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A sample may contain any or all of the following ions: \(\mathrm{Hg}_{2}{ }^{2+}\), \(\mathrm{Ba}^{2+}\), and \(\mathrm{Mn}^{2+}\). a. No precipitate formed when an aqueous solution of \(\mathrm{NaCl}\) was added to the sample solution. b. No precipitate formed when an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) was added to the sample solution. c. A precipitate formed when the sample solution was made basic with \(\mathrm{NaOH}\). Which ion or ions are present in the sample solution?

Short Answer

Expert verified
The ion present in the sample solution is Ba虏鈦.

Step by step solution

01

Analyze the results of adding NaCl

When NaCl is added to the solution and no precipitate forms, it means none of the ions in the sample solution form an insoluble compound with Cl鈦 ions. The potential compounds with Cl鈦 ions are: 1. \(Hg_2Cl_2\) 2. \(BaCl_2\) 3. \(MnCl_2\) Since no precipitate was formed, we know Ba虏鈦 is present as BaCl鈧 is soluble. If Hg鈧偮测伜 was present, \(Hg_2Cl_2\) would have formed which is insoluble. Therefore, we can conclude Hg鈧偮测伜 is not present in the sample solution.
02

Analyze the results of adding Na鈧係O鈧

After adding Na鈧係O鈧 to the sample solution, no precipitate forms as well. This means none of the ions in the sample form an insoluble compound with SO鈧劼测伝 ions. The potential compounds with SO鈧劼测伝 ions are: 1. \(Hg_2SO_4\) 2. \(BaSO_4\) 3. \(MnSO_4\) Since no precipitate was formed and we know that Ba虏鈦 is present, this means Mn虏鈦 is not present, as it would have formed an insoluble MnSO鈧. Now we are left with Ba虏鈦 as the only ion present from the initial ions.
03

Analyze the results of adding NaOH

When NaOH is added to the sample solution, a precipitate forms. This indicates that one or more of the ions in the sample solution forms an insoluble compound with OH鈦 ions. The potential compounds with OH鈦 ions are: 1. \(Hg_2(OH)_2\) 2. \(Ba(OH)_2\) 3. \(Mn(OH)_2\) Since Ba虏鈦 is present in the sample solution, the precipitate formed is likely due to the formation of Ba(OH)鈧, which is slightly soluble in water.
04

Conclusion

Based on the analysis of the given tests, we can conclude that the ion present in the sample solution is Ba虏鈦.

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Key Concepts

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

Solubility Rules
Solubility rules help us predict whether a compound will dissolve in water, or form a solid precipitate. These rules are like handy guidelines based on experimental observations.

For instance, most chloride (\( \text{Cl}^- \)) salts are soluble, meaning they dissolve in water easily. However, there are exceptions like \( \text{Hg}_2\text{Cl}_2 \), which is insoluble.

Sulfates (\( \text{SO}_4^{2-} \)) generally dissolve in water, but some, like \( \text{BaSO}_4 \), do not. This is why in our exercise, when no precipitate forms with \( \text{Na}_2\text{SO}_4 \) or \( \text{NaCl} \), we know some ions like \( \text{Ba}^{2+} \) may still be present.

Understanding solubility rules can save time by helping identify which elements or ions stay dissolved, guiding the outcome of reactions more predictably.
Chemical Precipitation
Chemical precipitation happens when two soluble salts react in a solution to form an insoluble compound, known as a precipitate. This solid compound then separates from the solution.

In our exercise scenario, a precipitate did not form when \( \text{NaCl} \) was added. This tells us that none of the ions reacted to form an insoluble chloride.

When \( \text{NaOH} \) was added, a precipitate formed, indicating a reaction occurred to form an insoluble hydroxide. This narrows down the possibilities of which ions are present because only specific ions will react this way in such conditions.

Chemical precipitation is a crucial process for identifying ions in a solution. It helps distinguish between different compounds based on their ability or inability to form solids when mixed.
Cation Identification
Cations are positively charged ions, such as \( \text{Ba}^{2+} \). Identifying them involves observing how they react with various anions.

For example, in the exercise, adding \( \text{Na}_2\text{SO}_4 \) and observing no precipitation suggests that certain cations like \( \text{Mn}^{2+} \), which would normally form precipitates, are absent.

The ability to identify cations through such reactions is vital for understanding the chemical makeup of a solution. It helps chemists decide which elements are present and predict the behavior of the solution during reactions.
Aqueous Solutions
An aqueous solution is simply a solution where water is the solvent. In chemistry, it is a common medium for reactions.

Within aqueous solutions, ions can move freely, allowing them to interact and react with each other. This mobility plays a key role in processes like precipitation.

In the exercise, the sample solution was aqueous. This setup is perfect for testing how different additives like \( \text{NaCl} \) or \( \text{NaOH} \) would affect the existing ions, helping deduce their presence based on whether a precipitate forms.

Thus, understanding how ions behave in aqueous solutions enables us to explore various chemical reactions and phenomena with precision.

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Most popular questions from this chapter

A \(1.42-\mathrm{g}\) sample of a pure compound, with formula \(\mathrm{M}_{2} \mathrm{SO}_{4}\), was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh \(1.36 \mathrm{~g}\). Determine the atomic mass of \(\mathrm{M}\), and identify \(\mathrm{M}\).

A \(10.00-\mathrm{mL}\) sample of sulfuric acid from an automobile battery requires \(35.08 \mathrm{~mL}\) of \(2.12 M\) sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid? Sulfuric acid contains two acidic hydrogens.

The unknown acid \(\mathrm{H}_{2} \mathrm{X}\) can be neutralized completely by \(\mathrm{OH}^{-}\) according to the following (unbalanced) equation: $$ \mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{X}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ The ion formed as a product, \(\mathrm{X}^{2-}\), was shown to have 36 total electrons. What is element \(\mathrm{X}\) ? Propose a name for \(\mathrm{H}_{2} \mathrm{X}\). To completely neutralize a sample of \(\mathrm{H}_{2} \mathrm{X}, 35.6 \mathrm{~mL}\) of \(0.175 \mathrm{M}\) \(\mathrm{OH}^{-}\) solution was required. What was the mass of the \(\mathrm{H}_{2} \mathrm{X}\) sample used?

Consider the reaction between oxygen \(\left(\mathrm{O}_{2}\right)\) gas and magnesium metal to form magnesium oxide. Using oxidation states, how many electrons would each oxygen atom gain, and how many electrons would each magnesium atom lose? How many magnesium atoms are needed to react with one oxygen molecule? Write a balanced equation for this reaction.

What mass of solid aluminum hydroxide can be produced when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) is added to \(200.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KOH} ?\)

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