(a)
Interpretation:
The optimal velocity is to be stated.
Concept introduction:
The Van Deemeter equation relates the HPLC plate height and velocity of plate. The Van Deemeter equation is as follows.
H=Bu+Cu=Bu+CSu+CMu ...... (I)
Here, the HPLC height is H, the plate velocity is u, the diffusion coefficient is B, the mass transfer coefficient in stationary phase is CS, the mass transfer coefficient in mobile phase is CM, the total mass transfer coefficient is C.

Answer to Problem 28.23QAP
The optimal velocity of HPLC plate is √BC_.
Explanation of Solution
Differentiate Equation (I) with respect to plate velocity.
dHdu=(−1)Bu2+C(1) ...... (II)
Substitute 0 for dHdu, uopt for u in the equation for optimal velocity.
0=−Bu2opt+Cuopt=√BC
(b)
Interpretation:
The minimum plate height is to be stated.
Concept introduction:
The Van Deemeter equation relates the HPLC plate height and velocity of plate. The Van Deemeter equation is as follows.
H=Bu+Cu=Bu+CSu+CMu
The relation between HPLC height is inversely proportional to plate velocity. So minimum height is obtained corresponding to optimal velocity in the Van Deemeter equation.

Answer to Problem 28.23QAP
The minimum plate height is 2√BC_.
Explanation of Solution
Substitute √BC for u in the Equation (I).
Hmin=B(√BC)+C√BC=√BC+√BC=2√BC
(c)
Interpretation:
The optimal velocity and minimum height of plate in terms of the diffusion coefficient in mobile phase, particle size and dimensional analysis is to be stated.
Concept introduction:
The mass transfer coefficient relates the particle size and diffusion coefficient in mobile phase. The diffusion coefficient relates the dimensional constant and diffusion coefficient in mobile phase.

Answer to Problem 28.23QAP
The optimal velocity is DMdP√2γω_ and the minimum height is 2dP√2γω_ of plate in terms of the diffusion coefficient in mobile phase, particle size and dimensional analysis is to be stated.
Explanation of Solution
Write the expression for mass transfer coefficient in mobile phase.
CM=ωd2pDM ...... (III)
Here, the diffusion coefficient in mobile phase is DM, the particle size is dP, the dimensional constant is ω.
Write the expression for diffusion coefficient.
B=2γDM ...... (IV)
Here, the dimensional coefficient is γ.
Write the expression for optimal velocity in mobile phase.
uopt=√BCM ...... (V)
Write the expression for minimum height in mobile phase.
Hmin=2√BCM ...... (VI)
Substitute ωd2pDM for CM, 2γDM for B in the Equation (V).
uopt=√(2γDM)(ωd2pDM)=√2γD2Mωd2p=DMdP√2γω ...... (VII)
Substitute ωd2pDM for CM, 2γDM for B in the Equation (VI).
Hmin=2√(2γDM)(ωd2pDM)=2dP√2γω ...... (VIII)
(d)
Interpretation:
The optimal velocity and minimum height of plate at unity order for dimensional constants.
Concept introduction:
The mass transfer coefficient relates the particle size and diffusion coefficient in mobile phase. The diffusion coefficient relates the dimensional constant and diffusion coefficient in mobile phase.

Answer to Problem 28.23QAP
The optimal velocity is DMdP_ and minimum height of plate is dP_ at unity order for dimensional constants.
Explanation of Solution
Write the expression for optimal velocity in terms of dimensional constants.
uopt=DMdP√2γω ...... (IX)
Substitute 1 for γ, and 1 for ω in the Equation (IX).
uopt=DMdP√2(1)(1)=√2DMdP≈DMdP
Write the expression for plate height in terms of dimensional constants.
Hmin=2dP√2γω ...... (X)
Substitute 1 for γ, and 1 for ω in the Equation (X).
Hmin=2dP√2(1)(1)=2√2dP≈dP
(e)
Interpretation:
The preceding conditions for reduced plate height to one third, the optimal velocity in these conditions and effect on number of theoretical plates is to be stated.
Concept introduction:
The plate height and number of theoretical plates directly depends on the particle size and optimal velocity inversely depends on particle size.

Answer to Problem 28.23QAP
For one third reduction in particle size the plate height is reduced to one third. The optimal velocity becomes 32 times to initial optimal velocity. The number of theoretical plates also reduced to one third of initial value.
Explanation of Solution
Write the expression for plate height.
H1dp1=H2dp2 ...... (XII)
Here, the initial plate height is H1, reduced plate height is H2, the initial particle size is dp1, the particle size at reduced plate height is dp2.
Substitute 23H1 for H2 in the equation (XII).
H1dp1=23(H1)dp2dp2=23dp1
The particle size is also reduced to one third to initial particle size for one third reduction in the plate height.
The optimal velocity is inversely proportional to the particle size, so optimal velocity become 32 times to initial optimal velocity.
Write the expression for number of theoretical plates.
N=HL
From the above equation it is clear that number of theoretical plates reduced to one third to initial number of theoretical plates.
(f)
Interpretation:
The condition to reduce the plate height to one third without reducing the number of theoretical plates is to be stated.
Concept introduction:
The number of theoretical plates directly depends on the height of the plate and inversely to the column length.

Answer to Problem 28.23QAP
The column length is also reduced to one third with plate height to maintain same number of theoretical plates.
Explanation of Solution
Write the expression for number of theoretical plate with one third reduced plate height
N=23HL ...... (XIII)
To maintain the same number of theoretical plate length of column also be reduced to one third. We have to substitute 23 in the equation (XIII).
(g)
Interpretation:
The two sources of band broadening which also contribute to overall width of HPLC peaks is to be stated.
Concept introduction:
The band broadening is defined as the reduction in separation of solute molecules in the column. This also reduces the quality and accuracy of the separation. The band broadening occurs due to longitudinal movement of particles in the column.

Answer to Problem 28.23QAP
The two sources of band broadening which also contribute to overall width of HPLC peaks are diameter of the particles and diffusion coefficient in mobile phase.
Explanation of Solution
The band broadening is the measure of column efficiency. It is inversely proportional to the mass transfer coefficient. If the mass transfer is slow in the column then wide band is obtained and narrow band is obtained when the mass transfer in the column is high. The sources of extra column band broadening which contributes to overall width of liquid chromatography peaks are following.
(1) The diameter of particles in column.
(2) The diffusion coefficient of mobile phase.
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Chapter 28 Solutions
PRINCIPLES OF INSTRUMENTAL ANALYSIS
- Using the bond energy values, calculate the energy that must be supplied or is released upon the polymerization of 755 monomers. If energy must be supplied, provide a positive number; if energy is released, provide a negative number. Hint: Avogadro’s number is 6.02 × 1023.arrow_forward-AG|F=2E|V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: Acidic solution -0.93 +0.38 -0.51 -0.06 H3PO4 →H4P206 H3PO3 H3PO2 → P→ PH3 -0.28 -0.50 → -0.50 Basic solution 3-1.12 -1.57 -2.05 -0.89 PO HPO →→H2PO2 P PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P2O6 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH, 0.0 -0.5- 2 3 9 3 -1.5 -2.0 Pa H,PO H,PO H,PO -3 -1 0 2 4 Oxidation state, N 2 b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) c) Elemental phosphorus tends to disproportionate under basic conditions. Use data in…arrow_forwardThese two reactions appear to start with the same starting materials but result in different products. How do the chemicals know which product to form? Are both products formed, or is there some information missing that will direct them a particular way?arrow_forward
- What would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 3 1 2 2. n-BuLi • Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Explanation Check Click and drag to start drawing a structure. 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Priva ×arrow_forwardPredict the products of this organic reaction: Explanation Check IN NaBH3CN H+ ? Click and drag to start drawing a structure. D 5 C +arrow_forwardPredict the products of this organic reaction: H3O+ + ? • Draw all the reasonable products in the drawing area below. If there are no products, because no reaction will occur, check the box under the drawing area. • Include both major and minor products, if some of the products will be more common than others. • Be sure to use wedge and dash bonds if you need to distinguish between enantiomers. No reaction. Click and drag to start drawing a structure. dmarrow_forward
- Iarrow_forwardDraw the anti-Markovnikov product of the hydration of this alkene. this problem. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for esc esc ☐ Explanation Check F1 1 2 F2 # 3 F3 + $ 14 × 1. BH THE BH3 2. H O NaOH '2 2' Click and drag to start drawing a structure. F4 Q W E R A S D % 905 LL F5 F6 F7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility < & 6 7 27 8 T Y U G H I F8 F9 F10 F11 F12 9 0 J K L P + // command option Z X C V B N M H H rol option commandarrow_forwardAG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, Narrow_forward
- 4. For the following complexes, draw the structures and give a d-electron count of the metal: a) Tris(acetylacetonato)iron(III) b) Hexabromoplatinate(2-) c) Potassium diamminetetrabromocobaltate(III) (6 points)arrow_forward2. Calculate the overall formation constant for [Fe(CN)6]³, given that the overall formation constant for [Fe(CN)6] 4 is ~1032, and that: Fe3+ (aq) + e = Fe²+ (aq) E° = +0.77 V [Fe(CN)6]³ (aq) + e¯ = [Fe(CN)6] (aq) E° = +0.36 V (4 points)arrow_forward5. Consider the compounds shown below as ligands in coordination chemistry and identify their denticity; comment on their ability to form chelate complexes. (6 points) N N A B N N N IN N Carrow_forward
- Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
