Pen-Problem Solutions for Analytical Chemistry and Quantitative Analysis 1st Edition. David Hage Jam

Pen-Problem Solu�ons for Analy�cal Chemistry and Quan�ta�ve Analysis 1e David Hage James Carr (Selected Chapters, 100% Original Verified, A+ Grade) Chapters Included in SM: 3, 4, 5, 6, 9, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21 and 23.

Analytical Chemistry Hage/Carr

Solution Pen-Problem Chapter three (1) Change 1.04 g/mL to grams per Liter: 1.04 g/mL x 1000 mL/1L = 1004 grams/L. Then determine the grams of CH2O in the liter of solution: 1004 x 0.0552 = 55.42 g Next, change the gram/L to moles/L: 55.42 x 1 mol/30.03g = 1.846 M = 1.85 M (2) Determine the grams of water in the liter of the solution: 1004 – 55.42 = 948.6 g.

Then convert grams of water to kg: 948.6 g x 1Kg/1000g = 0.9486 Kg. Divide moles in a liter by the kg of water in the liter: 1.85 moles/0.9486 Kg = 1.950 m

2

Pen Problem Chapter 4

SOLUTIONS TO PEN PROBLEM CHAPTER 4: (A) The value for five data points found for 99% confidence level in Table 4.9 is 0.959. The experimental value for r (0.99) exceeds the table value therefore there is a 99% confidence that the calculated line of best-fit does represent the data.

(B) Y1 = 0.0067 (410) – 2.4 = 0.347 Y2 = 0.0067 (420) – 2.4 = 0.414 Y3 = 0.0067 (440) – 2.4 = 0.548 Y4 = 0.0067 (450) – 2.4 = 0.615 Y5 = 0.0067 (500) – 2.4 = 0.950 (C) Residual values (Measured absorbance – calculated absorbance) 0.40 - .347 = +0.05 0.45 - .414 = +0.04 0.55 - .550 = 0 0.60 – 0.615 = - 0.02 1.0 – 0.95 = + 0.05 (D) The graph of this data (residual versus mg quercetin) reveals a non-random pattern. This suggested pattern indicates a problem, for at least this range of the data, in assuming that the linear line of best-fit should be used to represent this data. To be considered supportive of the linear fit model, the residual graph should show a random pattern.

Pen Problem Ch 5

SOLUTIONS FOR PEN PROBLEM Chapter 5 (A) sy = [(y – mx + b)2/n-2]1/2 (0.10 – 0.020(5.0) +0.0092)2 = 0.000085 (0.20 – 0.02(10.0) + 0.0092)2 = 0.000085 (0.39 – 0.02(20.0) + 0.0092)2 = 0.000369 (0.75 – 0.02(40.0) + 0.0092)2 = 0.00351 (1.60 – 0.020(80.0) + 0.0092)2 = 0.000085 0.00413; 0.00413/(5-2) = 0.001378 (0.00138)1/2 = 0.0371 (B) sb = ((xi2)/[nxi2 – (xi)2])1/2sy (See Table 4.8 for review of this equation) = ((8525)/[5(8525) – (155)2])1/2 (0.037) = 0.0163 (C) Determine limit of detection (LOD) = 3.3 (sb/m) = 3.3 (0.0163/0.020) = 2.7 ppb. Since the LOD for this technique is above 1 ppb this technique would not be suitable for determinations below 1 ppb.

(D) LOQ = 10 (sb/m) = 10 (0.037/0.020) = 8.2 ppb

Pen Problem Ch 6

SOLUTIONS FOR PEN PROBLEM Chapter 6 (A) log (Mg2+) =

 A  z2  I 1 a  b  I

=

(Mg2+) = 10(-0.436) = 0.365

(B) log (Cl-) =

 A  z2  I 1 a  b  I

(Cl-) = 10(-0.169) = 0.678

=

 0.51 (2)2  0.24  0.999 = = - 0.436  3 2 .

29 1  800  3.28x10  0.24

 0.51 (-1)2  0.24 = - 0.169  3 1  300  3.28x10  0.24

(C) (±MgCl2)3 = (Mg2+)2 (Cl-) = (0.366)2(0.677) = 0.0907 (±MgCl2) = (0.0907)1/3 = 0.449 = 0.45 (D) Mean activity coefficients for ionic solutes are important because every positive ion added to a solution will have a counter anion. Accurately measuring just one ion in such conditions is not possible because it will be influenced by the presence of the other ions. (The ion in question does not act independently.)

Pen Problem Ch 9

SOLUTIONS FOR PEN PROBLEM Chapter 9

K a1K a2 K a3 (A) αL3- = [H + ]3 + K [H + ]2 + K K [H + ] + K K K a1

a1

a2

a1

a2

a3

0.99

(1.0 x10-3 ) (2.0x10−6 )(1.0x10−11 ) = [H + ]3 + (1.0x10− 3 [H + ]2 ) + (1.0 x10 − 3 x 2.0 x10 − 6 [ H + ]) + ((1.0 x10- 3 ) ( 2.0 x10 − 6 ) (1.0 x10 −11 ))

(Using successive approximation, or calculator solver function, solve for [H+]) [H+] = 1.0 x10-13 M; pH = - log (1.0 x 10-13) = 13.00 (B) In the Kf expression substitute [L3-] for (αL3- CL3-); Kf =

[FeL] 3+

3−

[Fe ][L ]

Then use Kf’ = Kf (αL3-) =

=

[FeL] 3+

[Fe ]CL

[FeL] [Fe ][α L3− CL3− ]

3−

3+

= 0.99 x 5.0 x107 = 4.95 x 107

(C) At pH = 5.00 the fraction of species of the L3- form of the ligand would be smaller. This would make the conditional formation constant smaller so less metal would be complexed.

Pen Problem Ch 10

SOLUTIONS FOR PEN PROBLEM Chapter 10

(a )(a  ) 0.05916 (A) E = E log ( H 2 OH2 (a H 2 O ) 2 o

E = - 0.83 V –

2

)

(a )(a  ) 2 0.05916 log ( H 2 OH ) (activity of water = 1) 2 1

0.05916 0.05916 log (a H 2 ) – log (a OH - ) 2 (assume activity of H2 = 4.0 x 10-5) 2 2 0.05916 E = - 0.83 V + 0.13 – ( ) 2 log (a OH - ) 2 E = - 0.83 + 0.13 – (0.05916 log (a OH - ) )

(B) E = - 0.83 V –

E = - 0.83 + 0.13 + (0.05916 ( – log (a OH - ) ) ) * – log (a OH - ) = pOH = 14.00 – pH E = -0.83 + 0.13 + (0.05916 (14.00 – pH)) E = - 0.83 + 0.13 + 0.414 – (0.05916 pH) (C) At pH = 7.00; E = - 0.83 + 0.13 + 0.414 – (0.05916 (7.00)) = + 0.13V (D) At pH = 10.00; E = - 0.83 + 0.13 + 0.414 – (0.05916 (10.00)) E = -0.286 - (0.5916 ) = - 0.88V (E) At higher pH values this cell potential decreases. At higher pH values the [OH-] increases.

Since OH- is on the product side of this reaction, increasing the concentration of OH- favors the reactant side of the reaction thus lowering the cell potential as pH increases.

(F) Lowering the pH by one unit decreases the [OH-] therefore product formation will be favored. This will result in the cell potential becoming more positive by 0.059 volts.

Pen Problem Ch 11

SOLUTIONS FOR PEN PROBLEM Chapter 11

3OH H 2 NCONH2 (A) 1.80 X 10 M Fe x x = 2.70 x10-3 M needed for precipitation, with 3  2OH Fe 30% excess: 2.70 x10-3 x 130/100 = 3.51 x 10-3 M -3

3+

(B) Gravimetric factor =

2(55.85) 2Fe = = 0.6994 Fe2O3 [2(55.85)  3(15.999)]

grams precipitate x gravimetri c factor x 100 = mass percent of iron grams of sample 0.715 grams precipitate x 0.6994 x 100 = 0.500% 100.0 grams of sample

Pen Problem Ch 12

SOLUTIONS FOR PEN PROBLEM Chapter 12 (A) [H+] = 10-6.100 = 7.94 x10-7M; [OH-] = Kw/[H+] = 1.26 x10-8 M

3.90 x10 -6 Ka (B) A- = = = 0.907 [H  ]  Ka [3.98 x10 -7 ]  3.90 x10  6

[H  ]  [OH ] [3.98 x10 -7 ]  [2.51 x10 8 ] 0.907  CA 0.1771 = = 0.907 -7   [3.98 x10 ]  [2.51 x10 8 ] [H ]  [OH ] 1 1 0.0954 CB

A  

(C) F =

(D) VB = F x VE = 0.907 x 46.41 = 42.09 (E) Titration Error = VB – VE = 42.09 – 46.41 = – 4.32 mL; or 4.32/46.41 x 100 = 9.31% error

.

Pen Problem Ch 13

SOLUTIONS FOR PEN PROBLEM Chapter 13 (A) VCl- MCl- = VAg+ MAg+ 25.00 mL (MCl-) = (9.85 mL)(0.0150M) MCl- = 0.00591 M 0.00591 mol/L x 35.45 g/mol x 1000 mg/g = 209.5 mg/L Cl(B) Ksp = [Ag+][Cl-] = 1.78 x 10-10 At equilibrium; [Cl-] = [Ag+] X2 = 1.78 x 10-10 X = [Cl-] = 1.78x10 10 = 1.33 x10-5 M (C) In the Fajans method of determining chloride concentration, excess Ag+ causes a positive charge on the precipitate to attract the indicator in an adsorption process. Using the Mohr technique excess Ag+ causes a precipitation reaction that forms a red color as the Ksp of Ag2CrO4 is exceeded.

(A) (B)

Pen Problem Ch 15

SOLUTIONS FOR PEN PROBLEM Chapter 15 (A) Intial ICl added: VICl MICl = 0.0420L x 0.100 M = 0.00420 mole ICl Excess ICI = 1mole ICI/2 moles S2O32- x VS2O32- x M S2O32- = 1/2 x (0.0362 L x 0.200 M ) = 0.00362 mol ICl excess Moles of ICl reacting with the sample = 0.00420 – 0.00362 = 0.000580 moles ICI = 0.000580 mole I2

0.00580 moles I2 X ; X = 0.347 mole I2 reacting per 100 g of sample  0.167gram sample 100g sample 0.347 mole/100 g sample x 256 g/mol I2 = 88.9 Iodine number

(B) Table 15.6 reports an iodine number range for safflower oil as 122 – 141. Since this sample is far below that range it must be a more saturated oil.

Pen Problem Ch 16

SOLUTIONS FOR PEN PROBLEM Chapter 16 (A) Coulombs generated: (30.0 x 10-3 C/s)(635 s) = 19.05 coul Moles of electron generated: (19.05 coul)(1 mol e-/96500 coul) = 1.97 x10-4 mol eMoles of I3- generated = (1.97 x10-4 mole e-)(1 mole I3-/2 mol e-) = 9.87 x10-5 mol I3Moles vitamin C reacted = (9.87 x 10-5 mol I3-)( 1 mol Vit C/1 mol I3-) = 9.87 x10-5 mol Vit C Grams of Vit C = (9.87 x10-5 mol vit C)(176g/mol) = 0.0174 g vitamin C in 0.0500L = 0.347 g/L

(B) Coulombs generated: (62.5 x 10-3 C/s)(462 s) = 28.88 coul Moles of electron generated: (28.88 coul)(1 mol e-/96500 coul) = 2.99 x10-4 mol eMoles of I3- generated = (2.99 x10-4 mole e-)(1 mole I3-/2 mol e-) = 1.50 x10-4 mol I3Moles vitamin C reacted = (1.50 x10-4 mol I3-)( 1 mol Vit C/1 mol I3-) = 1.50 x10-4 mol Vit C Grams of Vit C = (1.50 x10-4 mol vit C)(176 g/mol) = 0.0527g vitamin C in 0.0500L = 0.528 g/L

.

Pen Problem Ch 17

SOLUTIONS FOR PEN PROBLEM Chapter 17

(P − PR ) (0.190 − 0.0100) = – log = (Po − PR ) (1.00 − 0.0100) = 0.740 -4 Determine the apparent concentration: A /ε = 0.740/6010 = 1.23 x 10 M

(A) Determine the observed absorbance: Aobs = – log

(B) Convert %T to A: A = – log T = - log (0.190) = 0.721 C = A/ε = 0.721/6010 = 1.20 x 10-4 M

(C) % difference in concentration of NADH in the 19.0%T measurement, with its 1.00% stray light assumption, compared to NADH concentration in the 19.0%T if there were no stray light: [(1.23 x 10-4 M – 1.20 x 10-4 M)/ 1.23 x 10-4 M] x 100 = 2.50%

Pen Problem Ch 18

SOLUTIONS FOR PEN PROBLEM Chapter 18 (A) The absorbance of the original sample and the spiked sample are compared as follows:

Ao C (V  Vs ) 0.721 Co (25.0  5.00) ; ; 0.858(25.00Co + 50.0) = 30.0Co;  o o  Asp Co Vo  Cs Vs 0.840 Co 25.00  10.0(5.00) Co = 5.02 ppm

(B) A = bC; 0.721 =  (1.00 cm) (5.02 ppm) ;  = 0.144 ppm-1 cm-1 (C) Use the smallest reliable absorbance reading and b to determine the concentration: A = bC; 0.05 = 0.144 ppm-1 cm-1 (1.00 cm) C 0.05/0.144 = C ~ 0.3 ppm

Pen Problem Ch 20

SOLUTIONS FOR PEN PROBLEM Chapter 20 (A) Convert pH to [H+] and pKa to Ka; 10-3.42 = [H+] = 3.8 x10-4; 10-5.82 = 1.5 x10-6 The distribution ratio of an acid, where the conjugate base does not extract, is: K D , HA [HA]phase2 Dc ~ = = 62 From this the partition ratio can be [HA]phase1  [A]phase1 1  Ka/[H  ] determined: Dc =

K D, HA

1  1.5 x10  6 /[3.8 x10 - 4 ]

;

KD,HA = 62.2

Now determine the value of Dc at pH = 5.00; [H+] = 1.0 x10-5 K D , HA 62.2 Dc ~ = = 54.1  1  Ka/[H ] 1  1.5 x10  6 /[1.0 x10 -5 ] Next, determine the fraction remaining in phase 1 (water portion) at pH = 5.00: n

1

    1 1 f phase 1 =   =   = 0.0299 1  54.1(30.0/50.0)  1  Dc (V2 /V1 )  Finally determine the fraction extracted to the chloroform phase: fphase2 = 1– fphase1 = 1 – 0.0299 = 0.97

(B) Determine the fraction remaining in phase 1 (water) after three 10.0 mL solvent portions: n

3

    1 1 fphase1 =   =   = 0.00061 1  54.1(10.0/ 50.0)  1  Dc (V2 /V1 ) 

The fraction in the chloroform phase : fphase2 = 1– fphase1 = 1 – 0.00061 = 0.999 ~ 1.0 (C) At pH = 6.00 the [H]+ = 1.0 x 10-6 K D , HA 62.2 Dc ~ = = 24.9  1  Ka/[H ] 1  1.5 x10  6 /[1.0 x10 -6 ] Next, determine the fraction remaining in phase 1 (water portion: n

1

    1 1 f phase 1 =  =    = 0.063 1  24.9(30.0/ 50.0)  1  Dc (V2 /V1 )  Finally determine the fraction extracted to the chloroform phase: fphase2 = 1– fphase1 = 1 – 0.063 = 0.94

(D) At lower pH values more of the weak acid would be protonated so a higher fraction would be extracted since that form has a higher solubility in chloroform. At lower pH values Dc would increase. In turn, an increase in Dc causes the fraction remaining in phase 1 to be smaller.

Pen Problem Ch 21

SOLUTIONS FOR PEN PROBLEM Chapter 21 (A) First, Determine the ratio of caffeine peak heights to deuterated peak heights for each standard: 0 ppm 5/155 = 0.0323 50 ppm 42/159 = 0.264 100 ppm 86/154 = 0.558 150 ppm 134/157 = 0.854 200 ppm 185/162 = 1.14 250 ppm 222/158 = 1.41 Next, apply a linear regression treatment with concentration as the x – axis and the peak ratio as the y – axis: y = ax + b; y = 0.00561x – 0.00882; correlation = 0.9995 (B) The caffeine peak height to deuterated caffeine peak height is (175/158) = 1.11 (C) 1.11 = 0.00561x – 0.00882 ; solving for “x” ; (1.11 –0.00882)/0.00561 = 196 ppm

Pen Problem Ch 23

SOLUTIONS FOR PEN PROBLEM Chapter 23 (A) Electrophoretic mobility () =

v (Ld /t m ) (44.0 cm/10.5 min) = 6.98 cm2 /kV min   E (V/L) (30.0 kV/50.0 cm)

(B) velocity = distance/time; 22.0 cm/10.5 min = 2.09 cm/min

(C) N =

EL d 2DL



(6.98 cm 2 /kVmin)(30 .0 kV)(44.0 cm)  2(1.8 x10 10 m 2 /sec)(60 s/min)(10 4 cm 2 /1m 2 )(50.0 cm) 8.53 x 105 theoretical plates