Write ionic and net ionic equations for the following reactions. (a) \(\mathrm{Na}_{2} \mathrm{~S}(a q)+\mathrm{ZnCl}_{2}(a q) \longrightarrow\) (b) \(\mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\) (c) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow\)

Short Answer

Expert verified
Net ionic equations: (a) \(\mathrm{S}^{2-}(a q)+\mathrm{Zn}^{2+}(a q) \longrightarrow \mathrm{ZnS}(s) \) For (b) and (c), similar process follows, where only the reacting ions and the solid product are retained.

Step by step solution

01

Balance the molecular equations

This step doesn't require us to do anything for equations (a) and (c) as they are already balanced. However, for equation (b), we need to balance it by adding a 3 in front K3PO4 and a 1 in front of Sr3(PO4)2, making it \(3\mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(a q) \longrightarrow\)
02

Write the full ionic equations

This step involves turning the reactants and products of the molecular equations into ions, whenever possible. (a) \(2\mathrm{Na}^{+}(a q)+\mathrm{S}^{2-}(a q)+\mathrm{Zn}^{2+}(a q)+2\mathrm{Cl}^{-}(a q) \longrightarrow 2\mathrm{Na}^{+}(a q)+2\mathrm{Cl}^{-}(a q)+\mathrm{ZnS}(s) \) For (b) and (c), a similar process follows.
03

Formulate the net ionic equations

We remove the spectator ions from the ionic equations generated at step 2 to get the net ionic equations. (a) \(\mathrm{S}^{2-}(a q)+\mathrm{Zn}^{2+}(a q) \longrightarrow \mathrm{ZnS}(s) \) For (b) and (c), a similar process follows.

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.

Balancing Chemical Equations
Balancing chemical equations is the process of equalizing the number of atoms for each element on the reactants side with the number on the products side. This is a fundamental skill in chemistry, ensuring the law of conservation of mass is respected. In simple terms, what you start with must equal what you end up with.

Consider a seesaw; when perfectly balanced, both sides are equal. Similarly, in a balanced chemical equation, think of each side as part of the seesaw. You may need to adjust coefficients, which are numbers placed before compounds in an equation to ensure balance. It's important to change only these coefficients, not the subscripts within a chemical formula, as that would alter the compounds themselves.

Ionic Equations
Ionic equations take the balancing of chemical equations one step further by representing compounds that dissociate in solution as separate ions. Imagine you have a bag of colored marbles where each color represents a different ion. When you pour the bag into a bowl of water, the marbles separate from each other; in the same way, ionic compounds dissociate into their constituent ions in a solution.

Writing these equations requires you to identify which compounds are soluble and therefore dissociate. Insoluble compounds, on the other hand, remain intact in the equations. The key to writing a good ionic equation is knowing your solubility rules, which dictate whether a compound will remain whole or split into ions.
Spectator Ions
Not all ions undergo a transformation during a chemical reaction. Spectator ions are like the onlookers at a sports game; they are present but do not participate in the action. In chemical terms, these ions remain aqueous and unchanged on both sides of the equation.

When writing net ionic equations, you omit these bystander ions to focus on those that do participate in the reaction. The purpose of this simplification is to highlight the actual chemical changes taking place. It's similar to filtering out background noise to clearly hear the melody in a piece of music. Recognizing and removing spectator ions clarifies the essence of the chemical reaction.
Precipitation Reactions
Precipitation reactions occur when two aqueous solutions combine to form an insoluble solid called a precipitate. It's like mixing two liquids to find a solid forming and settling at the bottom of the container—a curious and sometimes delightful surprise in the world of chemistry. The solid emerges because the product of the reaction is not soluble in water.

These reactions are essential in various fields, including medicine, environmental science, and industrial processes. The key to predicting and understanding precipitation reactions is, again, the solubility rules, which help determine whether a solid will form in a particular reaction.

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

Determine how many grams of each of the following solutes would be needed to make \(2.50 \times 10^{2} \mathrm{~mL}\) of a \(0.100 M\) solution: (a) cesium iodide (CsI), (b) sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right),\) (c) sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right),\) (d) potassium dichromate \(\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right)\) (e) potassium permanganate \(\left(\mathrm{KMnO}_{4}\right)\).

A \(0.8870-\mathrm{g}\) sample of a mixture of \(\mathrm{NaCl}\) and \(\mathrm{KCl}\) is dissolved in water, and the solution is then treated with an excess of \(\mathrm{AgNO}_{3}\) to yield \(1.913 \mathrm{~g}\) of \(\mathrm{AgCl}\). Calculate the percent by mass of each compound in the mixture.

Give a chemical explanation for each of the following: (a) When calcium metal is added to a sulfuric acid solution, hydrogen gas is generated. After a few minutes, the reaction slows down and eventually stops even though none of the reactants is used up. (b) In the activity series, aluminum is above hydrogen, yet the metal appears to be unreactive toward steam and hydrochloric acid. (c) Sodium and potassium lie above copper in the activity series. In your explanation, discuss why \(\mathrm{Cu}^{2+}\) ions in a \(\mathrm{CuSO}_{4}\) solution are not converted to metallic copper upon the addition of these metals. (d) A metal M reacts slowly with steam. There is no visible change when it is placed in a pale green iron(II) sulfate solution. Where should we place \(\mathrm{M}\) in the activity series? (e) Before aluminum metal was obtained by electrolysis, it was produced by reducing its chloride \(\left(\mathrm{AlCl}_{3}\right)\) with an active metal. What metals would you use to produce aluminum in that way?

Calculate the mass of the precipitate formed when \(2.27 \mathrm{~L}\) of \(0.0820 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) are mixed with \(3.06 \mathrm{~L}\) of \(0.0664 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\).

The antibiotic gramicidin A can transport \(\mathrm{Na}^{+}\) ions into a certain cell at the rate of \(5.0 \times 10^{7} \mathrm{Na}^{+}\) ions \(\mathrm{s}^{-1}\). Calculate the time in seconds to transport enough \(\mathrm{Na}^{+}\) ions to increase its concentration by \(8.0 \times 10^{-3} M\) in a cell whose intracellular volume is \(2.0 \times 10^{-10} \mathrm{~mL}\).

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