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Chapter 5: Problem 31

Chloroform is an internal standard in the determination of the pesticide DDT in a polarographic analysis in which each compound is reduced at an electrode surface. A mixture containing \(0.500 \mathrm{mM}\) chloroform and \(0.800 \mathrm{mM}\) DDT gave signals of \(15.3 \mu \mathrm{A}\) for chloroform and \(10.1 \mu\) A for DDT. An unknown solution \((10.0 \mathrm{~mL})\) containing DDT was placed in a \(100-\mathrm{mL}\) volumetric flask and \(10.2 \mu \mathrm{L}\) of chloroform (FM 119.39, density \(=1.484 \mathrm{~g} / \mathrm{mL}\) ) were added. After dilution to the mark with solvent, polarographic signals of \(29.4\) and \(8.7 \mu \mathrm{A}\) were observed for the chloroform and DDT, respectively. Find the concentration of DDT in the unknown.

Short Answer

Expert verified
The concentration of DDT in the unknown solution is 6.89 mM.

Step by step solution

01

Calculate the sensitivity factor

The sensitivity factor (SF) is the ratio of the observed polarographic signal to the concentration for both chloroform and DDT. First, calculate it for both the known standard signals and then again for the unknowns to establish a relationship. For chloroform in the known solution:\[SF_{CHCl_3}^{known} = \frac{15.3\, \mu \mathrm{A}}{0.500\, \mathrm{mM}} = 30.6\, \mu \mathrm{A}/\mathrm{mM}\]For DDT in the known solution:\[SF_{DDT}^{known} = \frac{10.1\, \mu \mathrm{A}}{0.800\, \mathrm{mM}} = 12.625\, \mu \mathrm{A}/\mathrm{mM}\]Now we have the sensitivity factors for the known mixture.
02

Determine Chloroform concentration in the unknown

To find the concentration of chloroform in the given unknown, convert the added volume in \(\mu L\) to moles using its molecular weight and density.First, calculate the mass of chloroform:\[\text{Mass (g)} = 1.484\, \text{g/mL} \times 10.2\, \mu L \times \frac{1}{1000}\, \text{mL/}\mu\text{L} = 0.0151368\, \text{g}\]Next, convert the mass to moles using the molar mass (FM 119.39):\[\text{Moles} = \frac{0.0151368\, \text{g}}{119.39\, \text{g/mol}} = 1.2671 \times 10^{-4}\, \text{mol}\]The volume of the solution is 100 mL or 0.1 L, so the concentration is:\[C_{CHCl_3} = \frac{1.2671 \times 10^{-4}\, \text{mol}}{0.1\, \text{L}} = 1.267\, \mathrm{mM}\]
03

Calculate Sensitivity Factor for Chloroform in Unknown

Now, we calculate the sensitivity factor for chloroform in the unknown solution using the observed signal.\[SF_{CHCl_3}^{unknown} = \frac{29.4\, \mu \mathrm{A}}{1.267\, \mathrm{mM}} = 23.2\, \mu \mathrm{A}/\mathrm{mM}\]This value is noted for comparing with the DDT's observed signal.
04

Calculate DDT Concentration in the Unknown

Using the sensitivity factor from the known DDT, determine the concentration of DDT in the unknown solution using the signal observed.\[C_{DDT}^{unknown} = \frac{8.7\, \mu \mathrm{A}}{12.625\, \mu \mathrm{A}/\mathrm{mM}} = 0.689\, \mathrm{mM}\]This is the concentration of DDT in the entire solution.
05

Adjustment for Dilution

As the unknown solution with DDT was initially in a 10 mL aliquot that was diluted to 100 mL, we adjust the concentration calculated to correspond to the original concentration before dilution.\[C_{DDT}^{original} = 0.689\, \mathrm{mM} \times \frac{100\, \text{mL}}{10\, \text{mL}} = 6.89\, \mathrm{mM}\]This is the original concentration of DDT in the unknown solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polarographic Analysis
Polarographic analysis is an important electrochemical method used for analyzing chemical substances by monitoring their reduction or oxidation currents at an electrode surface.
In this technique, the behavior of a compound in solution is studied as an electric potential is applied while the output current is measured. The current is related directly to chemical processes occurring at the electrode, such as the reduction or oxidation of compounds.
  • The technique provides highly specific information about a substance based on its electrochemical signature.
  • Polarography is especially useful for detecting and quantifying multiple compounds within a complex mixture.
In the context of the given exercise, polarographic analysis allows us to distinguish the separate polarographic signals of chloroform and DDT, which are used to calculate the concentration of these chemicals in solution.
Internal Standard
An internal standard is a chemical substance added in a constant amount to samples, blanks, and calibration standards in an analysis. It is used as a reference to account for variability and enhance the reliability and accuracy of analytical results.
In polarographic analysis, the internal standard should have a distinct and separate polarographic wave from the analytes of interest, so their signals do not interfere.
In the exercise, chloroform acts as the internal standard for the analysis of DDT. The known concentration of chloroform provides a basis for calculating the unknown concentrations of other components by comparing their relative signals. This internal standard helps correct for any variation in conditions that might affect the measurement.
Sensitivity Factor
The sensitivity factor (SF) is a crucial concept that quantifies how much signal is produced per unit of concentration for a given substance in an analytical technique.
The sensitivity factor is calculated by dividing the current (signal measurement) by the concentration of the substance, expressed in amperes per millimolar (碌A/mM).
  • The SF allows for direct comparison of different substances or different conditions, normalizing the observed signals for concentration.
  • It is calculated for both chloroform and DDT to ensure accurate concentration determination.
By computing the sensitivity factors for known concentrations of chloroform and DDT in both standard and unknown samples, we can determine the DDT concentration accurately by comparing the respective signals and their sensitivity factors.
Concentration Calculation
Calculating the concentration of a substance in a mixture involves several steps, particularly when using an internal standard in a complex sample.
Firstly, the sensitivity factor is established for each compound by comparing known concentrations and observed signals.
Next, for the unknown sample, the sensitivity factor helps determine the concentration of each compound using its respective signal.
  • In this exercise, you start by calculating chloroform's known concentration and mass before calculating its sensitivity factor.
  • Subsequently, DDT's concentration in the unknown solution is determined using its sensitivity factor and observed polarographic signal.
Finally, adjustments are made for any dilution during sample preparation, ensuring the calculated concentration reflects the original analyte levels before any dilution occurred. By applying these steps, the concentration of DDT in the original sample was effectively determined.

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Most popular questions from this chapter

imes Siandard addition graph. Students performed an experiment like that in Figure \(5.7\) in which each flask contained \(25.00 \mathrm{~mL}\) of serum, varying additions of \(2.640 \mathrm{M} \mathrm{NaCl}\) standard, and a total volume of \(50.00 \mathrm{~mL}\). \begin{tabular}{ccc} Flask & Volume of standard (mL) & \(\mathrm{Na}^{*}\) atomic emission sagnal (mV) \\ \hline 1 & 0 & \(3.13\) \\ 2 & \(1.000\) & \(5.40\) \\ 3 & \(2.000\) & \(7.89\) \\ 4 & \(3.000\) & \(10.30\) \\ 5 & \(4.000\) & \(12.48\) \\ \hline \end{tabular} (a) Prepare a standard addition graph and find \(\left[\mathrm{Na}^{+}\right]\)in the serum. (b) Find the standard deviation and \(95 \%\) confidence interval for \(\left[\mathrm{Na}^{+}\right]\)

Detection limit. Low concentrations of \(\mathrm{Ni}^{2+}\)-EDTA near the detection limit gave the following counts in a mass spectral measurement: \(175,104,164,193,131,189,155,133,151,176\). Ten measurements of a blank had a mean of 45 counts. A sample containing \(1.00 \mu \mathrm{M} \mathrm{Ni}^{2+}\)-EDTA gave 1797 counts. Estimate the detection limit for Ni- EDTA.

An unknown sample of \(\mathrm{Cu}^{2+}\) gave an absorbance of \(0.262\) in an atomic absorption analysis. Then \(1.00 \mathrm{~mL}\) of solution containing \(100.0 \mathrm{ppm}(=\mu \mathrm{g} / \mathrm{mL}) \mathrm{Cu}^{2+}\) was mixed with \(95.0 \mathrm{~mL}\) of unknown, and the mixture was diluted to \(100.0 \mathrm{~mL}\) in a volumetric flask. The absorbance of the new solution was \(0.500\). (a) Denoting the initial, unknown concentration as \(\left[\mathrm{Cu}^{2+}\right]_{\mathrm{i}}\), write an expression for the final concentration, \(\left[\mathrm{Cu}^{2+}\right]_{\mathrm{f}}\), after dilution. Units of concentration are ppm. (b) In a similar manner, write the final concentration of added standard \(\mathrm{Cu}^{2+}\), designated as \([\mathrm{S}]_{\mathrm{f}}\). (c) Find \(\left[\mathrm{Cu}^{2+}\right]_{\mathrm{i}}\) in the unknown.

Correlation coefficient and Excel graphing. Synthetic data are given below for a calibration curve in which random Gaussian noise with a magnitude of 80 was superimposed on \(y\) values that follow the equation \(y=26.4 x+1.37\). This exercise shows that a high value of \(R^{2}\) does not guarantee that data quality is excellent. (a) Enter concentration in column A and signal in column B of a spreadsheet. Prepare an XY (Scatter) chart of signal versus concentration without a line as described in Section 2-11. Use LINEST (Section 4-7) to find the least-squares parameters including \(R^{2}\). (b) Now insert the Trendline by following instructions in Section 4-9. In the Options window used to select the Trendline, select Display Equation and Display R-Squared. Verify that Trendline and LINEST give identical results. (c) Add \(95 \%\) confidence interval \(y\) error bars following the instructions at the end of Section 4-9. The 95\% confidence interval is \(\pm t s_{y}\), where \(s_{y}\) comes from LINEST and Student's \(t\) comes from Table 4-2 for \(95 \%\) confidence and \(11-2=9\) degrees of freedom. Also, compute \(t\) with the statement "= TINV \((0.05,9)^{n} .\) \begin{tabular}{cc|cc} Concentration \((x)\) & Signal \((y)\) & Concentration \((x)\) & Signal \((y)\) \\ \hline 0 & 14 & 60 & 1573 \\ 10 & 350 & 70 & 1732 \\ 20 & 566 & 80 & 2180 \\ 30 & 957 & 90 & 2330 \\ 40 & 1067 & 100 & 2508 \\ 50 & 1354 & & \\ \hline \end{tabular}

Verifying constant response for an internal standard. When we develop a method using an internal standard, it is important to verify that the response factor is constant over the calibration range. Data are shown below for a chromatographic analysis of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\), using deuterated naphthalene \(\left(\mathrm{C}_{10} \mathrm{D}_{8}\right.\) in which \(\mathrm{D}\) is the isotope \({ }^{2} \mathrm{H}\) ) as an internal standard. The two compounds emerge from the column at almost identical times and are measured by a mass spectrometer, which distinguishes them by molecular mass. From the definition of response factor in Equation 5-11, we can write $$ \frac{\text { Area of analyte signal }}{\text { Area of standard signal }}=F\left(\frac{\text { concentration of analyte }}{\text { concentration of standard }}\right) $$ Prepare a graph of peak area ratio \(\left(\mathrm{C}_{10} \mathrm{H}_{8} / \mathrm{C}_{10} \mathrm{D}_{8}\right)\) versus concentration ratio \(\left(\left[\mathrm{C}_{10} \mathrm{H}_{8}\right] /\left[\mathrm{C}_{10} \mathrm{D}_{8}\right]\right)\) and find the slope, which is the response factor. Evaluate \(F\) for each of the three samples and find the standard deviation of \(F\) to see how "constant" it is. $$ \begin{array}{ccccc} \text { Sample } & \begin{array}{c} \mathrm{C}_{10} \mathrm{H}_{8} \\ (\mathrm{ppm}) \end{array} & \begin{array}{c} \mathrm{C}_{10} \mathrm{D}_{8} \\ (\mathrm{ppm}) \end{array} & \begin{array}{c} \mathrm{C}_{10} \mathrm{H}_{8} \\ \text { peak area } \end{array} & \begin{array}{c} \mathrm{C}_{10} \mathrm{D}_{8} \\ \text { peak area } \end{array} \\ \hline 1 & 1.0 & 10.0 & 303 & 2992 \\ 2 & 5.0 & 10.0 & 3519 & 6141 \\ 3 & 10.0 & 10.0 & 3023 & 2819 \\ \hline \end{array} $$
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