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A solution containing an unknown number of metal ions is treated with dilute \(\mathrm{HCl}\); no precipitate forms. The \(\mathrm{pH}\) is adjusted to about 1 , and \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. Again, no precipitate forms. The pH of the solution is then adjusted to about 8 . Again, \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). No precipitate forms.

Short Answer

Expert verified
The tests performed on the unknown metal ion solution indicate that it contains Group 3 cations (Pb虏鈦, Fe虏鈦, Fe鲁鈦, Al鲁鈦, Cr鲁鈦, and Zn虏鈦) as precipitates formed at pH 8 when H鈧係 was bubbled through. No other group cations (Groups 1, 2, 4, and 5) were detected as they did not form precipitates during the tests.

Step by step solution

01

HCl test

The unknown solution is treated with dilute HCl, and no precipitate forms. This implies that the metal ions are not part of Group 1 cations (Ag鈦, Pb虏鈦, and Hg鈧偮测伜), which would form chlorides in the presence of HCl.
02

H鈧係 test at pH 1

The pH of the solution is adjusted to about 1, and H鈧係 is bubbled through. Again, no precipitate forms. This means that Group 2 cations (Bi鲁鈦, Cu虏鈦, As鲁鈦, Sb鲁鈦, Sn虏鈦, and Hg虏鈦) are also not present since they would form sulfides at pH 1.
03

H鈧係 test at pH 8

The pH of the solution is adjusted to about 8, then H鈧係 is bubbled through the solution again. This time, a precipitate forms. This indicates that Group 3 cations (Pb虏鈦, Fe虏鈦, Fe鲁鈦, Al鲁鈦, Cr鲁鈦, and Zn虏鈦), which form sulfides at pH 8, are present in the solution.
04

Filtrate test with (NH鈧)鈧侶PO鈧

The filtrate from the previous step is treated with (NH鈧)鈧侶PO鈧, and no precipitate forms. This indicates that Group 4 (Ca虏鈦, Ba虏鈦, Sr虏鈦, and Ra虏鈦) and Group 5 cations (Mg虏鈦 and NH鈧勨伜) are not present, as they form phosphate precipitates with (NH鈧)鈧侶PO鈧.
05

Conclusion

Based on these test results, the unknown metal ions in the solution belong to Group 3 cations (Pb虏鈦, Fe虏鈦, Fe鲁鈦, Al鲁鈦, Cr鲁鈦, and Zn虏鈦), which form sulfides at pH 8. Other groups of cations were not present, as they would have formed precipitates in the tests as described.

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Key Concepts

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

Qualitative Analysis in Chemistry
Qualitative analysis in chemistry refers to the process of determining the presence or absence of certain ions or compounds in a mixture, without precisely measuring the quantity. It's a bit like a detective solving a mystery, looking for clues (in this case, chemical reactions) that can help identify the substances involved.

Simple tests, such as adding a specific reagent to observe changes, like color change or precipitate formation, are part of qualitative analysis. In the exercise, different reagents were added under specific conditions to deduce which metal ions were present. This approach is systematic and eliminates possibilities one step at a time, moving from known reagent reactions to infer what must be in the solution. It's essential in fields like environmental science, medicine, and forensic science, and provides a solid foundation for more quantitative analyses.
Precipitation Reactions
Precipitation reactions are a staple in the chemist's toolkit. They occur when two aqueous solutions react to form an insoluble solid, known as a precipitate. This reaction is not just a party trick - it's pivotal in qualitative analysis for identifying various ions.

By adjusting conditions like pH, as seen in the exercise, chemists can manipulate which compounds will precipitate. For example, when the pH was raised to about 8 and H2S was introduced, it created an environment where certain metal sulfides could no longer stay dissolved, thereby crashing out as solids. These reactions help narrow down the list of possible ions present, separating them into visually distinguishable groups.
Properties of Metal Ions
Each metal ion has a unique set of properties that chemists can exploit to identify them. These properties include charge, ionic radius, and the ability to form precipitates or colored complexes.

For instance, in the exercise, the fact that no precipitate formed when (NH4)2HPO4 was added suggests that certain metals, with their particular charges and tendencies to form precipitates, were absent. Ions tend to show different behaviors under varying pH levels, which is why pH adjustments are a common tactic for coaxing out precipitates from certain metal ions while keeping others in solution.
Chemical Testing Procedures
Chemical testing procedures are the bread and butter of experimental chemistry, forming the base for analysis and identification of substances. A methodical approach, like the one followed in our textbook exercise, is crucial for accurate identification.

Introducing reagents and adjusting the conditions systematically allows for targeted tests that can sequentially eliminate or confirm the presence of specific ions. It is important to note that the correct interpretation of these test results relies on a solid understanding of the underlying chemical principles, such as solubility rules, reaction mechanisms, and complex-ion formation. Mastery of chemical tests leads to better insights in research, quality control in industry, and solving real-life problems where the composition of a sample is unknown.

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Most popular questions from this chapter

Predict whether the equivalence point of each of the following titrations is below, above, or at \(\mathrm{pH} 7\) : (a) formic acid titrated with \(\mathrm{NaOH}\), (b) calcium hydroxide titrated with perchloric acid, (c) pyridine titrated with nitric acid.

A \(35.0\) -mL sample of \(0.150 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is titrated with \(0.150 \mathrm{M} \mathrm{NaOH}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of base have been added: (a) \(0 \mathrm{~mL}\), (b) \(17.5 \mathrm{~mL}\), (c) \(34.5 \mathrm{~mL}\), (d) \(35.0 \mathrm{~mL}\), (e) \(35.5 \mathrm{~mL}\), (f) \(50.0 \mathrm{~mL}\).

(a) What is the common-ion effect? (b) Give an example of a salt that can decrease the ionization of \(\mathrm{HNO}_{2}\) in solution.

A hypothetical weak acid, HA, was combined with \(\mathrm{NaOH}\) in the following proportions: \(0.20 \mathrm{~mol}\) of \(\mathrm{HA}\), \(0.080 \mathrm{~mol}\) of \(\mathrm{NaOH}\). The mixture was diluted to a total volume of \(1.0 \mathrm{~L}\), and the pH measured. (a) If \(\mathrm{pH}=4.80\), what is the \(\mathrm{p} K_{a}\) of the acid? (b) How many additional moles of \(\mathrm{NaOH}\) should be added to the solution to increase the \(\mathrm{pH}\) to \(5.00 ?\)

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make \(1.00\) L of solution. Buffer A is prepared using \(1.00 \mathrm{~mol}\) each of formic acid and sodium formate. Buffer \(B\) is prepared by using \(0.010\) mol of each. (a) Calculate the pH of each buffer, and explain why they are equal. (b) Which buffer will have the greater buffer capacity? Explain. (c) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 M \mathrm{HCl}\). (d) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(10 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (e) Discuss your answers for parts (c) and (d) in light of your response to part (b).

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