Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Name the spectator ions in any reactions that may be involved when each of the following pairs of solutions are mixed. (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)\) and \(\mathrm{MgSO}_{4}(a q)\) (b) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\mathrm{Na}_{2} \mathrm{~S}(a q)\) (c) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}(a q)\) and \(\mathrm{CaCl}_{2}(a q)\)

Short Answer

Expert verified
The spectator ions in the given reactions are: (a) Sodium ion (Na\(^+\)) and sulfate ion (SO\(_4^{2-}\)) (b) Sodium ion (Na\(^+\)) and nitrate ion (NO\(_3^-\)) (c) Ammonium ion (NH\(_4^+\)) and chloride ion (Cl\(^-\))

Step by step solution

01

(a) Write the balanced chemical equation for the reaction between Na2CO3 and MgSO4

The balanced chemical equation for the reaction between Na2CO3 and MgSO4 is: \(Na_2CO_3(aq)+MgSO_4(aq)\rightarrow MgCO_3(s)+2NaSO_4(aq)\)
02

(a) Find the spectator ions for Na2CO3 and MgSO4 reaction

The spectator ions are the ones that do not participate in the reaction and remain in solution. In this case, the sodium ion (Na\(^+\)) and sulfate ion (SO\(_4^{2-}\)) are the spectator ions. They do not change during the reaction and remain in the solution.
03

(b) Write the balanced chemical equation for the reaction between Pb(NO3)2 and Na2S

The balanced chemical equation for the reaction between Pb(NO3)2 and Na2S is: \(Pb(NO_3)_2(aq)+Na_2S(aq)\rightarrow PbS(s)+2NaNO_3(aq)\)
04

(b) Find the spectator ions for Pb(NO3)2 and Na2S reaction

The spectator ions are the ones that do not participate in the reaction and remain in solution. In this case, the sodium ion (Na\(^+\)) and the nitrate ion (NO\(_3^-\)) are the spectator ions. They do not change during the reaction and remain in the solution.
05

(c) Write the balanced chemical equation for the reaction between (NH4)3PO4 and CaCl2

The balanced chemical equation for the reaction between (NH4)3PO4 and CaCl2 is: \((NH_4)_3PO_4(aq)+CaCl_2(aq)\rightarrow Ca(PO_4)_2(s)+6NH_4Cl(aq)\)
06

(c) Find the spectator ions for (NH4)3PO4 and CaCl2 reaction

The spectator ions are the ones that do not participate in the reaction and remain in solution. In this case, the ammonium ion (NH\(_4^+\)) and the chloride ion (Cl\(^-\)) are the spectator ions. They do not change during the reaction and remain in the solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Balanced Chemical Equations
Understanding balanced chemical equations is fundamental to mastering chemistry. A balanced chemical equation represents a chemical reaction where the number of atoms for each element is equal on both the reactant and product sides. This conservation of mass is expressed by ensuring that for every type of atom present, the same number of that atom exists before and after the reaction.

For instance, in the reaction where sodium carbonate reacts with magnesium sulfate to form magnesium carbonate and sodium sulfate, the equation is written as: \[Na_2CO_3(aq) + MgSO_4(aq) \rightarrow MgCO_3(s) + 2NaSO_4(aq)\].In this equation, all atoms balance out on both sides, abiding by the Law of Conservation of Mass. Students should verify this by counting the atoms of each element. Understanding and being able to balance these equations is crucial, as it provides a clear picture of the exact quantities of reactants and products involved in chemical reactions.
Ion Exchange Reactions
Chemical reactions that involve ions moving from one compound to another are known as ion exchange reactions. These reactions often occur in solutions where dissolved compounds exchange ions to form new substances. This kind of reaction is typical in many inorganic chemical processes and can be easily demonstrated in the lab using aqueous solutions.

Diving into the provided problem, when a sodium carbonate (Na2CO3) solution is mixed with a magnesium sulfate (MgSO4) solution, the two ionic compounds exchange ions. The Na+ ions pair up with SO42- ions, and the Mg2+ ions pair with CO32- ions, leading to the formation of a precipitate, magnesium carbonate (MgCO3), and sodium sulfate remains in the solution. These reactions are easily predicted by understanding the solubility rules and recognizing which combinations of ions will produce an insoluble compound, resulting in a precipitate.
Solubility of Compounds
The solubility of compounds is a critical concept in chemistry which dictates whether a substance will dissolve in a particular solvent, such as water. Soluble substances can dissolve to form a homogeneous mixture, while insoluble substances will not dissolve and oftentimes form a precipitate in a reaction.

In the reactions you're studying, the solubility rules help predict the behavior of ionic compounds in aqueous solutions. For example, carbonates are generally insoluble unless they're paired with alkali metal ions like sodium or potassium, or the ammonium ion. That is why when we mix sodium carbonate with magnesium sulfate, magnesium carbonate precipitates out of the solution since it’s insoluble in water, leaving behind the soluble sodium sulfate. Students should familiarize themselves with basic solubility guidelines to better predict the outcomes of ion exchange reactions and identify spectator ions effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You are presented with a white solid and told that due to careless labeling it is not clear if the substance is barium chloride, lead chloride, or zinc chloride. When you transfer the solid to a beaker and add water, the solid dissolves to give a clear solution. Next a \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) solution is added and a white precipitate forms. What is the identity of the unknown white solid? \([\) Section 4.2\(]\)

Write balanced molecular and net ionic equations for the reactions of (a) hydrochloric acid with nickel, (b) dilute sulfuric acid with iron, (c) hydrobromic acid with magnesium, (d) acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) with zinc.

Suppose you have \(5.00 \mathrm{~g}\) of powdered magnesium metal, \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M}\) potassium nitrate solution, and \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M}\) silver nitrate solution. (a) Which one of the solutions will react with the magnesium powder? (b) What is the net ionic equation that describes this reaction? (c) What volume of solution is needed to completely react with the magnesium? (d) What is the molarity of the \(\mathrm{Mg}^{2+}\) ions in the resulting solution?

Formic acid, \(\mathrm{HCOOH},\) is a weak electrolyte. What solute particles are present in an aqueous solution of this compound? Write the chemical equation for the ionization of \(\mathrm{HCOOH}\).

Glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3},\) is a substance used extensively in the manufacture of cosmetics, foodstuffs, antifreeze, and plastics. Glycerol is a water-soluble liquid with a density of \(1.2656 \mathrm{~g} / \mathrm{mL}\) at \(15^{\circ} \mathrm{C}\). Calculate the molarity of a solution of glycerol made by dissolving \(50.000 \mathrm{~mL}\) glycerol at \(15^{\circ} \mathrm{C}\) in enough water to make \(250.00 \mathrm{~mL}\) of solution.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free