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Consider the reaction of all pairs of the following compounds in water solution: \(\mathrm{Ba}(\mathrm{OH})_{2}, \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{H}_{2} \mathrm{SO}_{4},\) \(\mathrm{NaNO}_{3}, \mathrm{MgSO}_{4}\) (a) Which pair (or pairs) forms one insoluble compound and one soluble compound (not water)? (b) Which pair (or pairs) forms two insoluble compounds? (c) Which pair (or pairs) forms one insoluble compound and water?

Short Answer

Expert verified
(a) \(\mathrm{Pb(NO}_3)_2\) and \(\mathrm{H}_2\mathrm{SO}_4\). (b) No pair. (c) \(\mathrm{Ba(OH)}_2\) and \(\mathrm{H}_2\mathrm{SO}_4\).

Step by step solution

01

Identify Solubility Rules

Recall that sulfates are generally soluble except for those of Ba, Pb, and some others. Hydroxides are typically insoluble except for those of alkali metals and a few others like Ba. Nitrates are always soluble. Use these rules to predict solubility.
02

Pairing Compounds and Predicting Solubility

Pair each compound with the others and use solubility rules to predict the solubility. For each pair:1. \(\mathrm{Ba(OH)}_{2}\) and \(\mathrm{Pb(NO}_3)_2\): Reacting forms \(\mathrm{Pb(OH)}_{2}\) (insoluble) and \(\mathrm{Ba(NO}_3)_2\) (soluble).2. \(\mathrm{Ba(OH)}_{2}\) and \(\mathrm{H}_2\mathrm{SO}_4\): Reacting forms \(\mathrm{BaSO}_4\) (insoluble) and water (soluble).3. \(\mathrm{Pb(NO}_3)_2\) and \(\mathrm{H}_2\mathrm{SO}_4\): Reacting forms \(\mathrm{PbSO}_4\) (insoluble) and \(\mathrm{HNO}_3\) (soluble).4. Repeat this pairing process for the remaining other compound combinations.
03

Analyzing the Product Solubility for Scenarios

Using the predicted solubility data, check each pair against the given conditions in the problem:(a) Identify pairs like \(\mathrm{Pb(NO}_3)_2\) and \(\mathrm{H}_2\mathrm{SO}_4\) that produce one insoluble and one soluble product.(b) No given pair forms two insoluble compounds, as at least one product from any reaction is soluble.(c) Identify pairs like \(\mathrm{Ba(OH)}_2\) and \(\mathrm{H}_2\mathrm{SO}_4\) that form an insoluble product and water.

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Key Concepts

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

Insoluble Compounds
Insoluble compounds are those that do not dissolve significantly in water. This means that when they interact with water, they form a solid known as a precipitate. Knowing which compounds are insoluble is crucial for predicting the products of chemical reactions in water.
One of the basic solubility rules states that hydroxides (OH-) are largely insoluble except for those of alkali metals and barium, while sulfates (SO₄²⁻) are generally soluble with exceptions such as barium sulfate (BaSO₄) and lead sulfate (PbSO₄).
In our exercise, combinations like Ba(OH)₂ with Pb(NO₃)₂ result in insoluble lead hydroxide, Pb(OH)₂. Understanding these patterns is key for any chemistry student because predicting precipitates helps in identifying potential reactive hazards and industrial applications.
Soluble Compounds
Soluble compounds, unlike their insoluble counterparts, dissolve readily in water. Their ions dissociate completely in an aqueous solution, meaning they can contribute to chemical conductivity. This property makes them vital in various chemical processes and reactions.
A fundamental solubility rule to remember is that nitrates (NO₃⁻) are universally soluble. This rule makes compounds like sodium nitrate (NaNO₃) always a soluble participant in any aqueous reaction.
In the given problem, when barium hydroxide (Ba(OH)₂) reacts with lead(II) nitrate (Pb(NO₃)₂), barium nitrate (Ba(NO₃)₂) is soluble. This ensures at least one soluble product is present in the reaction. Recognizing such solubility helps in predicting reaction outcomes and balancing chemical equations effectively.
Chemical Reactions in Water
Chemical reactions in water, often referred to as aqueous reactions, are central in both laboratory and industrial chemistry. These can involve complex ion exchanges and formation of new compounds with unique properties.
One common type of reaction in water is double displacement or metathesis, where the ions of two compounds exchange partners. You'll notice that in the exercise given, several pairs of compounds react to form one insoluble compound and another soluble one through this kind of double displacement.
Understanding aqueous reactions includes predicting the formation of water as a product, such as when barium hydroxide reacts with sulfuric acid, forming water and an insoluble precipitate, barium sulfate. This ability to predict outcomes helps in planning chemical experiments and manufacturing processes.
Predicting Solubility
Predicting solubility involves using established rules and past observations to anticipate whether a compound will dissolve in water. It is a crucial skill in chemistry, allowing scientists to foresee the behavior of substances in reactions.
Some general solubility rules include detailing the solubility of common ions. For instance, while chlorides are mostly soluble, those of lead and silver are notable exceptions.
When we think about the task at hand, predicting solubility means accurately foreseeing which combinations of ions will form precipitates. For instance, combining Pb(NO₃)₂ and H₂SO₄ results in the prediction of an insoluble lead sulfate. Developing a keen understanding of these rules aids in predicting reaction types and managing chemical processes successfully.

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Most popular questions from this chapter

When using \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in aqueous reactions, many problems are presented or solved by assuming both of the hydrogen ions to be strongly ionized, as if they were both strong acid ionization processes. At the same time, the second \(\mathrm{H}^{+}\) only weakly ionizes in aqueous solution. 1\. Write molecular, complete ionic, and net ionic equations for the reaction between sulfuric acid and barium hydroxide (a strong base) using the following assumptions: (a) Both hydrogen ions of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are assumed to ionize \(100 \%\) like strong acids. (b) Only the first hydrogen ion of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(100 \%\) ionized, and the second is weakly ionized. 2\. Write molecular, complete ionic, and net ionic equations for the reaction between sulfuric acid and sodium acetate, a weakly basic salt, using the following assumptions: Only the first hydrogen ion of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(100 \%\) ionized, and the second is weakly ionized.

An aqueous solution contains \(3.75 \mathrm{~g}\) of iron(III) sulfate, \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3},\) per liter. What is the molarity of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} ?\) When the compound dissolves in water, the \(\mathrm{Fe}^{3+}\) ions and \(\mathrm{SO}_{4}{ }^{2-}\) ions in the crystal go into the solution. What is the molar concentration of each ion in the solution?

An antacid tablet contains sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\), and inert ingredients. A 0.465-g sample of powdered tablet was mixed with \(53.3 \mathrm{~mL}\) of \(0.190 \mathrm{M}\) \(\mathrm{HCl}\) (hydrochloric acid). The mixture was allowed to stand until it reacted. $$ \begin{aligned} \mathrm{NaHCO}_{3}(s)+\mathrm{HCl}(a q) \longrightarrow & \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) \end{aligned} $$ The excess hydrochloric acid was titrated with \(54.6 \mathrm{~mL}\) of \(0.128 \mathrm{M} \mathrm{NaOH}\) (sodium hydroxide). $$ \mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What is the percentage of sodium hydrogen carbonate in the antacid?

A transition metal \(X\) forms an oxide of formula \(\mathrm{X}_{2} \mathrm{O}_{3} .\) It is found that only \(50 \%\) of \(\mathrm{X}\) atoms in this compound are in the +3 oxidation state. The only other stable oxidation states of \(X\) are +2 and +5 . What percentage of \(\mathrm{X}\) atoms are in the +2 oxidation state in this compound?

An aqueous solution is made from \(0.798 \mathrm{~g}\) of potassium permanganate, \(\mathrm{KMnO}_{4}\). If the volume of solution is \(50.0 \mathrm{~mL},\) what is the molarity of \(\mathrm{KMnO}_{4}\) in the solution?

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