Fundamentals of Analytical Chemistry by Skoog et. al. 8th ed Chapter 17 answers

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

Chapter 17 17-1

(a) A chelate is a cyclic complex consisting of metal ion and a reagent that contains two or more electron donor groups located in such a position that they can bond with the metal ion to form a Heterocyclic ring structure. (b) A tetradentate chelating agent is a molecule that contains four pairs of donor electron located in such positions that they all can bond to a metal ion, thus forming two rings. (c) A ligand is a species that contains one or more electron pair donor groups that tend to form bonds with metal ions. (d) The coordination number is the number of covalent bonds that a cation tends to form with electron donor groups. (e) A conditional formation constant is an equilibrium constant for the reaction between a metal ion and a complexing agent that applies only when the pH and/or the concentration of other complexing ions are carefully specified. (f) NTA is the acronym for nitrilotriacetic acid, a tetradentate complexing agent that contains three carboxylate groups and one tertiary amine. As an electron donor, NTA has found applications in the titration of a variety of cations. (g) Water hardness is the concentration of calcium carbonate that is equivalent to the total concentration of all of the multivalent metal carbonates in the water. (h) In an EDTA displacement titration, an unmeasured excess of a solution containing the magnesium or zinc complex of EDTA is introduced into the solution of an analyte that forms a more stable complex that that of magnesium or zinc. The liberated magnesium or zinc ions are then titrated with a standard solution of EDTA. Displacement titrations are used for the determination of cations for which no good indicator exists.

Fundamentals of Analytical Chemistry: 8th ed. 17-2

Chapter 17

Three general methods for performing EDTA titrations are (1) direct titration, (2) back titration, and (3) displacement titration. Method (1) is simple, rapid, but requires one standard reagent. Method (2) is advantageous for those metals that react so slowly with EDTA as to make direct titration inconvenient. In addition, this procedure is useful for cations for which satisfactory indicators are not available. Finally, it is useful for analyzing samples that contain anions that form sparingly soluble precipitates with the analyte under analytical conditions. Method (3) is particularly useful in situations where no satisfactory indicators are available for direct titration.

17-3

Multidentate ligands offer the advantage that they usually form more stable complexes than do unidentate ligands. Furthermore, they often form but a single complex with the cation, which simplifies their titration curves and makes end-point detection easier.

17-4

(a) [ Ni(CN ) ] K1  2  [ Ni ][CN ] [ Ni(CN ) 2 ] Ni(CN )  CN   Ni(CN ) 2 K 2   [ Ni(CN ) ][CN ] Ni 2 CN   Ni(CN )  

Ni(CN ) 2 CN   Ni(CN ) 3  



Ni(CN ) 3 CN   Ni(CN ) 4 

2



[ Ni(CN ) 3 ] K3  [ Ni(CN ) 2 ][CN ] 2

[ Ni(CN ) 4 ] K4   [ Ni(CN ) 3 ][CN ]

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

(b)

[Cd(SCN ) ]  Cd 2 SCN   Cd ( SCN ) K  1  [Cd 2][SCN ] [Cd(SCN ) 2 ] Cd(SCN )  SCN   Cd ( SCN ) K  2 2  [Cd(SCN ) ][SCN ] Cd(SCN ) 2 SCN   Cd(SCN ) 3 

17-5

17-6





[Cd(SCN ) 3 ] K3  [Cd(SCN ) 2 ][SCN ]

(a)

hexaaminezinc(II), Zn(NH3)62+

(b)

dichloroargentate, Ag(Cl)2-

(c)

disulfatocuprate(II), Cu(SO4)22-

(d)

trioxalotoferrate(III), Fe(C2O4)33-

(e)

hexacyanoferrate(II), Fe(CN)64-

The overall formation constant n is equal to the product of the individual stepwise constants. Thus, the overall constant for formation of Cd(SCN)3- in Question 17-4(b) is 

[Cd(SCN ) 3 ] 3 K1 K 2 K 3  2 [Cd ][SCN ]3 which is the equilibrium constant for the reaction Cd 2 3SCN   Cd(SCN ) 3 



and

[Cd(SCN ) 2 ] 2 K1 K 2  2 [Cd ][SCN ]2 where the overall constant 2 is for the equation Cd 2 2SCN   Cd(SCN ) 2 

Fundamentals of Analytical Chemistry: 8th ed. 17-7

Chapter 17

(a) acetate (1)

CH 3COOH  CH 3COO  H   [CH 3COO ][ H ] cT [CH 3COOH ] [CH 3COO ] K a  [CH 3COOH ] [CH 3COO ][ H ] [CH 3COOH ]  Ka [ H ]  [CH 3COO ][ H ] [ H ] K a    cT  [CH 3COO ] [CH 3COO ]  1  [ CH COO ] 3 K   K Ka a  a   [CH 3COO ] [CH 3COO ] 1   cT [ H ] K a [CH 3COO ]  K a 

   

K   a  [ H ] K a   

(b) tartrate (2)

C 2 H 4 O 2 (COOH ) 2

 

[C H O (COOH )(COO )][ H ] C 2 H 4 O 2 (COOH )(COO ) H  K a1  2 4 2 [C 2 H 4 O 2 (COOH ) 2 ]  



C 2 H 4 O 2 (COOH )(COO )  C 2 H 4 O 2 (COO ) 2 H



Ka2

[C 2 H 4 O 2 (COO ) 2 ][ H ]  [C 2 H 4 O 2 (COOH )(COO )]

[C H O (COO ) 2 ][ H ] [C 2 H 4 O 2 (COOH )(COO )  2 4 2 Ka2 [C H O (COOH )(COO )][ H ] [C 2 H 4 O 2 (COO ) 2 ][ H ]2 [C 2 H 4 O 2 (COOH ) 2  2 4 2  K a1 K a1 K a 2 cT [C 2 H 4 O 2 (COOH ) 2 ] [C 2 H 4 O 2 (COOH )(COO )] [C 2 H 4 O 2 (COO ) 2 ] [C H O (COO ) 2 ][ H ]2 [C 2 H 4 O 2 (COO ) 2 ][ H ]  2 4 2  [C 2 H 4 O 2 (COO ) 2 ] K a1 K a 2 Ka2 [ H ]2 [ H ]  [ H ]2 K a1 [ H ] K a1 K a 2     [C 2 H 4 O 2 (COO ) 2 ]   1  [ C H O ( COO ) ] 2  K K  2 4 2   Ka2 K a1 K a 2  a1 a 2     [C H O (COO ) 2 ] 2  2 4 2 cT K a1 K a 2 [C 2 H 4 O 2 (COO ) 2 ]   2 2  [ H ] K a1 [ H ] K a1 K a 2  [ H ] K a1 [ H ] K a1 K a 2  [C 2 H 4 O 2 (COO ) 2 ]   K K a 1 a 2  

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

(c) phosphate (3) H 3 PO 4 H 2 PO 4 HPO 4

 



H 2 PO 4 H

 

HPO 4

2 

3

PO 4 

2



[ H 2 PO 4 ][ H ] K a1  [ H 3 PO 4 ]



H



2

Ka 2

[ HPO 4 ][ H ]   [ H 2 PO 4 ] 3

[ PO 4 ][ H ] H  K a 3  2 [ HPO 4 ] 3

[ PO 4 ][ H ] 2 [ HPO 4 ]  Ka3 2

3

[ HPO 4 ][ H ] [ PO 4 ][ H ]2  [ H 2 PO 4 ]   Ka2 Ka2 Ka3 

3

[ H 2 PO 4 ][ H ] [ PO 4 ][ H ]3 [ H 3 PO 4 ]   K a1 K a1 K a 2 K a 3 

2

3

cT [ H 3 PO 4 ] [ H 2 PO 4 ] [ HPO 4 ] [ PO 4 ] 3

3

3

[ PO 4 ][ H ]3 [ PO 4 ][ H ]2 [ PO 4 ][ H ] 3    [ PO 4 ] K a1 K a 2 K a 3 Ka 2 Ka3 Ka3 3 [ H ]2 [ H ]  3  [ H ] [ PO 4 ]   1 K K K  Ka3  a1 a 2 a 3 K a 2 K a 3 

[ H ]3 K a1 [ H ]2 K a1 K a 2 [ H ] K a1 K a 2 K a 2  3   [ PO 4 ]   K a1 K a 2 K a 3   3

3

[ PO 4 ] [ PO 4 ] 3   3 2 cT [ H ] K a1 [ H ] K a1 K a 2 [ H ] K a1 K a 2 K a 2  3   [ PO 4 ]   K a1 K a 2 K a 3   K a1 K a 2 K a 3  3 2 [ H ] K a1 [ H ] K a1 K a 2 [ H ] K a1 K a 2 K a 3 17-8

(a) Fe K

'

3

CH 3COO

Fe ( CH 3COO )

2

 

Fe (CH 3COO)

2

[ Fe(CH 3COO) 2] [ Fe(CH 3COO) 2] K Fe ( CH COO )2   3  3 [ Fe ][CH 3COO ] [ Fe 3]1cT

[ Fe(CH 3COO) 2] 1 K Fe ( CH COO )2  3 [ Fe 3]cT

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

(b) Fe 3 C 2 H 4 O 2 (COO ) 2 K Fe ( C H O 2

K

'

4

2 ( COO



)2 )





Fe ( C 2 H 4 O 2 ( COO ) 2 )



 

Fe (C 2 H 4 O 2 (COO ) 2 ) 

[ Fe(C H O (COO ) 2 ) ] [ Fe(C 2 H 4 O 2 (COO ) 2 ) ]  3 2 4 2  [ Fe ][C 2 H 4 O 2 (COO ) 2 ] [ Fe 3]2 cT [ Fe(C 2 H 4 O 2 (COO ) 2 ) ] 2 K Fe ( C H O ( COO) )   2 4 2 2 [ Fe 3]cT

(c)

Fe 3 PO 4

3  

FePO 4

[ FePO 4 ] [ FePO ] K FePO 4  3  3 4 3 [ Fe ][ PO 4 ] [ Fe ]3cT

[ FePO ] K ' FePO4 3 K FePO 4  3 4 [ Fe ]cT 17-9 Fe 3 3Ox 2  Fe(Ox ) 3 

3

3

3

3

[ Fe(Ox ) 3 ] [ Fe(Ox ) 3 ] [ Fe(Ox ) ] ' 3  3  3  3 3 3 ( 2 ) 3 3 2 3 3 [ Fe ][Ox ] [ Fe ]( 2 cT ) [ Fe ]( cT )

[Ox 2] 2  [Ox 2] 2 cT cT

17-10 Titrate the three ions in an aliquot of the sample that has been buffered to a pH of about 10. Buffer a second aliquot to a pH of about 4 and titrate the zinc and indium ions. Finally, titrate an aliquot that has been brought to a pH of about 1.5. Only the indium is complexed under these conditions. 17-11 [ ML n ] n  [ M ][ L]n Take the logarithm of both sides of the equation log n log[ML n ] log[M ] n log[L]

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

Now, write the right side of the equation as a pfunction (i.e., p[M] = -log[M])

log n pM npL pML n 17-12 The MgY2- is added to assure a sufficient analytical concentration of Mg2+ to provide a sharp end point with Eriochrome Black T indicator. 17-13

99.7 g Na 2 H 2 Y  2H 2O 1 mole EDTA 3.156 g reagent   100 g reagent 372.24 g Na 2 H 2 Y  2H 2 O 0.00845 M EDTA 1.000 L

17-14 0.004517 mmol Mg 2 1 mmol EDTA 50.00 mL   mL mmol Mg 2 0.007010 M EDTA 32.22 mL EDTA

17-15 (a)

0.0741 mmol Mg( NO 3 ) 2 1 mmol EDTA 1 mL EDTA 27.16 mL Mg( NO 3 ) 2   mL mmol Mg( NO 3 ) 2 0.0500 mmol 40.25 mL EDTA (b) 1000 mmol CaCO 3 1 mmol EDTA 1 mL EDTA 0.1973 g CaCO 3    39.42 mL EDTA 100.09 g mmol CaCO 3 0.0500 mmol

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

(c)

81.4 g CaHPO 4  2 H 2 O 1000 mmol CaHPO 4  2H 2O 0.5140 g sample   100 g sample 172.09 g 2.43123 mmol CaHPO 4  2H 2 O 1 mmol EDTA 1 mL EDTA 2.43123 mmol CaHPO 4  2H 2 O   48.63 mL EDTA mmol CaHPO 4  2 H 2 O 0.0500 mmol

(d) 1000 mmol 3MgCO 3 Mg(OH ) 2  3H 2 O 0.2222 g 3MgCO 3 Mg(OH ) 2  3H 2 O   365.3 g 4 mmol EDTA 1 mL EDTA  48.66 mL EDTA mmol 3MgCO 3 Mg(OH ) 2  3H 2 O 0.0500 mmol (e) 92.5 g CaCO 3  MgCO 3 1000 mmol CaCO 3  MgCO 3 0.1414 g sample    100 g sample 184.4 g 2 mmol EDTA 1 mL EDTA  28.37 mL EDTA mmol CaCO 3  MgCO 3 0.0500 mmol 17-16 First calculate the CoSO4 concentration, 1.694 mg CoSO 4 1 mmol CoSO 4  0.01093 M CoSO 4 mL 155.0 mg

In each part, 0.01093 mmol mmol CoSO4  25.00 mL 0.2732 mmol CoSO4 mL

(a) 1 mmol EDTA 1 mL EDTA 0.2732 mmol CoSO 4   31.62 mL EDTA mmol CoSO 4 0.008640 mmol

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

(b) 0.008640 mmol EDTA  mmol excess EDTA  50.00 mL  mL    1 mmol EDTA    0.2732 mmol CoSO4   0.1588 mmol EDTA mmol CoSO 4   1 mmol Zn 2 1 mL Zn 2 0.1588 mmol EDTA   16.80 mL Zn 2 mmol EDTA 0.009450 mmol (c) 1 mmol Zn 2 1 mmol EDTA 1 mL EDTA 0.2732 mmol CoSO 4    31.62 mL EDTA 2 mmol CoSO 4 mmol Zn 0.008640 mmol

17-17

0.01645 mmol EDTA 1 mmol Zn 2 65.39 g Zn 2     21 . 27 mL EDTA    mL mmol EDTA 1000 mmol   100% 0.7162 g sample 3.195% Zn 2

17-18

0.01768 mmol EDTA  mmol EDTA reacted  15.00 mL EDTA  mL   0.008120 mmol Cu 2 1 mmol EDTA   0.2303 mmol EDTA 4.30 mL Cu 2   mL mmol Cu 2     1 mmol Cr 51.996 mg Cr    0.2303 mmol EDTA     mmol EDTA mmol  0.998 mg Cr 2  3.00 4.00 cm cm 2

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

17-19

0.03560 mmol EDTA 1 mmol Tl 2SO 4 504.8 g Tl 2SO 4    13.34 mL EDTA    mL 2 mmol EDTA 1000 mmol   100% 9.76 g sample 1.228% Tl 2SO 4

17-20 (a) 0.7682 g MgCO 3 1000 mmol MgCO 3 1 mmol EDTA     50.0 mL MgCO 3   1000 mL  84 . 314 g mmol MgCO 3   42.35 mL EDTA 0.01076 M EDTA (b)   1.076 10 2 mmol  18.81 mL    mL  mmol CaCO 3 mmol MgCO 3   8.094 10 3 M  mL sample 25.00 mL

 1 mmol CaCO 3  1.076 10 2 mmol EDTA   31.54 mL EDTA   mL mmol EDTA  mmol CaCO 3    mL sample 50.00 mL 6.786 10 3 M mmol MgCO 3 8.094 10 3 M 6.786 10 3 M 1.308 10 3 M mL sample 6.786 10 3 mmol CaCO 3 100.09 g CaCO 3 1.000 mL sample   10 6 ppm mL sample 1000 mmol g sample 679.2 ppm CaCO 3 (c)

1.308 10 3 mmol MgCO 3 84.314 g MgCO 3 1.000 mL sample   10 6 ppm mL sample 1000 mmol g sample 110.3 ppm MgCO 3

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

17-21

0.01200 mmol EDTA 1 mmol Fe 3 mmol Fe 3  13.73 mL EDTA  0.16476 mmol Fe 3 mL mmol EDTA 0.01200 mmol EDTA 1 mmol Fe 2 mmol Fe 2   29.62 13.73mL EDTA  mL mmol EDTA 0.19068 mmol Fe 2 3   3 55.847 mg Fe   0.16476 mmol Fe    mmol  184.0 ppm Fe 3 L 50.00 mL  1000 mL 2   2  55.847 mg Fe   0 . 19068 mmol Fe    mmol  213.0 ppm Fe 2 L 50.00 mL  1000 mL

17-22

mmol Mg 2 mmol Ca 2  0.003960 mmol EDTA mmol Mg 2 mmol Ca 2 27.32 mL EDTA  0.108187 mmol mL mmol EDTA 0.003960 mmol EDTA 1 mmol Ca 2 mmol Ca 2  12.21 mL EDTA  0.048352 mmol Ca 2 mL mmol EDTA mmol Mg 2 0.108187 0.048352 0.059835 mmol Mg 2 2   2  40.08 mg Ca   0 . 048352 mmol Ca    mmol  387.6 ppm Ca 2 L 1 10.00 mL   1000 mL 2.000 L 2   2  24.305 mg Mg   0 . 059835 mmol Mg    mmol  290.9 ppm Mg 2 L 1 10.00 mL   1000 mL 2.000 L

Both values are within normal limits.

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

17-23 mmol Cd 2 mmol Pb 2  0.06950 mmol EDTA mmol Cd 2 mmol Pb 2 28.89 mL EDTA  2.00786 mmol mL mmol EDTA mmol Pb

2

0.06950 mmol EDTA 1 mmol Pb 2  11.56 mL EDTA  0.80342 mmol Pb 2 mL mmol EDTA

mmol Cd 2 2.00786 mmol 0.80342 mmol 1.20444 mmol Cd 2 2   2  207.2 g Pb   0 . 80342 mmol Pb   1000 mmol   100% 55.16% Pb 2 50.00 mL 1.509 g sample  250.0 mL 2   2  112.41 g Cd   1 . 204 mmol Cd   1000 mmol   100% 44.86% Cd 2 50.00 mL 1.509 g sample  250.0 mL

17-24 mmol Ni 2 mmol Cu 2  0.05285 mmol EDTA mmol Ni 2 mmol Cu 2 45.81 mL EDTA  2.42106 mmol mL mmol EDTA 0.07238 mmol Mg 2 1 mmol Cu 2 mmol Cu 2  22.85 mL Mg 2  1.65388 mmol Cu 2 mL mmol g 2 mmol Ni 2 2.42106 mmol 1.65388 mmol 0.76718 mmol Ni 2

2   2  58.693 g Ni   0 . 76718 mmol Ni   1000 mmol   100% 30.00% Ni 2 25.00 mL 0.6004 g sample  100.0 mL 2   2  63.546 g Cu   1 . 65388 mmol Cu   1000 mmol   100% 70.02% Cu 2 25.00 mL 0.6004 g sample  100.0 mL

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

17-25

0.01294 mmol EDTA 1 mmol ZnO 81.39 g ZnO    38.71 mL EDTA    mL mmol EDTA 1000 mmol   100% 10.00 mL 1.022 g sample  250.0 mL 99.7% ZnO 0.002727 mmol ZnY 2 1 mmol Fe 2 O 3 159.69 g Fe 2 O 3    2.40 mL ZnY 2    mL 2 mmol ZnY 2 1000 mmol   100% 50.00 mL 1.022 g sample  250.0 mL 0.256% Fe 2 O 3

17-26 1 mmol EDTA 1 mmol Ni2+ 2 mmol NaBr 2 mmol NaBrO3 For the 10.00 mL aliquot,

mmol NBr mmol NaBrO 3  mL sample solution 0.02089 mmol EDTA  2 mmol NaBr mmol NaBrO 3    21.94 mL EDTA    mL mmol EDTA  0.09166 M 10.00 mL

For the 25.00 mL aliquot,

mmol NaBr  mL sample solution 0.02089 mmol EDTA 2 mmol NaBr    26.73 mL EDTA   mL mmol EDTA   0.04467 M NaBr 25.00 mL mmol NaBrO 3 0.09166 0.04467 0.04699 M NaBrO 3 mL sample solution

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

0.04467 mmol NaBr 102.9 g NaBr    250.0 mL   mL 1000 mmol   100% 31.48% NaBr 3.650 g sample 0.04699 mmol NaBrO 3 150.9 g NaBrO 3    250.0 mL   mL 1000 mmol   100% 48.57% NaBrO 3 3.650 g sample 17-27 1 mmol Mg2+ 1 mmol EDTA ¼ mmol B(C6H5)4- ¼ K+ 0.05581 mmol Mg 2 1 mmol K  39.098 mg K   2    29 . 64 mL Mg     mL 4 mmol Mg 2 mmol  64.68 ppm K  L 250 mL  1000 mL

17-28 0.05182 mmol EDTA  mmol EDTA reacted in 50.00 mL  50.00 mL EDTA  mL   0.06241 mmol Cu 2 1 mmol EDTA   2.27208 mmol 5.11 mL Cu 2   mL mmol Cu 2    2.27208 mmol mmol EDTA reacted in 350.0 mL mmol Ni Fe Cr  11.3604 mmol 50.00 mL 250.0 mL

0.05182 mmol EDTA  36.28 mL EDTA   mL 9.4002 mmol mmol Ni mmol Fe  50.00 mL 250.0 mL mmol Cr 11.3604 mmol 9.4002 mmol 1.9603 mmol Ni 0.05182 mmol EDTA 1 mmol Ni    25.91 mL EDTA   mL mmol EDTA   6.7133 mmol Ni mmol Ni  50.00 mL 250.0 mL

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

mmol Fe 9.4002 mmol 6.7133 mmol 2.6869 mmol Fe 51.996 g Cr 1.9603 mmol Cr  1000 mmol %Cr  100% 15.75% Cr 0.6472 g 58.69 g Ni 6.7133 mmol Ni  1000 mmol % Ni  100% 60.88% Ni 0.6472 g 55.847 g Fe 2.6869 mmol Fe  1000 mmol % Fe  100% 23.19% Fe 0.6472 g

17-29

mmol EDTA mmol Pb Zn Cu  0.002500 mmol EDTA  37.56 mL EDTA   mL  4.6950 mmol 10.00 mL 500.0 mL 0.002500 mmol EDTA  27.67 mL EDTA   mL 1.3835 mmol mmol Pb mmol Zn  25.00 mL 500.0 mL mmol Cu 4.6950 mmol 1.3835 mmol 3.3115 mmol Cu 0.002500 EDTA 1 mmol Pb    10.80 mL EDTA   mL mmol EDTA   0.1350 mmol Pb mmol Pb  100.0 mL 500.0 mL mmol Zn 1.3835 mmol 0.1350 mmol 1.2485 mmol Zn 63.55 g Cu 3.3115 mmol Cu  1000 mmol %Cu  100% 64.08% Cu 0.3284 g

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

207.2 g Pb 0.1350 mmol Pb  1000 mmol % Pb  100% 8.518% Pb 0.3284 g 65.39 g Zn 1.2485 mmol Zn  1000 mmol % Zn  100% 24.86% Zn 0.3284 g %Sn 100% (64.07 8.518 24.87)% 2.54% Sn

17-30

A

B

C

D

E

F

G

H

2+

1

17-30 Conditional constants for Fe - EDTA complex

2

Note: The conditional constant K'MY is the product of 4 and KMY (Equation 17-25).

3

The value of KMY is found in Table 17-3.

4

KMY

2.10E+14

5

K1

1.02E-02

6.0

3.69E-17

2.25E-05

4.7E+09 Note that Excel does not

6

K2

2.14E-03

8.0

1.54E-19

5.39E-03

1.1E+12 follow the rounding rules

7

K3

6.92E-07

10.0

2.34E-21

3.55E-01

7.5E+13 developed in Section 6D-3.

8

K4

5.50E-11

9

Spreadsheet Documentation

pH

D

4

K'MY

10

D5=(10^-C5)^4+$B$5*(10^-C5)^3+$B$5*$B$6*(10^-C5)^2+$B$5*$B$6*$B$7*(10^-C5)+$B$5*$B$6*$B$7*$B$8

11

E5=$B$5*$B$6*$B$7*$B$8/D5

12

F5=E5*$B$4

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

17-31

A

B

C

D

E

F

G

H

2+

1

17-31 Conditional constants for Ba - EDTA complex

2

Note: The conditional constant K'MY is the product of 4 and KMY (Equation 17-25).

3

The value of KMY is found in Table 17-3

4

KMY

5.80E+07

5

K1

1.02E-02

6

K2

7

K3

8

K4

5.50E-11

9

Spreadsheet Documentation

pH

D

4

K'MY

7.0

1.73E-18

4.80E-04

2.8E+04

2.14E-03

9.0

1.60E-20

5.21E-02

3.0E+06

6.92E-07

11.0

9.82E-22

8.46E-01

4.9E+07

10

D5=(10^-C5)^4+$B$5*(10^-C5)^3+$B$5*$B$6*(10^-C5)^2+$B$5*$B$6*$B$7*(10^-C5)+$B$5*$B$6*$B$7*$B$8

11

E5=$B$5*$B$6*$B$7*$B$8/D5

12

F5==E5*$B$4

I

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

17-32

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 28 29 30 31 32 33 34 35 36 37 38 39

A B C D E F 2+ 17-32 Titration of 50.00 mL of 0.01000 M Sr with 0.02000 M EDTA Note: The conditional constant K'MY is the product of 4 and KMY (Equation 17-25). The value of KMY is found in Table 17-3. pH D K'MY 4 KMY 4.30E+08 EDTA K1 1.02E-02 11.0 9.82E-22 8.46E-01 3.64E+08 K2 2.14E-03 K3 6.92E-07 K4 5.50E-11 2+ Initial conc. Sr 0.01000 Initial Vol. 50.00 Initial conc. EDTA 0.02000 Vol. EDTA, mL cSr2+ cSrY2cT [Sr2+] [SrY2-] 0.00 0.01000 0 0.01000 10.00 0.00500 0.00333 0.00500 24.00 0.00027 0.00649 0.00027 24.90 0.00003 0.00665 0.00003 25.00 0.00000 0.00667 0.00667 4.28E-06 0.00667 25.10 0.00666 2.66E-05 6.87E-07 0.00666 26.00 0.00658 2.63E-04 6.87E-08 0.00658 30.00 0.00625 1.25E-03 1.37E-08 0.00625 Spreadsheet Documentation B12=($B$9*$D$9-$B$10*A12)/($D$9+A12) C12=($B$10*A12)/($D$9+A12) C16=($B$10*$A$16)/($D$9+A16) D17=($B$10*A17-$D$9*$B$9)/($D$9+A17) D16=($B$10*$A$16)/($D$9+A16) E12=B12 E16=SQRT(C16/$F$5) E17=C17/(D17*$F$5) F16=C16 H12=-LOG10(E12)

G

H

pSr 2.00 2.30 3.57 4.57 5.37 6.16 7.16 7.86

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

17-33

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 28 29 30 31 32 33 34 35 36 37 38 39

A B C D E F G H 2+ 17-33 Titration of 50.00 mL of 0.0150 M Fe with 0.0300 M EDTA Note: The conditional constant K'MY is the product of 4 and KMY (Equation 17-25). The value of KMY is found in Table 17-3. pH D K'MY 4 KMY 2.10E+14 EDTA K1 1.02E-02 7.0 1.73E-18 4.80E-04 1.01E+11 K2 2.14E-03 K3 6.92E-07 K4 5.50E-11 2+ Initial conc. Fe 0.0150 Initial Vol. 50.00 Initial conc. EDTA 0.0300 2+ 2Vol. EDTA, mL cFe2+ cFeY2cT [Fe ] [FeY ] pFe 0.00 0.01500 0 0.01500 1.82 10.00 0.00750 0.00500 0.00750 2.12 24.00 0.00041 0.00973 0.00041 3.39 24.90 0.00004 0.00997 0.00004 4.40 25.00 0.00000 0.01000 0.01000 3.15E-07 0.01000 6.50 25.10 0.00999 3.99E-05 2.48E-09 0.00999 8.61 26.00 0.00987 3.95E-04 2.48E-10 0.00987 9.61 30.00 0.009375 1.88E-03 4.96E-11 0.00938 10.30 Spreadsheet Documentation B12=($B$9*$D$9-$B$10*A12)/($D$9+A12) C12=($B$10*A12)/($D$9+A12) Note: The method is identical to Problem C16=($B$10*$A$16)/($D$9+A16) 17-32. D17=($B$10*A17-$D$9*$B$9)/($D$9+A17) D16=($B$10*$A$16)/($D$9+A16) E12=B12 E16=SQRT(C16/$F$5) E17=C17/(D17*$F$5) F16=C16 H12=-LOG10(E12)

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 17

17-34

0.01205 mmol EDTA  mmol Ca 2 mmol Mg 2  23.65 mL EDTA 0.2850 mmol mL   0.01205 mmol EDTA 1 mmol Ca 2   mmol Ca 2   14 . 53 mL EDTA  0.1751 mmol Ca 2   mL mmol EDTA   mmol Mg 2 0.2850 0.1751 0.1099 mmol Mg 2

(a) See discussion of water hardness in 17D-9. Water hardness ppm CaCO 3 ppm Ca 2 Mg 2  100.087 mg CaCO 3 0.2850 mmol  mmol 570.5 ppm CaCO 3 L 50.00 mL  1000 mL

(b)

 1 mmol CaCO 3 100.08 mg CaCO 3    0.1751 mmol Ca 2     mmol Ca 2 mmol  350.5 ppm CaCO 3 L 50.00 mL  1000 mL (c)  1 mmol MgCO 3 84.30 mg MgCO 3    0.1099 mmol Mg 2     mmol Mg 2 mmol  185.3 ppm MgCO 3 L 50.00 mL  1000 mL