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Chapter 19: Q15P (page 487)
Method of continuous variation. Make a graph of absorbance versus mole fraction of thiocyanate from the data in the table.
(c)Why does one solution containacid and the otheracid?
One solution contain acid and the other acid because ionic strength must be equal in both the solution.
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Chemical equilibrium and analysis of a mixture. (Warning! This is a long problem.) A remote optical sensor for in the ocean was designed to operate without the need for calibration.33
The sensor compartment is separated from seawater by a silicone membrane through which , but not dissolved ions, can diffuse. Inside the sensor, equilibrates with and . For each
measurement, the sensor is flushed with fresh solution containingbromothymol blue indicator. All indicator is in the formnear neutral pH, so we can
write two mass balances:
has an absorbance maximum at 434 nm andhas a maximum at 620 nm. The sensor measures the absorbance ratio reproducibly without need for calibration. From this ratio, we can findin the seawater as outlined here:
(a).From Beer’s law for the mixture, write equations forin terms of the absorbance at 620 and 434 nmThen show that
(A)
(b) From the mass balance (1) and the acid dissociation constant
, show that
(B)
(C)
(c) Show that (D)
(d) From the carbonic acid dissociation equilibria, show that
(e) Write the charge balance for the solution in the sensor compartment. Substitute in expressions B, C, E, and F forHln,
(f) Suppose that the various constants have the following values:
From the measured absorbance ratio=2.84, findin the seawater.
(g) Approximately what is the ionic strength inside the sensor compartment? Were we justified in neglecting activity coefficients in working this problem?
Explain how signal amplification is achieved in enzyme linked immunosorbent assays.
Simulating a Job’s plot. Consider the reaction . Suppose that the following mixtures of A and B at a fixed total concentration of
10-4 M are prepared:
(b)Prepare a graph by the method of continuous variation in which you plotversus mole fraction offor each equilibrium constant. Explain the shapes of the curves.
19-C. 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 \(\left( {{X_0}} \right)\) 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 \((\Delta A)\) 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 \({\varepsilon _X}{X_{0 - }}\). The increase in absorbance when P is added is
The spreadsheet uses Solver to vary K and \(\Delta E\) in cells B10:B11 to minimize the sum of squares of differences between observed and calculated \(\Delta A\) in 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 \(\Delta {A_{calc}} = \Delta E(PX)\)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\({X_0} = 5.7\mu M\). If half is reacted, then \((X) = (PX) = 2.85\mu M\) and\((P) = {P_0} - (PX) = 40.4 - 2.85 = 37.55\mu M\). The binding
constant is \(K = (PX)/(P)(X)) = (2.85\mu M)/(37.55\mu M)(2.85\mu M)) = \)\(2.7 \times 1{0^4}\) which we enter as our guess for K in cell B10. We estimate \(\Delta \varepsilon \)in cell B11 by supposing that 50% of X has reacted in row 20. In Equation B on line 7, \(\Delta A = \Delta \varepsilon (PX)\).The measured
value of \((\Delta A)\)in row 20 is 0.0291 and we just estimated that \((PX) = \)\(2.85\mu M\). Therefore, our guess for \(\Delta \varepsilon \)in cell B11 is \(\Delta \varepsilon = \Delta A/\)\((PX) = (0.0291)/(2.85\mu M) = 1.0 \times 1{0^4}\)
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 and \(\Delta \varepsilon \)in cells B10:B11 to minimize \(\Sigma {\left( {{A_{oths\;}} - {A_{calc\;}}} \right)^2}\)in cell I21.
When are isosbestic points observed and why?
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