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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
Based on the series of tests, the metal ions in the unknown solution most likely belong to Group 3, which includes Al鲁+, As鲁+/5+, Cr鲁+, Co虏+, Ni虏+, Mn虏+, Fe鲁+, and Zn虏+. This is determined through the analysis of their responses to HCl, H2S at pH 1 and pH 8, and (NH4)2HPO4 treatments, where the only precipitate was formed after adjusting the pH to 8 and treating it with H2S.

Step by step solution

01

Analyze the response to dilute HCl

As no precipitate is formed upon the addition of dilute HCl, this rules out the metal ions from Group 1 (Ag+, Hg2虏+, Pb虏+) as they typically form precipitates when in contact with chloride ions.
02

Analyze the response to H2S at pH 1

At pH 1 and after bubbling H2S, no precipitate forms. This indicates that the solution does not contain metal ions from Group 2 (Bi鲁+, Cu虏+, Hg虏+, Cd虏+, Sb鲁+/5+) as they usually form precipitates with sulfide ions (S虏鈦) in acidic conditions.
03

Analyze the response to H2S at pH8

The formation of a precipitate after adjusting the pH to 8 and bubbling H2S suggests that the metal ions in the unknown solution are from Group 3 (Al鲁+, As鲁+/5+, Cr鲁+, Co虏+, Ni虏+, Mn虏+, Fe鲁+, Zn虏+), as they form precipitates with sulfide ions at higher pH levels.
04

Analyze the response to (NH4)2HPO4

No precipitate was formed after treating the filtrate from the last step with (NH4)2HPO4, which means the unknown solution does not contain metal ions from Group 4 (Ba虏+, Sr虏+, Ca虏+) nor Group 5 (Mg虏+, K+, Na+), as they typically form precipitates with phosphate ions (PO4鲁鈦).
05

Conclusion

Based on the analysis of the results from the series of tests, the metal ions in the unknown solution are most likely to belong to Group 3 (Al鲁+, As鲁+/5+, Cr鲁+, Co虏+, Ni虏+, Mn虏+, Fe鲁+, Zn虏+). This is because the solution had a specific response to each test, with the only precipitate forming after adjusting the pH to 8 and treating it with H2S.

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Key Concepts

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

Precipitation Reactions
Precipitation reactions are a fundamental process in qualitative analysis, used to determine the presence of certain ions in a solution. In essence, when two soluble salts react, their ions may form a new compound that is insoluble in water, leading to the formation of a solid precipitate. In the exercise, when no precipitate formed after the addition of dilute HCl, we infer that ions forming precipitates with chlorides at low pH (Group 1) are absent.

Subsequently, the lack of precipitate formation after introducing H2S at a low pH indicates that ions reacting with sulfide ions under acidic conditions (Group 2) are not present either. However, a precipitate does form at a raised pH level of 8 after H2S is added, pointing toward the presence of metal ions from Group 3. These ions are more likely to precipitate with sulfide ions in less acidic to basic conditions, showcasing how a systematic process of elimination can lead to the identification of certain metal ions.
pH Level Adjustments
Adjusting the pH level of a solution is a critical step in qualitative analysis, particularly when testing for metal ions with sulfide. Metal ions vary in their tendencies to form precipitates with sulfide ions at different pH levels. In the given exercise, no precipitate formed at a pH of 1 when H2S was bubbled through the solution. This implies that a higher pH is necessary for certain sulfide precipitates to form.

When the pH was increased to approximately 8, the conditions became favorable for the metals of Group 3 to react with the sulfide ions to form a precipitate. This change in solubility is heavily influenced by the pH, and analysts exploit this by incrementally adjusting it to narrow down the possible ions in the solution. It's essential for the student to understand how pH affects the chemistry of the solution and thus the outcomes of the analysis.
Sulfide Ion Solubility
The solubility of sulfide ions (S2-) plays a crucial role in the formation of metal sulfides and the resultant analysis of metal ions. The solubility of sulfides is highly pH-dependent, with many metal sulfides being much less soluble in alkaline conditions as opposed to acidic ones.

In this exercise, when the solution's pH was elevated to around 8, a precipitate formed upon the introduction of H2S, indicating the presence of Group 3 metal ions which are typically more reactive to sulfide ions under less acidic conditions. It's an important note for students to understand that the presence or absence of a precipitate upon bubbling H2S can tell us a great deal about which metal ions are in the solution based on their solubility properties at different pH levels. With no precipitate formed in earlier acidic conditions and one forming at a higher pH, we observe the relative insolubility of certain metal sulfides in neutral to basic environments, which is a vital part of concluding the metal composition of an unknown solution.

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Most popular questions from this chapter

A buffer contains a weak acid, \(\mathrm{HX}\), and its conjugate base. The weak acid has a \(\mathrm{pK}_{a}\) of \(4.5\), and the buffer solution has a pH of 4.3. Without doing a calculation, predict whether \([\mathrm{HX}]=\left[\mathrm{X}^{-}\right],[\mathrm{HX}]>\left[\mathrm{X}^{-}\right]\), or \([\mathrm{HX}]<\left[\mathrm{X}^{-}\right]\) Explain. [Section 17.2]

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid \(\mathrm{HA}\) and a base \(B\) are mixed. The \(p H\) of the resulting solution is 9.2. (a) Write the equilibrium equation and equilibriumconstant expression for the reaction between \(\mathrm{HA}\) and \(\mathrm{B}\). (b) If \(K_{a}\) for HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and B? (c) What is the value of \(K_{b}\) for \(B\) ?

The acid-base indicator bromcresol green is a weak acid. The yellow acid and blue base forms of the indicator are present in equal concentrations in a solution when the \(\mathrm{pH}\) is \(4.68\). What is the \(\mathrm{p} K_{a}\) for bromcresol green?

(a) Precipitation of the group 4 cations (Figure 17.22) requires a basic medium. Why is this so? (b) What is the most significant difference between the sulfides precipitated in group 2 and those precipitated in group \(3 ?\) (c) Suggest a procedure that would serve to redissolve the group 3 cations following their precipitation.

How many microliters of \(1.000 \mathrm{M} \mathrm{NaOH}\) solution must be added to \(25.00 \mathrm{~mL}\) of a \(0.1000 \mathrm{M}\) solution of lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to produce \(\mathrm{a}\) buffer with \(\mathrm{pH}=3.75 ?\)

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