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Chapter 19: Q1P (page 484)

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.

Short Answer

Expert verified

The solution for [X] is $[x] = 8.034 ‐ 10^{- 5} M$

The solution for [Y] is $[y] = 2.62 ‐ 10^{- 4} M$

Step by step solution

01

Find the solution for [X] :

$[X] = \frac{|\begin{matrix} 0.233 387 \\ 0.200 642 \end{matrix}|}{|\begin{matrix} 1640 387 \\ 399 642 \end{matrix}|} [X] = \frac{(0.233 ‐ 642) - (387 ‐ 0.200)}{(1640 ‐ 642) - (399 ‐ 387)} [X] = \frac{72.186}{898467} [X] = 8.034 ‐ 10^{- 5} M$

02

Find the solution for [Y]:

The solution of [Y] can be calculated as follows

$[y] = \frac{|\begin{matrix} 1640 0.233 \\ 399 0.200 \end{matrix}|}{|\begin{matrix} 1640 387 \\ 399 642 \end{matrix}|} [y] = \frac{(1640 \cdot 0.200) - (399 \cdot 0.233)}{(1640 \cdot 642) - (399 \cdot 387)} [y] = \frac{235.033}{898467} [y] = 2.6 . 10^{- 4} M$

03

Spreadsheet:

The formula use for the cell F5 and F6

= MMULT(MINVERSE(B5:C6); D5:D6)

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Most popular questions from this chapter

When are isosbestic points observed and why?

The end of Box 19-2 states that “the advantage of up conversion for biomedical probes is that low energy near-infrared (800 to 1000 nm) incident radiation stimulates little background emission from the complex biological matrix that can be highly fluorescent under visible radiation.” Suggest why near-infrared radiation stimulates less emission than visible radiation and why this behavior is useful.

Challenging your acid-base prowess. A solution was prepared by mixing 25.00mL of 0.800Maniline, 25.00mLsulfanilic acid, andand then diluting to 100.0mL. (stands for protonated indicator.)

The absorbance measured at550nmin 5.00 - cmwas 0.110.Find the concentrations of $HIn and In and p_{a} for HIn$

Chemical equilibrium and analysis of a mixture. (Warning! This is a long problem.) A remote optical sensor for ${CO}_{2}$ 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 ${CO}_{2}$ , but not dissolved ions, can diffuse. Inside the sensor, ${CO}_{2}$ equilibrates with ${HCO}_{3}$ and ${CO}_{3}^{2}$ . 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:

$[{HIn}^{-}] + [\ln^{2 -}] = F_{In} = 50 .0 μMand [{Na}^{+}] = F_{Nα} = 50 .0 μM + 42 .0 μM = 92 .0 μM$

has an absorbance maximum at 434 nm andhas a maximum at 620 nm. The sensor measures the absorbance ratio $R_{A} = A_{620} / A_{434}$ 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

$\frac{[\ln^{2 -}]}{[Hln -]} = \frac{R_{A} ε_{434}^{{HHn}^{-}} - ε_{6, 20}^{Hln -}}{ε_{620}^{\ln^{2 -}} - R_{A} ε_{434}^{\ln^{2 -}}} = R_{\ln}$ (A)

(b) From the mass balance (1) and the acid dissociation constant

, show that

$[{Hln}^{-}] = \frac{F_{1 n}}{R_{\ln} + 1}$ (B)

$[\ln^{2 -}] = \frac{K_{\ln} F_{\ln}}{[H^{+}] (R_{\ln} + 1)}$ (C)

(c) Show that $[H^{+}] = K_{\ln} / R_{\ln}$ (D)

(d) From the carbonic acid dissociation equilibria, show that

$\begin{aligned} [{HCO}_{3}^{-}] = \frac{K_{1} [CO (aq)] E}{[H^{+}]} \\ [{CO}_{3}^{2 -}] = \frac{K_{1} K_{2} [CO (aq)] F}{{[H^{+}]}^{2}} \end{aligned}$

(e) Write the charge balance for the solution in the sensor compartment. Substitute in expressions B, C, E, and F forHln, ${In}^{2 -}, [{HCO}_{3}], and [{CO}_{3}^{2 -}]$

(f) Suppose that the various constants have the following values:

$\begin{array}{ll} ε_{4344}^{{HHn}^{-}} = 8 .00 \times 10^{3} M^{- 1} {cm}^{- 1} K_{1} = 3 .0 \times 10^{- 7} \\ ε_{6620}^{Hn -} = 0 K_{2} = 3 .3 \times 10^{- 11} \\ ε_{434}^{\ln^{2}} = 1 .90 \times 10^{3} M^{- 1} {cm}^{- 1} K_{\ln} = 2 .0 \times 10^{- 7} \\ ε_{620}^{\ln^{2 -}} = 1 .70 \times 10^{4} M^{- 1} {cm}^{- 1} K_{w} = 6 .7 \times 10^{- 15} \end{array}$

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?

Two ways to analyze a mixture. Figure 19-5 shows the spectrum of the indicator bromothymol blue adjusted to several pH values. The spectrum at pHis that of the pure blue form and the spectrum at pH 1.8is that of the pure yellow form. At other pHvalues, there is a mixture of the two forms. The total concentration isand the path length isin all spectra. For the purpose of calculation, assume that there are more than two significant digits in concentration and path length. Absorbance at the dots on three of the curves in Figure 19-5 is given in the table.

(a) Prepare a spreadsheet like Figure 19-3 to use absorption at all six wavelengths to find $[{In}^{-}] and [HIn]$ in the mixture. Comment on the sum $[{In}^{-}] + [HIn]$ .

(b) From $[{In}^{-}]$ in the mixture, and from $p K_{a} = 7.10$ for HIn,calculate theof the mixture. (This calculation is the source of pH labels in the figure.)

(c) Use Equations 19-6 at the peak wavelengths ofto findin the mixture. Compare your answers to those in (a). Which answers, (a) or (c), are probably more accurate? Why?

See all solutions

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