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

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{FeSO}_{4}(a q)+\mathrm{KCl}(a q)\) b. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q)\) c. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) d. \(\mathrm{K}_{2} \mathrm{~S}(a q)+\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

Short Answer

Expert verified
Precipitates formed in each case are: - a. No precipitate - b. Al(OH)₃ - c. CaSO₄ - d. NiS

Step by step solution

01

a. FeSO4(aq) + KCl(aq)

1. Analyze the chemical equation: The given solutions are: FeSO4(aq) - containing Fe²⁺ and SO₄²⁻ ions, and KCl(aq) - containing K⁺ and Cl⁻ ions. 2. Determine the possible combinations: The possible combinations of these ions are: FeCl₂ and K₂SO₄. 3. Check solubility: According to the solubility rules, both FeCl₂ and K₂SO₄ are soluble, so no precipitate will form in this case.
02

b. Al(NO₃)₃(aq) + Ba(OH)₂(aq)

1. Analyze the chemical equation: The given solutions are: Al(NO₃)₃(aq) - containing Al³⁺ and NO₃⁻ ions, and Ba(OH)₂(aq) - containing Ba²⁺ and OH⁻ ions. 2. Determine the possible combinations: The possible combinations of these ions are: Al(OH)₃ and Ba(NO₃)₂. 3. Check solubility: According to the solubility rules, Al(OH)₃ is insoluble, and Ba(NO₃)₂ is soluble. 4. Write down the balanced chemical equation: When these two solutions are combined, a precipitate will form as Al(OH)₃. \[ Al^{3+}(aq) + 3OH^{-}(aq) \rightarrow Al(OH)_3(s) \]
03

c. CaCl₂(aq) + Na₂SO₄(aq)

1. Analyze the chemical equation: The given solutions are: CaCl₂(aq) - containing Ca²⁺ and Cl⁻ ions, and Na₂SO₄(aq) - containing Na⁺ and SO₄²⁻ ions. 2. Determine the possible combinations: The possible combinations of these ions are: CaSO₄ and NaCl. 3. Check solubility: According to the solubility rules, CaSO₄ is slightly soluble (which can still form a precipitate), and NaCl is soluble. 4. Write down the balanced chemical equation: When these two solutions are combined, a precipitate will form as CaSO₄. \[ Ca^{2+}(aq) + SO_{4}^{2-}(aq) \rightarrow CaSO_4(s) \]
04

d. K₂S(aq) + Ni(NO₃)₂(aq)

1. Analyze the chemical equation: The given solutions are: K₂S(aq) - containing K⁺ and S²⁻ ions, and Ni(NO₃)₂(aq) - containing Ni²⁺ and NO₃⁻ ions. 2. Determine the possible combinations: The possible combinations of these ions are: KNO₃ and NiS. 3. Check solubility: According to the solubility rules, KNO₃ is soluble, and NiS is insoluble. 4. Write down the balanced chemical equation: When these two solutions are combined, a precipitate will form as NiS. \[ Ni^{2+}(aq) + S^{2-}(aq) \rightarrow NiS(s) \] In summary, precipitates will form in the following cases: - b. Al(NO₃)₃(aq) + Ba(OH)₂(aq) will form Al(OH)₃ - c. CaCl₂(aq) + Na₂SO₄(aq) will form CaSO₄ - d. K₂S(aq) + Ni(NO₃)₂(aq) will form NiS

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.

Solubility Rules
Solubility rules are guidelines that help determine whether a substance will dissolve in water, forming an aqueous solution, or remain as a solid, forming a precipitate. These rules are essential in predicting the outcomes of chemical reactions, especially when dealing with ionic compounds in water.

- **Common soluble substances:** Nitrates ( O_3^-), acetates, and ammonium compounds are typically soluble. Most alkali metal salts, such as those containing sodium or potassium, also dissolve in water.
- **Common insoluble substances:** Carbonates, phosphates, sulfides, and hydroxides often do not dissolve, with exceptions like calcium and barium hydroxides being slightly soluble. By using these rules, you can predict if a combination of ions will stay soluble in an aqueous solution or form a precipitate.

For example, in one of the cases from the exercise, we found that Al(OH)_3 is an insoluble hydroxide, which formed a precipitate when Al(NO_3)_3 and Ba(OH)_2 were mixed.
Chemical Equation Balancing
Balancing chemical equations is crucial to accurately represent chemical reactions.Every balanced equation reflects the Law of Conservation of Mass, where the number of each type of atom is conserved from reactants to products.

A balanced equation provides the correct stoichiometric quantities for each substance involved and predicts how much product will form or how much reagent is needed. To balance a chemical equation:
  • Write the unbalanced equation based on the reactant and product formulas.
  • Count the number of atoms of each element on both sides.
  • Add coefficients (numbers in front of compounds) to match the number of atoms on each side.
Using this approach, the reaction between Al(NO_3)_3 and Ba(OH)_2 can be balanced as:\[ Al^{3+}(aq) + 3OH^{-}(aq) \rightarrow Al(OH)_3(s) \]
Here, one aluminum ion reacts with three hydroxide ions to form insoluble aluminum hydroxide.
Aqueous Solutions
An aqueous solution is formed when a substance is dissolved in water. In chemistry, they are often used to carry out reactions due to water's excellent solvent properties. When ionic compounds dissolve, they dissociate into their component ions, which then freely move in the solution, allowing for reactions.

When iron(II) sulfate and potassium chloride are mixed, they form an aqueous solution containing Fe²⁺, SO₄²⁻, K⁺, and Cl⁻ ions. These ions can move around freely in the solution, but they do not produce any new insoluble compounds, as shown by the solubility rules.
- **Formation of aqueous solutions** allows reactions to occur more readily due to increased ion mobility.
- **Dissociation**: The process in which ionic compounds separate into their individual ions in solution is key to the behavior of ionic reactions.
Chemical Reactions Analysis
Analyzing chemical reactions involves not only recognizing reactants and products but also understanding how they change and what new substances might form. This requires a comprehension of ionic transformations and the formation of any precipitates.

- **Combining Ions:** When aqueous solutions of CaCl_2 and Na_2SO_4 are mixed, new combinations of ions can occur, such as Ca^{2+} and SO_4^{2-}, which precipitate as CaSO_4(s).
- **Predicting Outcomes:** By understanding solubility rules and ion reactivity, one can predict when a reaction will result in a precipitate, like NiS from Ni(NO_3)_2 and K_2S.
- **Reaction Pathways:** It's crucial to identify what is occurring in a solution at a molecular level, ensuring recognition of the solid precipitates forming from soluble ions. Be mindful of which ions remain in solution as spectator ions, as they don't participate in the chemical change.

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

How would you prepare \(1.00 \mathrm{~L}\) of a \(0.50-M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" ( \(18 \mathrm{M}\) ) sulfuric acid b. HCl from "concentrated" (12 \(M\) ) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO \(_{3}\) from "concentrated" ( \(16 M\) ) reagent e. Sodium carbonate from the pure solid

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 \(1.0\) 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. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

Complete and balance each acid-base reaction. a. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) Contains three acidic hydrogens b. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) Contains two acidic hydrogens c. \(\mathrm{H}_{2} \mathrm{Se}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\) Contains two acidic hydrogens d. \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) Contains two acidic hydrogens

Consider the reaction between sodium metal and fluorine \(\left(\mathrm{F}_{2}\right)\) gas to form sodium fluoride. Using oxidation states, how many electrons would each sodium atom lose, and how many electrons would each fluorine atom gain? How many sodium atoms are needed to react with one fluorine molecule? Write a balanced equation for this reaction.

A \(30.0-\mathrm{mL}\) sample of an unknown strong base is neutralized after the addition of \(12.0 \mathrm{~mL}\) of a \(0.150 \mathrm{M} \mathrm{HNO}_{3}\) solution. If the unknown base concentration is \(0.0300 M\), give some possible identities for the unknown base.

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