Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 52

When the following solutions are mixed together, what precipitate (if any) will form? a. $\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CuSO}_{4}(a q)$ b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{MgI}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\mathrm{AlBr}_{3}(a q)\)

Short Answer

Expert verified
In summary, a precipitate will form in reactions a and c: a. \(Hg_{2}(NO_{3})_{2} + CuSO_{4}\) will form \(Hg_{2}SO_{4}\) as a precipitate. c. \(K_{2}CO_{3} + MgI_{2}\) will form \(MgCO_{3}\) as a precipitate.

Step by step solution

01

a. Hg₂(NO₃)₂(aq) + CuSO₄(aq)

In this case, let's consider the possible double replacement reaction products: - Hg₂SO₄ and Cu(NO₃)₂ Now, let's apply the solubility rules: - Hg₂SO₄ is insoluble (rule: most sulfates are soluble, except for BaSO₄, PbSO₄, Hg₂SO₄, and CaSO₄) - Cu(NO₃)₂ is soluble (rule: all nitrates are soluble) Since Hg₂SO₄ is insoluble, a precipitate will form in this case.
02

b. Ni(NO₃)₂(aq) + CaCl₂(aq)

The possible double replacement reaction products are: - NiCl₂ and Ca(NO₃)₂ Let's use the solubility rules: - NiCl₂ is soluble (rule: all chlorides are soluble, except for AgCl, PbCl₂, and Hg₂Cl₂) - Ca(NO₃)₂ is soluble (rule: all nitrates are soluble) Since both products are soluble, no precipitate will form in this case.
03

c. K₂CO₃(aq) + MgI₂(aq)

The possible double replacement reaction products are: - KI and MgCO₃ Let's apply the solubility rules: - KI is soluble (rule: all alkali metal salts are soluble) - MgCO₃ is insoluble (rule: most carbonates are insoluble, except for those of alkali metals and ammonium) Since MgCO₃ is insoluble, a precipitate will form in this case.
04

d. Na₂CrO₄(aq) + AlBr₃(aq)

The possible double replacement reaction products are: - NaBr and Al₂(CrO₄)₃ Let's apply solubility rules: - NaBr is soluble (rule: all alkali metal salts are soluble) - Al₂(CrO₄)₃ is soluble (rule: most chromates are insoluble, except for those of alkali metals and ammonium) Since both products are soluble, no precipitate will form in this case. In summary, a precipitate will form in reactions a, and c: a. Hg₂(NO₃)₂ + CuSO₄ will form Hg₂SO₄ as a precipitate. c. K₂CO₃ + MgI₂ will form MgCO₃ as a precipitate.

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.

Start your free trial

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

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

A 0.500 -L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0 -mL aliquot and adding 50.0 \(\mathrm{mL}\) of 0.213 \(\mathrm{M}\) NaOH. After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required 13.21 \(\mathrm{mL}\) of 0.103 \(\mathrm{M}\) \(\mathrm{HCl}\) for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) . Sulfuric acid has two acidic hydrogens.

A sample may contain any or all of the following ions: \(\mathrm{Hg}_{2}^{2+}\) \(\mathrm{Ba}^{2+}\) and \(\mathrm{Mn}^{2+}\) a. No precipitate formed when an aqueous solution of NaCl was added to the sample solution. b. No precipitate formed when an aqueous solution of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ was added to the sample solution. c. A precipitate formed when the sample solution was made basic with NaOH. Which ion or ions are present in the sample solution?

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid–base reactions. a. \(\mathrm{HNO}_{3}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) b. $\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \rightarrow$ c. \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HCl}(a q) \rightarrow\)

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$\mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }}$$ In the case of very dilute aqueous solutions, a concentration of 1.0 \(\mathrm{ppm}\) is equal to 1.0\(\mu \mathrm{g}\) of solute per 1.0 \(\mathrm{mL}\) , which equals 1.0 \(\mathrm{g}\) solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. 5.0 \(\mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. 1.0 \(\mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. 10.0 \(\mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. 0.10 $\mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{C} 1_{5}\right)\( in \)\mathrm{H}_{2} \mathrm{O}$

Assign oxidation numbers to all the atoms in each of the following. a. \(\mathrm{SrCr}_{2} \mathrm{O}_{7} \quad\) g. \(\mathrm{PbSO}_{3}\) b. \(\mathrm{CuCl}_{2} \quad \quad\) h. \(\mathrm{PbO}_{2}\) c. \(\mathrm{O}_{2} \quad\quad\quad\) i. $\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$ d. \(\mathrm{H}_{2} \mathrm{O}_{2} \quad\quad \mathrm{j} . \mathrm{CO}_{2}\) e. \(\mathrm{MgCO}_{3} \quad\) k. $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}$ f. \(\mathrm{Ag} \quad\quad\quad \)l. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Nuclear 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.

Sign-up for free