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Which solutions would you use to precipitate \(\mathrm{Mg}^{2+}\) from an aqueous solution of \(\mathrm{MgCl}_{2} ?\) Explain your choice. (a) \(\mathrm{KNO}_{3}(\mathrm{aq}) ;\) (b) \(\mathrm{NH}_{3}(\mathrm{aq}) ;\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq});\) (d) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(\mathrm{aq})\).

Short Answer

Expert verified
Among the given options, only \(\mathrm{NH}_{3}\) would precipitate \(\mathrm{Mg}^{2+}\) from the aqueous solution of \(\mathrm{MgCl}_{2}\)

Step by step solution

01

Analyze solubilities

Certain salts have certain solubility rules. Magnesium salts are generally soluble, so a solution that responds with a magnesium ion to make an insoluble compound is required.
02

Analyze reactant (a): \(\mathrm{KNO}_{3}\)

Firstly look at \(\mathrm{KNO}_{3}\). The nitrate (\(\mathrm{NO}_{3}^{-}\)) ion is a part of a group of common ions that do not react to form precipitates. Therefore, \(\mathrm{KNO}_{3}\) would not precipitate \(\mathrm{Mg}^{2+}\).
03

Analyze reactant (b): \(\mathrm{NH}_{3}\)

Next, consider \(\mathrm{NH}_{3}\). Ammonia in water acts as a base and forms ammonium ion (\(\mathrm{NH}_{4}^{+}\)) and hydroxide ion (\(\mathrm{OH}^{-}\)). The hydroxide ion can react with \(\mathrm{Mg}^{2+}\) to form \(\mathrm{Mg(OH)}_{2}\), which is insoluble in water. Therefore, \(\mathrm{NH}_{3}\) would precipitate \(\mathrm{Mg}^{2+}\).
04

Analyze reactant (c): \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

Next is \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The sulfate (\(\mathrm{SO}_{4}^{2-}\)) ion will react with \(\mathrm{Mg}^{2+}\) to form \(\mathrm{MgSO}_{4}\) which is soluble in water according to solubility rules, hence will not precipitate out of solution. Therefore, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) would not precipitate \(\mathrm{Mg}^{2+}\).
05

Analyze reactant (d): \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (acetic acid) is a weak acid and does not have a significant effect on magnesium ion in water, as it does not provide any ion that can form a precipitate with \(\mathrm{Mg}^{2+}\). Hence \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) would not precipitate \(\mathrm{Mg}^{2+}\).

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Key Concepts

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

Magnesium Ion Precipitation
To successfully precipitate magnesium ions (\(\mathrm{Mg}^{2+}\)) from a solution, a reactant is needed that combines with \(\mathrm{Mg}^{2+}\) to form an insoluble compound. When working with chemical precipitation, it is crucial to identify interactions that result in a solid form that settles out of the solution.
In the given exercise, ammonia (\(\mathrm{NH}_{3}\)) is one reactant capable of precipitating \(\mathrm{Mg}^{2+}\). When ammonia dissolves in water, it can produce hydroxide ions (\(\mathrm{OH}^{-}\)). These hydroxide ions interact with \(\mathrm{Mg}^{2+}\) to form magnesium hydroxide (\(\mathrm{Mg(OH)}_{2}\)), which is insoluble in water. This means magnesium hydroxide will form a solid precipitate, effectively removing \(\mathrm{Mg}^{2+}\) ions from the solution.
Understanding this process can help in various practical applications, including the treatment of wastewater or in carrying out laboratory experiments where magnesium's presence needs to be minimized.
Solubility Rules in Chemistry
In chemistry, solubility rules are essential for predicting whether a compound will dissolve in water or form a precipitate. These rules are based on the nature of the ions involved and their ability to interact in a way that either allows them to remain in solution or forces them to come out as a solid.
Some key solubility rules include:
  • Nitrate ions (\(\mathrm{NO}_{3}^{-}\)) always form soluble compounds, meaning they never precipitate with any metal ions, including \(\mathrm{Mg}^{2+}\).
  • Sulfate ions (\(\mathrm{SO}_{4}^{2-}\)) generally form soluble salts, but they can form insoluble compounds with certain cations, although \(\mathrm{MgSO}_{4}\) remains soluble in water.
These rules help predict the behavior of ionic compounds in aqueous solutions. By applying these solubility principles, you can deduce which sequences of ions will result in precipitation and which will not, streamlining experiments and solving chemical problems efficiently.
Insoluble Compounds
Insoluble compounds are those that do not dissolve significantly in a solvent, such as water, and thus remain as a solid in the mixture. The precipitate formation stems from the inability of a compound to be evenly distributed and interact with the solvent molecules.
One common insoluble compound is magnesium hydroxide (\(\mathrm{Mg(OH)}_{2}\)), which forms when hydroxide ions from a base, such as ammonia, react with magnesium ions. Despite water being a universal solvent, certain ionic bonds are too strong to be overcome by water molecules, resulting in precipitation.
In precipitation reactions, knowing the insolubility of particular compounds allows chemists to selectively remove or isolate ions from solutions. This is a critical insight in contexts ranging from everyday chemical processes to more advanced laboratory techniques used in research and industry.

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Most popular questions from this chapter

Household ammonia, used as a window cleaner and for other cleaning purposes, is \(\mathrm{NH}_{3}(\mathrm{aq}) .\) The \(\mathrm{NH}_{3}\) present in a 5.00 \(\mathrm{mL}\) sample is neutralized by 28.72 \(\mathrm{mL}\) of \(1.021 \mathrm{M} \mathrm{HCl}\). The net ionic equation for the neutralization is \(\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4}^{+}(\mathrm{aq})\) What is the molarity of \(\mathrm{NH}_{3}\) in the sample?

A \(0.4324 \mathrm{g}\) sample of a potassium hydroxide-lithium hydroxide mixture requires \(28.28 \mathrm{mL}\) of \(0.3520 \mathrm{M} \mathrm{HCl}\) for its titration to the equivalence point. What is the mass percent lithium hydroxide in this mixture?

When concentrated \(\mathrm{CaCl}_{2}(\mathrm{aq})\) is added to \(\mathrm{Na}_{2} \mathrm{HPO}_{4}(\mathrm{aq}),\) a white precipitate forms that is \(38.7 \%\) Ca by mass. Write a net ionic equation representing the probable reaction that occurs.

The titration of \(5.00 \mathrm{mL}\) of a saturated solution of sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) at \(25^{\circ} \mathrm{C}\) requires \(25.8 \mathrm{mL}\) of \(0.02140 \mathrm{M} \mathrm{KMnO}_{4}\) in acidic solution. What mass of \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in grams would be present in \(1.00 \mathrm{L}\) of this saturated solution? \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{MnO}_{4}^{-} \longrightarrow_{\mathrm{Mn}^{2+}}+\mathrm{CO}_{2}(\mathrm{g}) \quad\) (not balanced)

An unknown white solid consists of two compounds, each containing a different cation. As suggested in the illustration, the unknown is partially soluble in water. The solution is treated with \(\mathrm{NaOH}(\mathrm{aq})\) and yields a white precipitate. The part of the original solid that is insoluble in water dissolves in \(\mathrm{HCl}(\mathrm{aq})\) with the evolution of a gas. The resulting solution is then treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})\) and yields a white precipitate. (a) Is it possible that any of the cations \(M g^{2+}, C u^{2+}\) \(\mathrm{Ba}^{2+}, \mathrm{Na}^{+},\) or \(\mathrm{NH}_{4}^{+}\) were present in the original unknown? Explain your reasoning. (b) What compounds could be in the unknown mixture (that is, what anions might be present)?

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