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Chapter 19: Q18P (page 488)

A study was conducted with derivatives of the DNA nucleotide bases adenine and thymine bound inside micelles () in aqueous solution.
Sodium dodecyl sulfate forms micelles with the hydrocarbon tails pointed inward and ionic headgroups exposed to water. It was hypothesized that the bases would form ahydrogen-bonded complex inside the micelle as they do in DNA:
To test the hypothesis, aliquots of 5.0 mMadenine derivative were mixed with aliquots of 5.0 mMthymine derivative in proportions shown in the table. Each solution also contained 20mMsodium dodecyl sulfate. The concentration of product measured by nuclear magnetic resonance also is shown in the table. Are the results consistent with formation of a 1:1complex? Explain your answer.

Short Answer

Expert verified

The mole fraction of the complex formation is 0.5

Step by step solution

01

Define mole fraction.

It is the ratio of number of mole of one component to the total number of moles in given mixture.

02

Determine spreadsheet.

The column E is calculates as:

$\frac{n (thymine)}{n (thymine) + n (adenine)}$

03

Draw graph.

If we see, the peak is at mole fraction 0.5, which will fit in 1:1 complex formation as:

$\frac{1}{1 + 1} = 0.5$

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Most popular questions from this chapter

The protein bovine serum albumin can bind several molecules of the dye methyl orange. To measure the binding constant K for one dye molecule, solutions were prepared with a fixed concentration $(x_{0})$ of dye and a larger, variable concentration of protein (P). The equilibrium is Reaction 19-18, with X 5 methyl

orange.

Experimental data are shown in cells A16-D20 in the spreadsheet on the next page. The authors report the increase in absorbance $∆ A$ at 490 nm as P is added to X. X and PX absorb visible light, but P does not. Equilibrium expression 19-20 applies and [PX] is given by Equation 19-21. Before P is added, the absorbance is. The increase in absorbance when P is added is

The spreadsheet uses Solver to vary K and $∆ E$ in cells B10:B11 to minimize the sum of squares of differences between observed and calculatedin solutions with different amounts of P. Cell E16 computes [PX] from Equation 19-21, which is Equation A on line 6 of the spreadsheet. Cells F16 and G16 find [X] and [P] from mass balances. Cell H16 computes ${ΔA}_{calc} = ΔE [PX]$ which is Equation B on line 7.

To estimate a value of K in cell B10, suppose that 50% of X has reacted in row 20 of the spreadsheet. The total concentration of X is. If half is reacted, then $[X] = [PX] = 2 .85 μM and [P] = P_{0} - [PX] = 40 .4 - 2 .85 = 37 .55 μM$ The binding

constant is $K = [PX] \land P] [X]) = [2 .85 μM] \land 37 .55 μM] [2 .85 μM]) = 2 .7 \times 10^{4}$ which we enter as our guess for K in cell B10. We estimatein cell B11 by supposing that 50% of X has reacted in row 20. In Equation B on line 7, $ΔA = Δ ε [PX]$ .The measured

value ofin row 20 is 0.0291 and we just estimated that. Therefore, our guess for localid="1668328314124" $Δε in cell B 11 is Δε = ΔA / [PX] = (0 .0291) (2 .85 μM) = 1 .0 \times 10^{4}$ in cell B11 is

Your assignment is to write formulas in columns E through J of the spreadsheet to reproduce what is shown and to find values in cells E17:J20. Then use Solver to find K andin cells B10:B11 to minimize $Σ {(A_{oths} - A_{calc})}^{2}$ in cell I21.

Explain what is done in flow injection analysis and sequential injection. What is the principal difference between the two techniques? Which one is called “lab-on-a-valve”?

This problem can be worked by calculator or with the spreadsheet in Figure 19-4. Consider compounds X and Y in the example labeled “Analysis of a Mixture, Using Equations 19-6” on page 464. Find [X] and [Y] in a solution whose absorbance is 0.233 at 272 nm and 0.200 at 327 nm in a 0.100-cm cell.

Find the concentration of [X] if the absorbance are 0.700 at 272 nm and

0.550 at 327 nm.

Now we use Solver to find Kfor the previous problem. The only absorbing species at 332 nmis the complex, so, from Beer’s law $[complex] = A / ε (becausepathlength = 1.000 cm) . I_{2}$ is either free or bound in the complex,so $[I_{2}] = [I_{2}]_{tot} - [complex] .$ There is a huge excess of mesitylene, so $[mesitylene]$ $\approx [mesitylene]_{tot}$

$K = \frac{[c o m p l e x]}{[l_{2}] [m e s i t y l e n e]} = \frac{A / ε}{({[l_{2}]}_{t o t} - A / ε) {[m e s t i t y l e n e]}_{t o t}}$

The spreadsheet shows some of the data. You will need to use all the data. Column A contains [mesitylene] and column B contains ${[l_{2}]}_{tot}$ _. Column C lists the measured absorbance. Guessa value of the molar absorptivity of the complex, $ε, in cell A 7 .$ Then compute the concentration of the complex $(= A / ε)$ in column D. The equilibrium constant in column $E is given by E 2 =$ $[complex] / ([I_{2}] [mesitylene]) = (D 2) / ((B 2 - D 2) * A 2) .$

should we minimize with Solver? We want to varyεin cell A7 until the values of Kin column E are as constant as possible. We would like to minimize a function like $\sum (K_{i} - K_{average})^{2}$ , where K_iis the value in each line of the table and Kaverage is the average of all computed values. The problem with $\sum (K_{i} - K_{average})^{2}$ is that we can minimize this function simply by making $K_{i}$ very small, but not necessarily constant. What we really want is for all the $K_{i}$ to be clustered around the mean value. A good way to do this is to minimize the relative standard deviationof the K, which is (standard deviation)/average. In cell E5we compute the average value of Kand in cell E6the standard deviation. Cell E7contains the relative standard deviation. Use Solver to minimize cell E7by varying cell A7. Compare your answer with that of Problem 19-13.

See all solutions

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