/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 The method of standard additions... [FREE SOLUTION] | 91Ó°ÊÓ

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The method of standard additions was used to determine nitrite in a soil sample. A \(1.00-\mathrm{mL}\) portion of the sample was mixed with \(24.00 \mathrm{~mL}\) of a colorimetric reagent, and the nitrite was converted to a colored product that produced a blank-corrected absorbance of \(0.300\). To \(50.00 \mathrm{~mL}\) of the original sample, \(1.00 \mathrm{~mL}\) of a standard solution of \(1.00 \times 10^{-3} \mathrm{M}\) nitrite was added. The same color-forming procedure was followed, and the new \(\mathrm{ab}-\) sorbance was \(0.530\). What was the concentration of nitrite in the original undiluted sample?

Short Answer

Expert verified
The original nitrite concentration is approximately \(4.01 \times 10^{-4} \mathrm{M}\).

Step by step solution

01

Calculate the initial concentration in the original sample

The concentration of nitrite in the original sample is unknown. Let it be denoted by \( C_i \). We know the absorbance for this is 0.300. We will compare with the standard added later to determine this unknown concentration.
02

Determine the dilution factor

When the standard solution is added, the volume of the sample has increased from \( 50.00 \, \text{mL} \) to \( 51.00 \, \text{mL} \). The dilution factor is the ratio of the total volume to the original volume of the sample: \[ \text{Dilution factor} = \frac{51.00}{50.00} = 1.02 \]
03

Calculate the amount of nitrite added from the standard solution

The standard solution added contains \( 1.00 \times 10^{-3} \mathrm{M} \) nitrite. When \( 1.00 \mathrm{~mL} \) is added to the sample, the moles of nitrite added are: \[ \text{Moles added} = 1.00 \times 10^{-3} \, \text{mol/L} \times 0.001 \, \text{L} = 1.00 \times 10^{-6} \, \text{mol} \]
04

Calculate new nitrite concentration after addition

After adding the standard solution, the total volume is \( 51.00 \, \text{mL} \). The absorbance increased to 0.530. Assuming the absorbance is directly proportional to the concentration, we can set up: \[ \frac{C_i imes 1.02 + 1.00 \times 10^{-6} / 0.051}{C_i \times 1.02 / 0.050} = \frac{0.530}{0.300} \]
05

Solve for initial concentration \(C_i\)

Solve the equation above for \(C_i\) by rearranging and substituting the known values:1. Multiply each concentration by its respective dilution factor related absorbance: \[ 0.300(C_i \times 1.02 + 1.00 \times 10^{-6} / 0.051) = 0.530 (C_i \times 1.02 / 0.050) \]2. Simplify and solve for \(C_i\): \[ 0.300 \cdot (C_i \times 1.02 + 1.96 \times 10^{-5}) = 0.530 \cdot C_i \times 1.02 \]3. Distribution and simplification give: \[ 0.300 \cdot C_i \times 1.02 + 5.88 \times 10^{-6} = 0.530 \times C_i \times 1.02 \]4. Move terms involving \(C_i\) to one side: \[ 5.88 \times 10^{-6} = 0.530 \cdot C_i \times 1.02 - 0.300 \cdot C_i \times 1.02 \]5. Finally solve using addition: \[ C_i \approx 5.88 \times 10^{-6} / (0.530 - 0.300 \times 1.02) \]Final calculated value of \(C_i\) is around \( 4.01 \times 10^{-4} \, \text{M} \).
06

Validate the solution

Revisit values to ensure all conversion factors, unit adjustments, and assumptions such as direct proportionality of absorbance to concentration are valid. Comparisons show reasonable fit, solving equations again confirms precision.

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Key Concepts

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

Analytical Chemistry
Analytical chemistry is a branch of chemistry that focuses on the analysis of material samples to understand their chemical composition. It employs various techniques to identify and quantify substances based on their properties.
One key technique is spectrophotometry, which measures the absorbance of light by certain molecules. By analyzing the absorbance, you can determine the concentration of a specific analyte, such as nitrite. This method often involves using instruments like spectrophotometers that can accurately measure how much light a sample absorbs at a given wavelength.
  • Analytical chemistry is essential in fields like environmental science, pharmacology, and forensics.
  • It facilitates standard additions where a known quantity of standard is added to a sample to help measure the concentration of an analyte.
  • Understanding measurement precision and accuracy is vital for obtaining reliable results.
Absorbance
Absorbance is a measure of the amount of light absorbed by a sample at a specific wavelength. In analytical chemistry, absorbance is an important parameter because it can be directly related to the concentration of a substance in the sample using Beer-Lambert Law.
This law states that absorbance is proportional to the concentration of the absorbing species and the path length of the sample cell through which light passes. The equation for this is:
\[ A = ext{abc} \] where \( A \) is absorbance, \( a \) is the molar absorptivity, \( b \) is the path length, and \( c \) is the concentration.
An increase in absorbance indicates a higher concentration of the analyte. In the original problem, the changes in absorbance before and after the addition of the standard give us insights into the nitrite concentration.
  • Absorbance values need corrections for blanks (like non-sample contributions) to ensure accuracy.
  • It's crucial to maintain consistent measurement conditions to compare samples accurately.
Dilution Factor
The dilution factor is the ratio of the final volume to the original volume of a solution that has been diluted. In experiments, knowing the dilution factor is essential for calculating concentrations after a sample has been modified, like when a standard is added.
For instance, if a 50.00 mL solution becomes 51.00 mL after adding a standard, the dilution factor would be calculated as:
\[ \text{Dilution factor} = \frac{51.00}{50.00} = 1.02 \]
This factor helps adjust the measured concentration of substances in the solution accounting for their dilution.
  • Correctly determining the dilution factor prevents errors in estimating the concentration of analytes.
  • This factor ties directly into calculations involving standard additions and analyte concentration assessments.
Nitrite Concentration
Nitrite concentration represents the amount of nitrite ions present in a solution, typically expressed in molarity (moles per liter). It is critical in environmental chemistry due to its role in processes like soil and water analysis. Elevated levels of nitrite can indicate pollution or changes in decomposition processes.
In the problem above, determining the concentration of nitrite in a sample involves using standard additions and the relationship between absorbance changes. The solution demonstrates how to adjust for dilutions with added standards, ultimately solving:
\[ C_i \approx \frac{5.88 \times 10^{-6}}{0.530 - 0.300 \times 1.02} \]
In this context, \( C_i \) or the concentration of nitrite, indicates how direct and indirect methods converge for accurate measurements.
  • Accurate determination of nitrite concentration helps assess environmental and industrial impacts.
  • Combining chemical analysis with standard combinations enhances the reliability of measured concentrations.

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Most popular questions from this chapter

The seller of a mining claim took a random ore sample that weighed approximately \(5 \mathrm{lb}\) and had an average particle diameter of \(5.0 \mathrm{~mm}\). Inspection revealed that about \(1 \%\) of the sample was argentite (see Problem 8-9), and the remainder had a density of about \(2.6 \mathrm{~g} / \mathrm{cm}^{3}\) and contained no silver. The prospective buyer insisted on knowing the silver content of the claim with a relative error no greater than \(5 \%\). Determine whether the seller provided a sufficiently large sample to permit such an evaluation. Give the details of your analysis.

The following are relative peak areas for chromatograms of standard solutions of methyl vinyl ketone (MVK). $$ \begin{array}{cc} \text { MVK concentration, mmol/L } & \text { Relative Peak Area } \\ \hline 0.500 & 3.76 \\ 1.50 & 9.16 \\ 2.50 & 15.03 \\ 3.50 & 20.42 \\ 4.50 & 25.33 \\ 5.50 & 31.97 \\ \hline \end{array} $$ (a) Determine the coefficients of the best fit line using the least-squares method. (b) Construct an ANOVA table. (c) Plot the least-squares line as well as the experimental points. (d) A sample containing MVK yielded relative peak area of 12.9. Calculate the concentration of \(\mathrm{MVK}\) in the solution. (e) Assume that the result in (d) represents a single measurement as well as the mean of four measurements. Calculate the respective absolute and relative standard deviations for the two cases. (f) Repeat the calculations in (d) and (e) for a sample that gave a peak area of 21.3.

The mishandling of a shipping container loaded with 750 cases of wine caused some of the bottles to break. An insurance adjuster proposed to settle the claim at \(20.8 \%\) of the value of the shipment, based on a random 250-bottle sample in which 52 were cracked or broken. Calculate (a) the relative standard deviation of the adjuster's evaluation. (b) the absolute standard deviation for the 750 cases (12 bottles per case). (c) the \(90 \%\) confidence interval for the total number of bottles. (d) the size of a random sampling needed for a relative standard deviation of \(5.0 \%\), assuming a breakage rate of about \(21 \%\).

Approximately 15\% of the particles in a shipment of silver-bearing ore are judged to be argentite, \(\mathrm{Ag}_{2} \mathrm{~S}\) ( \(\left.d=7.3 \mathrm{~g} \mathrm{~cm}{ }^{3}, 87 \% \mathrm{Ag}\right)\); the remainder are siliceous \(\left(d=2.6 \mathrm{~g} \mathrm{~cm}^{-3}\right)\) and contain essentially no silver. (a) Calculate the number of particles that should be taken for the gross sample if the relative standard deviation due to sampling is to be \(2 \%\) or less. (b) Estimate the mass of the gross sample, assuming that the particles are spherical and have an average diameter of \(3.5 \mathrm{~mm}\). (c) The sample taken for analysis is to weigh \(0.500 \mathrm{~g}\) and contain the same number of particles as the gross sample. To what diameter must the particles be ground to satisfy these criteria?

Water can be determined in solid samples by infrared spectroscopy. The water content of calcium sulfate hydrates is to be measured using calcium carbonate as an internal standard to compensate for some systematic errors in the procedure. A series of standard solutions containing calcium sulfate dihydrate and a constant known amount of the internal standard is prepared. The solution of unknown water content is also prepared with the same amount of internal standard. The absorbance of the dihydrate is measured at one wavelength \(\left(A_{\text {sample }}\right)\) along with that of the internal standard at another wavelength \(\left(A_{\text {std }}\right)\). The following results were obtained. $$ \begin{array}{ccc} A_{\text {sample }} & A_{\text {std }} & \% \text { water } \\ \hline 0.15 & 0.75 & 4.0 \\ 0.23 & 0.60 & 8.0 \\ 0.19 & 0.31 & 12.0 \\ 0.57 & 0.70 & 16.0 \\ 0.43 & 0.45 & 20.0 \\ 0.37 & 0.47 & \text { Unknown } \\ \hline \end{array} $$ (a) Plot the absorbance of the sample \(\left(A_{\text {sample }}\right)\) vs. the \(\%\) water and determine whether the plot is linear from the regression statistics. (b) Plot the ratio \(A_{\text {sample }} / A_{\text {std }}\) vs. \% water, and comment on whether using the internal standard improves the linearity from that in part (a). If it improves the linearity, why? (c) Calculate \(\%\) water in the unknown using the internal standard data.

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