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

For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, (b) \(\mathrm{PbF}_{2}\), (c) \(\mathrm{AuCl}_{3}\) (d) \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (e) \(\mathrm{CuBr}\).

Short Answer

Expert verified
The net ionic equations for the reaction of slightly soluble salts with a strong acid (HCl) are: (a) \( MnS (s) + 2H^{+} (aq) \rightarrow Mn^{2+} (aq) + H_{2}S (g) \) (b) \( PbF_{2} (s) + 2H^{+} (aq) \rightarrow Pb^{2+} (aq) + 2HF (aq) \) (c) No net ionic equation for \(\mathrm{AuCl}_{3}\) as both products are in solution. (d) \( Hg_{2}C_{2}O_{4} (s) + 2H^{+} (aq) \rightarrow Hg_{2}^{2+} (aq) + H_{2}C_{2}O_{4} (aq) \) (e) \( Br^{-} (aq) + H^{+} (aq) \rightarrow HBr (aq) \)

Step by step solution

01

(Identify the Strong Acid)

To write the net ionic equations, we first need to select a strong acid. A common strong acid is hydrochloric acid (HCl). We will use HCl as our strong acid for all the reactions.
02

(Reaction of MnS with HCl)

When manganese(II) sulfide (MnS) reacts with hydrochloric acid (HCl), it forms manganese(II) chloride (MnCl2) and hydrogen sulfide (H2S): \( MnS + 2HCl \rightarrow MnCl_{2} + H_{2}S \) To write the net ionic equation, we first write down the ions formed: \( MnS (s) \rightarrow Mn^{2+} (aq) + S^{2-} (aq) \) \( HCl (aq) \rightarrow H^{+} (aq) + Cl^{-} (aq) \) Then, we cancel out the spectator ions, which are the ions that do not participate in the reaction. In this case, the spectator ion is Cl-. The net ionic equation is: \( MnS (s) + 2H^{+} (aq) \rightarrow Mn^{2+} (aq) + H_{2}S (g) \)
03

(Reaction of \(\mathrm{PbF}_{2}\) with HCl)

When lead(II) fluoride (PbF2) reacts with hydrochloric acid (HCl), it forms lead(II) chloride (PbCl2) and hydrogen fluoride (HF): \( PbF_{2} + 2HCl \rightarrow PbCl_{2} + 2HF \) To write the net ionic equation, we first write down the ions formed: \( PbF_{2} (s) \rightarrow Pb^{2+} (aq) + 2F^{-} (aq) \) \( HCl (aq) \rightarrow H^{+} (aq) + Cl^{-} (aq) \) Canceling out the spectator ions (Cl-), the net ionic equation is: \( PbF_{2} (s) + 2H^{+} (aq) \rightarrow Pb^{2+} (aq) + 2HF (aq) \)
04

(Reaction of \(\mathrm{AuCl}_{3}\) with HCl)

Gold(III) chloride (AuCl3) is soluble in water, so when it reacts with hydrochloric acid (HCl) there is no reaction since both products are in solution. Thus, there is no net ionic equation.
05

(Reaction of \(\mathrm{Hg}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) with HCl)

When mercury(I) oxalate (Hg2C2O4) reacts with hydrochloric acid (HCl), it forms mercury(I) chloride (Hg2Cl2) and oxalic acid (H2C2O4): \( Hg_{2}C_{2}O_{4} + 2HCl \rightarrow Hg_{2}Cl_{2} + H_{2}C_{2}O_{4} \) To write the net ionic equation, we first write down the ions formed: \( Hg_{2}C_{2}O_{4} (s) \rightarrow Hg_{2}^{2+} (aq) + C_{2}O_{4}^{2-} (aq) \) \( HCl (aq) \rightarrow H^{+} (aq) + Cl^{-} (aq) \) Canceling out the spectator ions (Cl-), the net ionic equation is: \( Hg_{2}C_{2}O_{4} (s) + 2H^{+} (aq) \rightarrow Hg_{2}^{2+} (aq) + H_{2}C_{2}O_{4} (aq) \)
06

(Reaction of \(\mathrm{CuBr}\) with HCl)

When copper(I) bromide (CuBr) reacts with hydrochloric acid (HCl), it forms copper(I) chloride (CuCl) and hydrobromic acid (HBr): \( CuBr + HCl \rightarrow CuCl + HBr \) To write the net ionic equation, we first write down the ions formed: \( CuBr (s) \rightarrow Cu^{+} (aq) + Br^{-} (aq) \) \( HCl (aq) \rightarrow H^{+} (aq) + Cl^{-} (aq) \) Canceling out the spectator ions (Cu+), the net ionic equation is: \( Br^{-} (aq) + H^{+} (aq) \rightarrow HBr (aq) \)

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.

Slightly Soluble Salts
In chemistry, slightly soluble salts are compounds that do not dissolve completely in water. When placed in water, these salts form a dynamic equilibrium between the dissolved ions and the solid salt.
However, some of the salt can dissolve to a minor extent. This is an important concept because the partial dissolution allows for certain reactions to occur, such as reactions with strong acids.
For example, manganese(II) sulfide (MnS) is a slightly soluble salt. When it reacts with a strong acid, like
hydrochloric acid, it partially dissolves, allowing it to participate in a chemical reaction.
  • MnS(s) dissociates into Mn2+(aq) and S2-(aq), although only a small amount integrates into the solution.
  • Understanding the limited solubility aids in predicting possible outcomes when going through chemical reactions in a solution.
Reaction with Strong Acid
A strong acid, such as hydrochloric acid (HCl), is one that completely ionizes in water. This characteristic allows strong acids to effectively interact with slightly soluble salts, influencing the rate and extent of reactions.
The strong acid donates protons (H+) which can then react with the negatively charged ions of a slightly soluble salt.
This concept takes center stage in various reactions where the salt reacts and new molecules form along with a potential evolution of gases or precipitates, like hydrogen sulfide from MnS and HCl.
  • Example reaction: MnS(s) + 2H+(aq) → Mn2+(aq) + H2S(g).
  • Reactions like these require breaking down the involved compounds into their ionic constituents to properly map out the changes occurring in the reaction.
Spectator Ions
Spectator ions are ions that appear on both sides of a chemical equation without undergoing any change. They do not participate in the actual chemical bonding or reaction process and can be removed while writing net ionic equations.
In our exercises with slightly soluble salts and strong acids, recognizing and removing spectator ions is crucial to solving net ionic equations.
Typically, these ions are the cations and anions from the strong acid which remain in the aqueous phase and do not get consumed in the reactive process.
  • For instance, when manganese(II) sulfide reacts with hydrochloric acid, the chloride ions (Cl-) act as spectator ions.
  • Removing spectator ions accordingly yields the concise net ionic equation: MnS(s) + 2H+(aq) → Mn2+(aq) + H2S(g).
Chemical Reactions in Aqueous Solutions
Chemical reactions in aqueous solutions involve substances that are either dissolved in water or capable of interacting under these conditions. This encompasses a vast array of reactions including precipitation, acid-base neutralization, and redox reactions.
The examination of chemical reactions in aqueous environments often involves identifying the products of reactions between ions. The solvent, usually water, facilitates the dissociation of compounds, making these reactions more manageable to study.
  • These reactions include experiments like that of PbF2 with HCl where lead ions remain in solution while fluoride ions undergo reactions to form new products.
  • Understanding the behavior of ions in aqueous solutions clarifies mechanisms and outcomes of such reactions efficiently.

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

Assume that \(30.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of a weak base B that accepts one proton is titrated with a \(0.10 \mathrm{M}\) solution of the monoprotic strong acid HA. (a) How many moles of HA have been added at the equivalence point? (b) What is the predominant form of B at the equivalence point? (c) Is the pH 7, less than 7, or more than 7 at the equivalence point? (d) Which indicator, phenolphthalein or methyl red, is likely to be the better choice for this titration?

An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered off, the \(\mathrm{pH}\) is adjusted to about 1 and \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled in; a precipitate again forms. After filtering off this precipitate, the pH is adjusted to 8 and \(\mathrm{H}_{2} \mathrm{~S}\) is again added; no precipitate forms. No precipitate forms upon addition of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). (See Figure 7.23.) The remaining solution shows a yellow color in a flame test (see Figure 7.22). Based on these observations, which of the following compounds might be present, which are definitely present, and which are definitely absent: CdS, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{HgO}_{3}, \mathrm{ZnSO}_{4}, \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\), and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) ?

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid \(\mathrm{HA}\) and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is \(9.2\). (a) Write the chemical equation and equilibrium-constant expression for the reaction between HA and B. (b) If \(K_{a}\) for \(\mathrm{HA}\) is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between \(\mathrm{HA}\) and \(\mathrm{B}\) ? (c) What is the value of \(K_{\mathrm{b}}\) for \(\mathrm{B}\) ?

\( A buffer is prepared by adding \)20.0 \mathrm{~g}\( of sodium acetate \)\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\( to \)500 \mathrm{~mL}\( of a \)0.150 \mathrm{M}\( acetic acid \)\left(\mathrm{CH}_{3} \mathrm{COOH}\right)$ solution. (a) Determine the pH of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the

A buffer contains 0.15 mol of propionic acid $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)$ and 0.10 mol of sodium propionate \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COONa}\right)\) in 1.20 \(\mathrm{L}\) . (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.01 \(\mathrm{mol}\) of \(\mathrm{NaOH}\) ? (c) What is the pH of the buffer after the addition of 0.01 \(\mathrm{mol}\) of \(\mathrm{HI} ?\)

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