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Chapter 10: Problem 51

Use the solubility rules to predict the products of the following reactions. In each case, complete and balance the equation and write the corresponding net ionic equation. If no precipitate forms, then write "no reaction." (a) \(\mathrm{CuCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \rightarrow\) (b) \(\mathrm{MgBr}_{2}(a q)+\mathrm{K}_{2} \mathrm{CO}_{3}(a q) \rightarrow\) (c) \(\mathrm{BaCl}_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) (d) \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{KCl}(a q) \rightarrow\)

Short Answer

Expert verified
(a) CuS(s), (b) MgCO鈧(s), (c) BaSO鈧(s), (d) Hg鈧侰l鈧(s) form precipitates.

Step by step solution

01

Identify Possible Products

For each reaction, determine the possible products by swapping the cations and anions in the reactants. For (a), the possible products are CuS and NaCl; for (b), the products are MgCO鈧 and KBr; for (c), the products are BaSO鈧 and KCl; for (d), the products are Hg鈧侰l鈧 and KNO鈧.
02

Apply Solubility Rules

Use the solubility rules to determine whether any of the products form precipitates. In (a), CuS is insoluble. In (b), MgCO鈧 is usually insoluble. In (c), BaSO鈧 is insoluble. In (d), Hg鈧侰l鈧 is insoluble.
03

Write Balanced Equations

For reactions that form precipitates, write and balance the chemical equations. (a) CuCl鈧(aq) + Na鈧係(aq) 鈫 CuS(s) + 2NaCl(aq). (b) MgBr鈧(aq) + K鈧侰O鈧(aq) 鈫 MgCO鈧(s) + 2KBr(aq). (c) BaCl鈧(aq) + K鈧係O鈧(aq) 鈫 BaSO鈧(s) + 2KCl(aq). (d) Hg鈧(NO鈧)鈧(aq) + 2KCl(aq) 鈫 Hg鈧侰l鈧(s) + 2KNO鈧(aq).
04

Write Net Ionic Equations

For the reactions forming precipitates, remove spectator ions to write the net ionic equations. (a) Cu虏鈦(aq) + S虏鈦(aq) 鈫 CuS(s). (b) Mg虏鈦(aq) + CO鈧兟测伝(aq) 鈫 MgCO鈧(s). (c) Ba虏鈦(aq) + SO鈧劼测伝(aq) 鈫 BaSO鈧(s). (d) Hg鈧偮测伜(aq) + 2Cl鈦(aq) 鈫 Hg鈧侰l鈧(s).

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Key Concepts

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

Solubility Rules
In chemistry, solubility rules help us predict whether a compound will dissolve in water or form a precipitate. These guidelines are derived from empirical evidence and provide a quick reference to anticipate solubility. Here are some general principles:
  • Sodium (Na鈦), potassium (K鈦), and ammonium (NH鈧勨伜) salts are usually soluble in water.
  • Nitrates (NO鈧冣伝), acetates (CH鈧僀OO鈦), and most perchlorates are soluble.
  • Chlorides (Cl鈦), bromides (Br鈦), and iodides (I鈦) are soluble except with silver (Ag鈦), lead (Pb虏鈦), and mercury (Hg鈧偮测伜).
  • Sulfates (SO鈧劼测伝) are generally soluble, except for those of barium (Ba虏鈦), strontium (Sr虏鈦), and lead (Pb虏鈦).
  • Carbonates (CO鈧兟测伝), phosphates (PO鈧劼斥伝), sulfides (S虏鈦), and hydroxides (OH鈦) are typically insoluble, except when paired with the always-soluble ions such as Na鈦, K鈦, and NH鈧勨伜.
By using these rules, we can determine the fate of ionic compounds when mixed in aqueous solutions. When a product is insoluble, it forms a precipitate, making solubility rules an essential tool for predicting chemical reactions.
Precipitation Reactions
A precipitation reaction occurs when two solutions are mixed, and an insoluble solid, known as a precipitate, forms. This process often involves ionic compounds in aqueous solutions, where the cations and anions swap partners. Let's explore this concept:
  • Precipitation reactions are a type of double displacement reaction. In these reactions, the ions exchange their partners, leading to the formation of new compounds.
  • These reactions are typically characterized by their fast occurrence and visible evidence (a solid forms in the solution).
  • Identifying a precipitate depends on understanding solubility rules. If the products of an ionic exchange result in an insoluble compound, a precipitate will form.
In the given exercise: - (a) CuCl鈧(aq) and Na鈧係(aq) combine to form CuS(s), a precipitate. - (b) MgBr鈧(aq) and K鈧侰O鈧(aq) produce MgCO鈧(s), another insoluble solid. Precipitation reactions not only demonstrate chemical compound interactions but also have practical applications in fields like chemistry and environmental science.
Chemical Equations
Chemical equations are shorthand notations that describe chemical reactions, illustrating how substances interact to form new products. Here's what you need to know:
  • A chemical equation has reactants (starting materials) on the left and products (formed substances) on the right, separated by an arrow.
  • Balancing chemical equations ensures the law of conservation of mass is respected, meaning the number of each type of atom is the same on both sides of the equation.
  • States of matter, shown in parentheses, indicate the physical form of each substance (e.g., (aq) for aqueous, (s) for solid).
For example, in the chemical equation for the reaction of CuCl鈧(aq) and Na鈧係(aq):\[ \mathrm{CuCl}_{2}(aq) + \mathrm{Na}_2\mathrm{S}(aq) \rightarrow \mathrm{CuS}(s) + 2\mathrm{NaCl}(aq) \]This equation represents the reactants converting into products, visually encompassing the essence of chemical transformations.
Spectator Ions
In chemical reactions, especially in aqueous solutions, some ions do not participate in the actual formation of a precipitate or other significant change. These ions are called spectator ions.
  • Spectator ions appear unchanged on both sides of a chemical equation. Their role is simply to maintain charge balance in the solution.
  • The removal of these ions leads to the formation of net ionic equations, which show only the ions that are directly involved in the formation of new products.
For instance, in the given exercise, the net ionic equations for each precipitation reaction exclude spectator ions: - For the formation of CuS: Cu虏鈦(aq) + S虏鈦(aq) 鈫 CuS(s) - All other ions, like Na鈦 and Cl鈦 in this case, remain unchanged and are not included in the net ionic equation. Understanding spectator ions helps in simplifying ionic equations and focusing on the chemical change, providing clearer insight into reaction mechanisms.

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Most popular questions from this chapter

Predict the products and write balanced chemical equations and net ionic equations for the following gas-forming reactions: (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{NaHCO}_{3}(s) \rightarrow\) (b) \(\mathrm{HNO}_{5}(a q)+\mathrm{CaS}(s) \rightarrow\) (c) \(\mathrm{HCl}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{3}(a q) \rightarrow(\) Hint \(:\) Analogous to the reaction of acids with the carbonate ion.)

Predict the products and write balanced chemical equations and net ionic equations for the following gas-forming reactions: (a) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(a q)+\mathrm{NaOH}(a q) \rightarrow\) (Hint: Water is one of the products.) (b) \(\mathrm{HNO}_{5}(a q)+\mathrm{BaCO}_{3}(s) \rightarrow\) (c) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow(\) Hint \(:\) Water is one of the products of this decomposition reaction.)

Name the following ionic compounds, which are used in silver-based photography: (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{~S}_{2} \mathrm{O}_{5}(s)\) (fixer) (b) \(\mathrm{Na}_{2} \mathrm{SO}_{3}(s)\) (preservative) (c) \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\) (activator) (d) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{5}(s)\) (fixer)

Indicate which element is oxidized and which is reduced in the reactions described by the following chemical equations: (a) \(\mathrm{Ca}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CaCl}_{2}(s)\) (b) \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\) (c) \(2 \mathrm{Rb}(s)+\mathrm{Br}_{2}(l) \rightarrow 2 \mathrm{RbBr}(s)\) (d) \(2 \mathrm{Na}(s)+\mathrm{S}(s) \rightarrow \mathrm{Na}_{2} \mathrm{~S}(s)\)

Predict the products of the following reactions and write the balanced chemical equation. If no reaction occurs, then write "no reaction." (a) An aqueous solution of potassium chromate is mixed with an aqueous solution of lead(II) nitrate. (b) Hydrochloric acid is added to an aqueous solution of sodium sulfide. (c) An aqueous solution of barium hydroxide is mixed with an aqueous solution of zinc sulfate. (d) Solid calcium oxide is added to an aqueous solution of nitric acid.
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