Why are silica stationary phase height limited to operating in the pH range 2-8?why does the silica in figure25-8have improved stability at low pH?

Short Answer

Expert verified

Because silica dissolves above pH=8And the siloxane bond hydrolyzes below pH=2

As a result of the interference between the approach of H+.And the siloxane bond, acid-catalyzed hydrolysis occurs is to avoid.

Step by step solution

01

definition of silica

Silica is also used in the grinding and polishing of glass and stone, as well as in the production of glass, ceramics, silicon carbide, ferrosilicon, and silicone.

Because silica dissolves above pH=8And the siloxane bond hydrolyzes below pH=2,

silica stationary phases are generally limited to operating in the pH range

2-8.Figure 25-8shows that silica has improved stability at low pH due to the presence of bulky isobutyl groups.

02

Result of approach

As a result of the interference between the approach ofH+

And the siloxane bond, acid-catalyzed hydrolysis occurs is to avoid.

Hence,

Because silica dissolves above pH=8And the siloxane bond hydrolyzes below pH=2, silica stationary phases are generally limited to operating in the pH range

2-8 . Figure 25-8 shows that silica has improved stability at low pH due to the presence of bulky isobutyl groups. As a result of the interference between the approach of H+.And the siloxane bond, acid-catalyzed hydrolysis occurs is to avoid.

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Most popular questions from this chapter

The antitumor drug gimatecan is available as nearly pure (S)-enantiomer. Neither pure (R)-enantiomer nor a racemic (equal) mixture of the two enantiomers is available. To measure small quantities of (R)-enantiomer in nearly pure (S)-gimatecan, a preparation was subjected to normal-phase chromatography on each of the enantiomers of a commercial, chiral stationary phase designated (S,S)- and (R,R)-DACH-DNB. Chromatography on the (R,R)-stationary phase gave a slightly asymmetric peak at tr 5 6.10 min with retention factor k 5 1.22. Chromatography on the (S,S)- stationary phase gave a slightly asymmetric peak at tr 5 6.96 min with k 5 1.50. With the (S,S) stationary phase, a small peak with 0.03% of the area of the main peak was observed at 6.10 min.

Chromatography of gimatecan on each enantiomer of a chiral stationary phase. Lower traces have enlarged vertical scale. [Data from E. Badaloni, W. Cabri, A. Ciogli, R. Deias, F. Gasparrini, F. Giorgi, A. Vigevani, and C. Villani, “Combination of HPLC ‘Inverted Chirality Columns Approach’ and MS/MS Detection for Extreme Enantiomeric Excess Determination Even in Absence of Reference Samples.” Anal. Chem. 2007, 79, 6013.]

(a) Explain the appearance of the upper chromatograms. Dashed lines are position markers, not part of the chromatogram. What Problems 709 would the chromatogram of pure (R)-gimatecan look like on the same two stationary phases?

(b) Explain the appearance of the two lower chromatograms and why it can be concluded that the gimatecan contained 0.03% of the (R)-enantiomer. Why is the (R)-enantiomer not observed with the (R,R)-stationary phase?

(c) Find the relative retention (a) for the two enantiomers on the (S,S)-stationary phase.

(d) The column provides N 5 6 800 plates. What would be the resolution between the two equal peaks in a racemic (equal) mixture of (R)- and (S)-gimatecan? If the peaks were symmetric, does this resolution provide baseline separation in which signal returns to baseline before the next peak begins?

(a). Sketch a graph of the van Deemnter equation (plate height versus linear flow rate).what would the curve look like if the multiple path term were 0? If the longtitundinal diffusinal diffusion term were 0?

(b). Explain why the van Deemter curve for 1.8μmparticles in figure 25-3is nearly flat at high flow rate.what can you say about each of the terms in the van Deemeter equation for 1.8μmparticles.

(c). Explain why the 2.7μmsuperficially porous particle enables separations similar to those achieved by 1.8μmtotally porous particles,but the superficially porous particle requires lower pressure.

(a) List ways in which the resolution between two closely spaced peaks might be changed.

(b) After optimization of an isocratic elution with several solvents, the resolution of two peaks is 1.2How might you improve the resolution without changing solvents or the kind of stationary phase?

The rate at which heat is generated inside a chromatographycolumn from friction of flowing liquid is power (watts, W=J/s)= volume flow raterole="math" localid="1656474760665" (m3/s)×pressure drop (pascals, role="math" localid="1656474828044" Pa=kg/[m?s2]).

(a) Explain the analogy between heat generated in a chromatography column and heat generated in an electric circuit (power = current×voltage).

(b) At what rate (watts = J/s) is heat generated for a flow of 1 mL/minwith a pressure difference of3500 bar between the inlet and outlet?

You will need to convert mL/min tom3/s.Also 1 bar=105Pa.

(a) Why does mobile phase strength increase as solvent

becomes less polar in reversed-phase chromatography, whereas

mobile phase strength increases as solvent becomes more polar in

normal-phase chromatography?

(b) What kind of gradient is used in supercritical fluid chromatography?

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