Equation (2) K 3 [AsO 34- ][H 2 AsO -4 ]  = 2.910-5 K2 [HAsO 24- ]2

15-30 HOAc + H2O  H3O+ + OAcNH4+ + H2O  H3O+ + NH3

KHOAc = 1.7510-5

K NH = 5.7010-10 4

Subtracting the first reaction from the second and rearranging give NH4+ + OAc-  NH3 + HOAc [ NH 3 ][HOAc] 5.70  10 10  = 3.2610-5  5 [ NH 4 ][OAc ] 1.75  10

K = K NH / KHOAc 4

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

15-31 A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 2 29

(a)

(b)

(c)

(d)

(e)

(f)

B pH 2.00 6.00 10.00 2.00 6.00 10.00 2.00 6.00 10.00 2.00 6.00 10.00 2.00 6.00 10.00 2.00 6.00 10.00

C +

[H3O ] 1.00E-02 1.00E-06 1.00E-10 1.00E-02 1.00E-06 1.00E-10 1.00E-02 1.00E-06 1.00E-10 1.00E-02 1.00E-06 1.00E-10 1.00E-02 1.00E-06 1.00E-10 1.00E-02 1.00E-06 1.00E-10

D

E

K1 1.12E-03

K2 3.91E-06

7.11E-03

7.45E-04

5.8E-03

3E-02

5.60E-02

F

G

0 0.899 1.82E-04 2.28E-12 6.32E-08 4.5E-13 0.584 1.32E-04 2.21E-11 1.73E-05 4.02E-07 0.931 5.31E-05 1.93E-16 1.1E-07 3.2E-12 0.633 1.55E-04 1.52E-11 1.62E-07 0.250 2.87E-05 2.06E-12 5.42E-05 0.151 3.23E-07 3.29E-15 K3

H

I

J

1 0.101 0.204 2.56E-05 0.416 0.940 1.57E-03 0.069 3.96E-02 1.44E-09 0.367 0.901 8.80E-04 0.750 0.861 6.17E-04 0.845 0.018 1.85E-06

2 3.94E-05 7.96E-01 1.00E+00 2.63E-06 5.94E-02 9.94E-01 1.20E-04 6.85E-01 2.49E-04 4.04E-06 9.91E-02 9.68E-01 1.21E-05 1.39E-01 9.99E-01 4.58E-03 9.82E-01 1.00E+00

3

Spreadsheet Documentation C2 = 10^(-C2) G2 = C2^2/(C2^2+$D$2*C2+$D$2*$E$2) H2 = C2*$D$2/(C2^2+$D$2*C2+$D$2*$E$2) I2 = $D$2*$E$2/(C2^2+$D$2*C2+$D$2*$E$2) G5 = C5^3/(C5^3+$D$5*C5^2+$D$5*$E$5*C5+$D$5*$E$5*$F$5) H5 = C5^2*$D$5/(C5^3+$D$5*C5^2+$D$5*$E$5*C5+$D$5*$E$5*$F$5) I5 = C5*$D$5*$E$5/(C5^3+$D$5*C5^2+$D$5*$E$5*C5+$D$5*$E$5*$F$5) J5 = $D$5*$E$5*$F$5/(C5^3+$D$5*C5^2+$D$5*$E$5*C5+$D$5*$E$5*$F$5)

1.18E-16 2.67E-08 4.47E-03 4.82E-09 2.75E-01 1.00E+00 1.29E-15 3.17E-07 3.10E-02

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 15

15-32 Letting H3A = H3AsO4 we write [H 3 O  ][H 2 A - ] K1 = (1) [H 3 A]

K1K2 =

[H 3 O  ]2 [HA 2- ] [H 3 A]

K2 =

[H 3 O  ][HA 2- ]

(2)

K3 =

[H 2 A - ]

K1K2K3 =

[H 3 O  ][A 3- ] [HA 2- ]

[H 3 O  ]3 [A 3- ] [H 3 A]

(3)

By definition 0 =

[H 3 A] cT

1 =

[H 2 A - ] cT

2 =

[HA 2- ] cT

3 =

[ A 3- ] cT

where cT = [H3A] + [H2A-] + [HA2-] + [A3-]

(4)

Substituting Equations (1), (2) and (3) into (4) yields cT = [H3A] +

K1[H 3 A] K1 K 2 [H 3 A] K1 K 2 K 3 [H 3 A]   [H 3 O  ] [H 3 O  ] 2 [ H 3 O  ]3

KK K cT K1 K1 K 2 = 1+   1 2  33   2 [H 3 A] [H 3 O ] [H 3 O ] [H 3 O ] Multiplying the numerator and denominator of the right side of this equation by [H3O+]3 gives [ H 3 O  ]3  K 1 [ H 3 O  ] 2  K 1 K 2 [ H 3 O  ]  K 1 K 2 K 3 cT = [ H 3 O  ]3 [H 3 A]

Letting D = [H3O+]3 + K1[H3O+]2 + K1K2[H3O+] + K1K2K3 gives

cT D = [ H 3 O  ]3 [H 3 A] and inverting the two terms leads to 0

(5)

Fundamentals of Analytical Chemistry: 8th ed.

[H 3 A] [ H 3 O  ]3 = = 0 cT D

Chapter 15

(6)

Substituting Equation (1) into (6) gives [H 3 O  ][H 2 A - ] [ H 3 O  ]3 = K 1c T D [H 2 A - ] K1 [H 3 O  ] 2 = = 1 cT D

In the same way substituting Equation (2) into (6) gives upon rearrangement

K K [H O  ] [HA 2- ] = 1 2 3 = 2 cT D Similarly, substituting Equation (3) into (6) yields KK K [ A 3- ] = 1 2 3 = 3 cT D