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Chapter 4: Problem 16

An aqueous sodium hydroxide solution mixed with an aqueous magnesium nitrate solution yields which of the following products? a. magnesium hydroxide \((a q)\) b. magnesium dihydroxide(s) c. magnesium hydroxide \((s)\) d. dimagnesium hydroxide \((s)\) e. sodium nitrate \((l)\)

Short Answer

Expert verified
c. magnesium hydroxide \((s)\) and sodium nitrate \((aq)\) are the products.

Step by step solution

01

Identify the reactants

The reactants given are aqueous sodium hydroxide (N\(\text{aOH}\)) and aqueous magnesium nitrate (M\(\text{g(NO}_{3}\text{)}_{2}\)).
02

Predict the possible reaction

When sodium hydroxide reacts with magnesium nitrate, it involves a double replacement reaction where the cations and anions switch places.
03

Write the balanced chemical equation

Write the balanced equation for the reaction. Sodium hydroxide reacts with magnesium nitrate as follows: \( 2\,\text{NaOH}\,(aq) + \text{Mg(NO}_3)_2\,(aq) \to \text{Mg(OH)}_2\,(s) + 2\,\text{NaNO}_3\,(aq) \).
04

Identify the products

The products formed are magnesium hydroxide \((s)\), which is a precipitate, and aqueous sodium nitrate \(\text{NaNO}_{3}(aq)\).
05

Determine the state of magnesium hydroxide

In aqueous solutions, magnesium hydroxide is insoluble, so it precipitates out of the solution, which makes it a solid \((s)\).

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Key Concepts

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

Chemical Equations
Chemical equations are essential for understanding reactions in chemistry. They show the transformation of reactants into products, using chemical formulas to represent these compounds. In our double replacement reaction,
  • Sodium hydroxide (NaOH) is one reactant, with its chemical formula representing sodium (Na) and hydroxide (OH) ions.
  • Magnesium nitrate (Mg(NO鈧�)鈧�) is the other reactant, representing magnesium (Mg) ions and nitrate (NO鈧�) ions.
To write a balanced equation, we must ensure equal numbers of each type of ion on both sides of the equation, maintaining the law of conservation of mass. In this reaction:
The chemical equation is:
\[ 2 ext{NaOH (aq)} + ext{Mg(NO}_3)_2 ext{(aq)} \rightarrow ext{Mg(OH)}_2 ext{(s)} + 2 ext{NaNO}_3 ext{(aq)} \]
This shows that two molecules of sodium hydroxide react with one molecule of magnesium nitrate to form magnesium hydroxide and sodium nitrate.
Precipitation Reactions
Precipitation reactions occur when two soluble substances in solution combine to form an insoluble product, called a precipitate. These reactions are a fascinating aspect of chemical processes because they visibly demonstrate a change, like cloudiness forming in a clear liquid.
In the sodium hydroxide and magnesium nitrate reaction, a double replacement occurs. This means that the cations and anions in the reactants switch partners. The solid magnesium hydroxide, represented as Mg(OH)鈧�, is the precipitate formed. When the hydroxide ions ( ext{OH}鈦�) from NaOH react with magnesium ions ( ext{Mg}虏鈦�) from Mg(NO鈧�)鈧�, they form the insoluble solid, magnesium hydroxide:
  • Hydroxide ions ( ext{OH}鈦�) combine with magnesium ions ( ext{Mg}虏鈦�) to form Mg(OH)鈧�.
Precipitation reactions are important for separating substances, and they are frequently used in chemical analysis and industrial processes. In our reaction, the formation of a solid precipitate confirms the occurrence of a chemical reaction.
Solubility Principles
Solubility principles explain why some compounds dissolve easily in water while others do not. A compound's solubility is affected by the nature of the ions forming the compound and the solvent used. These principles are particularly important in precipitation reactions.
Sodium nitrate, a product of our reaction, remains soluble because it meets the solubility principle for nitrates, which states that most nitrates are soluble in water. Therefore, sodium nitrate (Sodium ions ( ext{Na}鈦�) and nitrate ions ( ext{NO鈧儅鈦�) stay dissolved in water.
However, magnesium hydroxide does not follow this principle as it is considered insoluble. Its formation as a solid precipitate is due to the low solubility of hydroxides of certain metals, including magnesium, in water. This insolubility is crucial because it drives the precipitation reaction, resulting in the solid formation.
By understanding solubility principles, we predict and control the outcomes of chemical reactions and processes. This knowledge helps chemists create desired chemical states, such as solutions or precipitates, efficiently.

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Most popular questions from this chapter

Iron(III) chloride can be prepared by reacting iron metal with chlorine. What is the balanced equation for this reaction? How many grams of iron are required to make \(3.00 \mathrm{~L}\) of aqueous solution containing \(9.00 \%\) iron(III) chloride by mass? The density of the solution is \(1.067 \mathrm{~g} / \mathrm{mL}\).

Copper has compounds with copper(I) ion or copper(II) ion. A compound of copper and chlorine was treated with a solution of silver nitrate, \(\mathrm{AgNO}_{3}\), to convert the chloride ion in the compound to a precipitate of AgCl. A \(59.40-\mathrm{mg}\) sample of the copper compound gave \(86.00 \mathrm{mg} \mathrm{AgCl}\). a. Calculate the percentage of chlorine in the copper compound. b. Decide whether the formula of the compound is \(\mathrm{CuCl}\) or \(\mathrm{CuCl}_{2}\)

A 414-mL sample of \(0.196 M \mathrm{MgBr}_{2}\) solution is prepared in a large flask. A \(43.0-\mathrm{mL}\) portion of the solution is then placed into an empty \(100.0-\mathrm{mL}\) beaker. What is the concentration of the solution in the beaker?

A sample of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) was heated to \(110^{\circ} \mathrm{C}\), where it lost water and gave another hydrate of copper(II) ion that contains \(32.50 \%\) Cu. A \(98.77\) -mg sample of this new hydrate gave \(116.66 \mathrm{mg}\) of barium sulfate precipitate when treated with a barium nitrate solution. What is the formula of the new hydrate?

A water-soluble compound of gold and chlorine is treated with silver nitrate to convert the chlorine completely to silver chloride, \(\mathrm{AgCl}\). In an experiment, \(328 \mathrm{mg}\) of the compound gave \(464 \mathrm{mg}\) of silver chloride. Calculate the percentage of \(\mathrm{Cl}\) in the compound. What is its empirical formula? 4.122 A solution of scandium chloride was treated with silver nitrate. The chlorine in the scandium compound was converted to silver chloride, AgCl. A \(58.9\) -mg sample of scandium chloride gave \(167.4 \mathrm{mg}\) of silver chloride. What are the mass percentages of \(\mathrm{Sc}\) and \(\mathrm{Cl}\) in scandium chloride? What is its empirical formula?
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