Qualitative analysis of water supply
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Qualitative analysis of water supply
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[ID] => 558840
[post_author] => 12815
[post_date] => 2025-01-14 05:16:41
[post_date_gmt] => 2025-01-14 10:16:41
[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
A variety of chronic disorders may develop from the accumulation of non-essential trace elements. Humans have been exposed to such elements only recently due to extensive mining. Because humans did not encounter these elements during the course of evolution, they have not developed homeostatic mechanisms to accommodate or eliminate these elements from their bodies. For example, cadmium has been linked to the development to itai-itai disease which has symptoms such as renal failure, weak bones, and joint and spine pain. Studies indicate that elevated cadmium levels, due to inefficient removal from the body by the kidneys ,leads to mitochondrial damage and eventually renal failure.
Patients all originating from one manufacturing town and its environs have been reporting at various hospitals with wide-ranging symptoms, including headache, vomiting, muscle pain, memory loss, constipation, and high blood pressure. To determine what may be responsible, an investigator can use qualitative analysis to identify the ions present in a sample by systematically conducting different chemical tests. The most common test is selective precipitation, which separates cations into groups based on their solubility under specific conditions. The reagents are added sequentially to the same solution throughout testing. After separation, the individual ions can be identified by confirmatory tests, which can include observing their color, odor, appearance, and solubility.
A systematic qualitative analysis is conducted on a water sample at a local lab, and the results are in the table below:
Table 1. Qualitative Analysis of Water Sample by Precipitating Cation Groups with Reagents
Many of the tests for identifying cations have been superseded by techniques like mass spectrometry and absorption spectrophotometry due to the range of chemicals the instruments can detect. However, traditional qualitative analysis can be still useful in field work or other situations where access is not expedient. A neighboring facility conducted a spectroscopic analysis of a sample with another suspected pollutant, and the results are below:
Figure 1. Absorption Spectrogram of Water Sample from Adjacent Town
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[question] => The addition of nitric acid to the water sample produces a colorless and odorless gas. What is the likely reaction when nitric acid is added to the sample?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
The question stem states that a colorless, odorless gas is produced.
Chlorine gas (Cl2) has the odor of bleach, and sulfur dioxide (SO2) has the rotten odor of sulfur compounds. Both are distinctive.
Carbon dioxide (CO2) and hydrogen gas (H2) are both odorless, as well as most of the nitrogen oxides including nitric oxide (NO). Nitrogen dioxide (NO2) has only a faint odor as the major component of air pollution.
Color is imparted to an atom whenever there are electronic transitions, absorption and emission, in the visible wavelengths. Cl2 has a yellow-green color, while NO2 has a reddish-brown hue, while CO2, H2, and NO are colorless.
When nitric acid is added to the water sample, carbonate 
becomes carbonic acid (H2CO3), which quickly decomposes to produce CO2 and water. Hence CO2 is the colorless and odorless gas in question.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. 
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[1] => Array
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[each_answer] => B. 
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[2] => Array
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[each_answer] => C. 
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[3] => Array
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[each_answer] => D. 
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[1] => Array
(
[quiz_unique_key] => 3873426850
[question] => Which of the following can be best detected by UV-Vis absorption spectroscopy?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
Hint: UV-Vis absorption spectroscopy is only useful for detecting electronic transitions corresponding to 220-400 nm in the UV range and 400-800 nm in the visible range.
Transition metal ions are one of the species detected because of their
d electrons. Transitional metals in their elemental states will not be detected.
A complex ion often contains a transition metal ion as its center. Those that do not include a transition metal ion will not be detected.
Compounds with double bonds are another species detected because of their electrons. While all double bonds absorb UV light, only species with conjugated systems will fall into the range of detection for UV-Vis spectroscopy.
The bicyclic compound cis-decalin only has sigma bonds and would not be detected.
Anthracene has three conjugated benzene rings for its structure and would be detected by UV-Vis spectroscopy. In fact, anthracene exhibits blue fluorescence under UV light.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. 
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[1] => Array
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[each_answer] => B. 
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[2] => Array
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[each_answer] => C. 
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[3] => Array
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[each_answer] => D. Cr (s)
)
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => The Category 6 cations consists of the Group IA elements and ammonium, which cannot be further partitioned by selective precipitation. Based on Table 1, what is the identity of the Category 5 precipitate?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
The reagents are added sequentially to the same solution. Category 1 or insoluble chlorides precipitate out first.
The solubility rule states that halides (Cl–, Br–, I–) are soluble except for those of mercury 
, silver (Ag+), and lead (Pb2+). Pb2+ would have precipitated out as PbCl in Category 1 before reaching Category 5.
There are no Category 2 or acid-insoluble sulfides. Category 3 or insoluble hydroxides will precipitate out next. The solubility rule states that hydroxides are insoluble except for those of the Group IA cations, calcium (Ca2+), barium (Ba2+), and strontium (Sr2+).
Zinc (Zn2+) and chromium (Cr3+) would have also precipitated out as Zn(OH)2 and Cr(OH)3 in Category 3 before reaching Category 5.
Category 4 or base-insoluble sulfides will precipitate out next. The solubility rule states that sulfides are insoluble except for those of the Group IA and IIA cations and ammonium 
Since strontium (Sr2+) is the only ion remaining in solution, Sr3(PO4)2 must be the precipitate.
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[answers] => Array
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[0] => Array
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[each_answer] => A. CrPO4
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[1] => Array
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[each_answer] => B. Pb3(PO4)2
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[2] => Array
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[each_answer] => C. Sr3(PO4)2
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[each_answer] => D. Zn3(PO4)2
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[3] => Array
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[quiz_unique_key] => 2261298308
[question] => Which of the following statements correctly explains the role of the reagents in the qualitative analysis?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
For Category 2, hydrogen sulfide (H2S) is a very weak acid, and essentially all of it will remain protonated. For any weak acid, the following equilibrium, H2S(g) + H2O(I) → 2H3O+(aq) + S2-(aq), exists.
Nitric acid (HNO3) ensures that the hydronium ion levels are high, which shifts the equilibrium to the left. The goal is to keep sulfide (S2-) concentrations as low as possible to precipitate the most insoluble or acid-insoluble sulfides, which include copper (Cu2+), cadmium (Cd2+), and tin (Sn2+).
For Category 3, the buffer solution creates the following equilibrium, 
Ammonia hydrolyzes weakly to produce hydroxide, and the presence of ammonium suppresses that even further. The goal is to keep the hydroxide concentration as low as possible to precipitate only the most insoluble hydroxides, which include iron (Fe3+), chromium (Cr3+), and aluminum (Al3+).
For Category 5, concentrated ammmonia creates very basic conditions. Phosphate 
is a strong base and would easily become hydrogen phosphate 
so ammonia ensures that the phosphate levels are high enough to precipitate the insoluble phosphates, which include magnesium (Mg2+), calcium (Ca2+), barium (Ba2+), and strontium (Sr2+).
For Category 4, hydrogen sulfide is used again, and the same equilibrium, H2S(g) + H2O(I) → 2H3O+(aq) + S2-(aq), exists. For this round, ammonia reacts with the hydronium ion to drive the equilibrium to the right.
The goal is to create a greater concentration of sulfide than is possible under acidic or neutral conditions to precipitate the base-insoluble sulfides, which include cobalt (Co2+), manganese (Mn2+), nickel (Ni2+), and zinc (Zn2+).
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[each_answer] => A. H2S is placed in HNO3 to increase the S2- ion available to precipitate Category 2 cations.
)
[1] => Array
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[each_answer] => B. The buffer solution increases the OH– ion available to precipitate the Category 3 cations.
)
[2] => Array
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[each_answer] => C. Concentrated NH3 decreases the ion available to precipitate the Category 5 cations.
)
[3] => Array
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[each_answer] => D. The buffer solution increases the S2- ion available to precipitate the Category 4 cations when NH3 reacts with H3O+.
)
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[4] => Array
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[quiz_unique_key] => 574431310
[question] => Based on the absorption spectrum in Figure 1, what was the color of the original sample?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
Take note of the labels on the two axes. On the y axis, there is absorption, and a higher peak means a greater absorption.
The complementary color is observed whenever a color is absorbed. On the color wheel, the complementary color is the one diametrically opposite, so if a sample absorbs blue, then orange is observed.
The x-axis is in units of wavenumber, which is the reciprocal of wavelength. Use 400 nm for the violet end of the spectrum. 400 nm is the same as 400 x 10-9 m and can be converted to 
Use 700 nm for the red end of the spectrum. 700 nm is the same as 700 x 10-9 m and can be converted to 
Using those two calculated values, it can be determined that absorption occurs in the red end of the spectrum.
Since the sample is absorbing in the red end of the spectrum, the sample should appear green. Green does not appear in any of the answer choices, but blue-green is the closest answer choice. In fact, this is the spectrum of a complex ion containing copper.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. Violet
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[each_answer] => B. Orange
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[each_answer] => C. Red
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[each_answer] => D. Blue-Green
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