/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Will precipitation occur when th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 23

Will precipitation occur when the following solutions are mixed? If so, write a balanced chemical equation for the reac- tion. (a) \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\) and \(\mathrm{NaOH},(\mathbf{b}) \mathrm{K}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\), (c) \(\mathrm{Na}_{2} \mathrm{~S}\) and \(\mathrm{FeCl}_{3}\)

Short Answer

Expert verified
(a) Precipitation occurs, the balanced chemical equation is: \[\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}+2\mathrm{NaOH}\longrightarrow\mathrm{Ca(OH)_2}+\mathrm{2Na}\left(\mathrm{CH}_{3} \mathrm{COO}\right)\] (b) No precipitation occurs (c) Precipitation occurs, the balanced chemical equation is: \[\mathrm{2Na}_{2} \mathrm{S}+3\mathrm{FeCl}_{3}\longrightarrow6\mathrm{NaCl}+\mathrm{2FeS}\]

Step by step solution

01

Identifying the potential products

We will begin by predicting the products formed from the reaction of the given solutions in part (a) \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\) and \(\mathrm{NaOH}\), part (b) \(\mathrm{K}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\), and part (c) \(\mathrm{Na}_{2} \mathrm{~S}\) and \(\mathrm{FeCl}_{3}\). Note that these are double displacement reactions.
02

Check the solubility rules

Now, we will apply the solubility rules to determine if any of the potential products formed are insoluble (forming a precipitate). (a) Starting with \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\) and \(\mathrm{NaOH}\): The potential products are: \(\mathrm{Ca(OH)_2}\) and \(\mathrm{Na}\left(\mathrm{CH}_{3} \mathrm{COO}\right)\) Solubility rules: \(\mathrm{Ca(OH)_2}\): calcium hydroxide is considered slightly soluble in water, thus forming a precipitate \(\mathrm{Na}\left(\mathrm{CH}_{3} \mathrm{COO}\right)\): sodium salts are soluble in water The balanced chemical equation is: \[\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}+2\mathrm{NaOH}\longrightarrow\mathrm{Ca(OH)_2}+\mathrm{2Na}\left(\mathrm{CH}_{3} \mathrm{COO}\right)\] (b) Moving on to \(\mathrm{K}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\): The potential products are: \(\mathrm{KNO_3}\) and \(\mathrm{NH_4}\mathrm{(CO_{3})}\) Solubility rules: \(\mathrm{KNO_3}\): potassium salts are soluble in water \(\mathrm{NH_4}\mathrm{(CO_{3})}\): ammonium salts are soluble in water In this case, no precipitation occurs. (c) Finally, \(\mathrm{Na}_{2} \mathrm{S}\) and \(\mathrm{FeCl}_{3}\): The potential products are: \(\mathrm{NaCl}\) and \(\mathrm{FeS}\) Solubility rules: \(\mathrm{NaCl}\): sodium salts are soluble in water \(\mathrm{FeS}\): Metal sulfides, in general are insoluble, including \(\mathrm{FeS}\) which forms a precipitate The balanced chemical equation is: \[\mathrm{2Na}_{2} \mathrm{S}+3\mathrm{FeCl}_{3}\longrightarrow6\mathrm{NaCl}+\mathrm{2FeS}\]
03

Summarize the results

Now we can summarize the results for the given parts: (a) Precipitation occurs, the balanced chemical equation is: \[\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}+2\mathrm{NaOH}\longrightarrow\mathrm{Ca(OH)_2}+\mathrm{2Na}\left(\mathrm{CH}_{3} \mathrm{COO}\right)\] (b) No precipitation occurs (c) Precipitation occurs, the balanced chemical equation is: \[\mathrm{2Na}_{2} \mathrm{S}+3\mathrm{FeCl}_{3}\longrightarrow6\mathrm{NaCl}+\mathrm{2FeS}\]

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 91影视!

Key Concepts

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

Chemical Equations
When substances react chemically, a new set of compounds is formed from the initial reactants, and this transformation is expressed through chemical equations. These equations represent what happens during a chemical reaction. Each side of the equation has chemical formulas that denote substances involved in the reaction.
On the left are the reactants, and on the right, the products are displayed.
  • Reactants: Substances that start the reaction.
  • Products: Compounds that are formed as a result of the reaction.
To comply with the law of conservation of mass, chemical equations need to be balanced. This means that the number of atoms for each element should be equal on both sides of the equation.
In the original exercise, we see this in action with balanced equations for reactions, such as:\[\text{Ca(CH}_3\text{COO)}_2 + 2\text{NaOH} \rightarrow \text{Ca(OH)}_2 + 2\text{Na(CH}_3\text{COO)}\]
Understanding chemical equations helps us visualize and predict the outcome of chemical reactions.
Solubility Rules
Solubility rules are like guidelines that help predict whether a compound will dissolve in water or form a precipitate. They鈥檙e crucial when determining outcomes of reactions involving dissolved salts and are essential for forecasting precipitation reactions.
With solubility rules, we can generally assess if the new products in a reaction remain dissolved or form a solid. Helpful solubility guidelines include:
  • Most nitrate (\( ext{NO}_3^-\)) salts are soluble.
  • Most sodium (\( ext{Na}^+\)), potassium (\( ext{K}^+\)), and ammonium (\( ext{NH}_4^+\)) salts are soluble.
  • Salts containing carbonate (\( ext{CO}_3^{2-}\)) are generally insoluble, except with alkaline metals (like potassium and sodium) and ammonium.
  • Hydroxides, such as calcium hydroxide (\( ext{Ca(OH)}_2\)), are generally considered slightly soluble.

In the step-by-step solution, solubility rules were used effectively to understand that in reactions (b), the products potassium nitrate (\( ext{KNO}_3\)) and ammonium carbonate (\( ext{NH}_4 ext{CO}_3\)) are both soluble, leading to no precipitate.
Precipitation Reactions
Precipitation reactions occur when two aqueous solutions are mixed, and an insoluble solid, called a precipitate, forms and separates from the solution. This type of reaction is a subset of double displacement reactions and is typical in situations involving salts.
The driving force of these reactions is the formation of a product that does not dissolve in water. In the textbook exercise, we saw:
  • The combination of \( ext{Na}_2 ext{S}\) and \( ext{FeCl}_3\) forming \( ext{FeS}\), a classic precipitate, follows this rule.
  • Another example is \( ext{Ca(OH)}_2\), which is slightly soluble, formed from \( ext{Ca(CH}_3 ext{COO)}_2\) and \( ext{NaOH}\).

To predict the occurrence of a precipitation reaction, it is important to refer to solubility rules: a precipitate is likely to form if any of the possible products of your aqueous reactants are insoluble in water. This understanding allows chemists to plan reactions thoughtfully and utilize separation techniques based on solubility.

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

Which ions remain in solution, unreacted, after each of the following pairs of solutions is mixed? (a) potassium carbonate and magnesium sulfate (b) lead nitrate and lithium sulfide (c) ammonium phosphate and calcium chloride

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. (a) \(\mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{FeCl}_{3}(a q) \longrightarrow\) (b) \(\mathrm{ZnCl}_{2}(a q)+\mathrm{Cs}_{2} \mathrm{CO}_{3}(a q) \longrightarrow\) (c) \(\mathrm{Na}_{2} \mathrm{~S}(a q)+\operatorname{CoSO}_{4}(a q) \longrightarrow\)

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{MNH}_{3} ?\) (b) If you take a \(10.0-\mathrm{mL}\) portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L},\) what will be the concentration of the final solution?

(a) How many grams of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), should you dissolve in water to make \(1.00 \mathrm{~L}\) of vodka (which is an aqueous solution that is \(6.86 \mathrm{M}\) ethanol)? (b) Using the density of ethanol \((0.789 \mathrm{~g} / \mathrm{mL})\), calculate the volume of ethanol you need to make \(1.00 \mathrm{~L}\) of vodka.

Would you expect that an anion would be physically closer to the oxygen or to the hydrogens of water molecules that surround it in solution?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation
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.