In your own words, define or explain the terms or symbols \((\mathrm{a}) \rightleftharpoons(\mathrm{b})[] ;(\mathrm{c})\) spectator ion; (d) weak acid.

Short Answer

Expert verified
'$(\mathrm{a}) \rightleftharpoons (\mathrm{b})[]$' indicates a dynamic equilibrium in a chemical reaction. A 'spectator ion' does not participate directly in the reaction but exists in the same form on both sides of the equation. The symbol '$(\mathrm{c})$' generally refers to a part of a reaction, but its meaning can vary. A 'weak acid' doesn't fully dissociate in water and generally the equilibrium favors the non-ionized form.

Step by step solution

01

Define or explain $(\mathrm{a}) \rightleftharpoons (\mathrm{b})[]$

$(a) \rightleftharpoons (b)[]$ is a symbol used in chemistry to indicate a dynamic equilibrium. This symbol shows that the forward and reverse reactions are happening at the same rate. In this equilibrium, $(a)$ and $(b)$ can be any reactants and products.
02

Define or explain 'spectator ion'

A 'spectator ion' is an ion that exists in the same form on both the reactant and product sides of a chemical equation. Thus, it does not directly participate in the reaction, but may be required for the reaction to occur, such as providing a charge balance.
03

Define or explain $(\mathrm{c})$

The symbol $(c)$ in the chemistry context normally refers to a part of a reaction or equation. The exact meaning can vary based on its location and the accompanying elements or compounds in the reaction.
04

Define or explain 'weak acid'

A 'weak acid' is an acid that does not completely dissociate into its ions in water. It has a relatively low acid dissociation constant (Ka), and usually the equilibrium favors the unionized form. Examples of weak acids include acetic acid (CH3COOH) and formic acid (HCOOH)

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.

Dynamic Equilibrium
Think of a dynamic equilibrium as a perfectly balanced seesaw that moves up and down without ever favoring one side. In the realm of chemistry, this concept refers to a state where the forward and reverse reactions occur at the same rate in a closed system. Imagine substances A and B are reacting to form C and D. At dynamic equilibrium, A and B transform into C and D just as quickly as C and D revert back to A and B.

The reactions are ongoing, much like a busy highway with cars (reactants and products) moving in opposite directions at equal rates. There's no net change in the concentration of reactants or products over time, hence the name 'dynamic' because it's an active, yet balanced, situation. This is crucial in chemistry because it helps us understand that at equilibrium, reactions haven't stopped; they are simply in a state of balance.
Spectator Ion
In chemistry, particularly in solutions and ionic compounds, not all players on the field are actively engaged in the game. 'Spectator ions' are like the fans in the stands; they're present at the event but don't influence the outcome of the game directly.

They are ions that remain unchanged on both sides of a chemical equation. Spectator ions do not participate in the actual chemical reaction, but their presence is important for maintaining electrical neutrality in the reaction mixture. For instance, in a reaction between sodium chloride and silver nitrate, Na⁺ and NO₃⁻ are spectator ions. They do not take part in the reaction but are essential parts of the ionic compounds involved.
Weak Acid
While strong acids are like a wrecking ball that completely breaks down into ions, a 'weak acid' is more like a door that only opens partway - it doesn't completely dissociate in water. These acids only partially release their hydrogen ions into the solution. Because not all of the acid molecules give up their hydrogen ions, the solution contains a mix of ions and undissociated acid molecules.

This incomplete dissociation is what we observe with acetic acid in vinegar or citric acid in citrus fruits. These weak acids have a taste that's sour, yet not as overwhelming as their stronger counterparts. The property that measures just how 'weak' they are is encapsulated in a unique value: the acid dissociation constant.
Acid Dissociation Constant
The acid dissociation constant, represented by the symbol Ka, is like a scorecard that tells us how strong or weak an acid is. The weaker the acid, the smaller the Ka value. This constant is a quantifiable measure of the extent to which an acid can donate hydrogen ions in an aqueous solution.

It's calculated from the concentrations of the products (the ions) divided by the concentration of the reactants (the undissociated acid), not including the solvent water. A larger value of Ka indicates that more hydrogen ions are released, signifying a stronger acid. For example, hydrochloric acid (HCl), a strong acid, has a much larger Ka than acetic acid (CH₃COOH), which is a weak acid.

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

When concentrated \(\mathrm{CaCl}_{2}(\mathrm{aq})\) is added to \(\mathrm{Na}_{2} \mathrm{HPO}_{4}(\mathrm{aq}),\) a white precipitate forms that is \(38.7 \%\) Ca by mass. Write a net ionic equation representing the probable reaction that occurs.

Assign oxidation states to the elements involved in the following reactions. Indicate which are redox reactions and which are not. (a) $\mathrm{MgCO}_{3}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow$ $\mathrm{Mg}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g})$ (b) $\mathrm{Cl}_{2}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Br}_{2}(\mathrm{aq})$ (c) $\mathrm{Ag}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow$ $\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NO}_{2}(\mathrm{g})$ (d) $2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{CrO}_{4}^{2-}(\mathrm{aq}) \longrightarrow \mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})$

What are the oxidizing and reducing agents in the following redox reactions? (a) \(5 \mathrm{SO}_{3}^{2-}+2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+} \longrightarrow\) \(5 \mathrm{SO}_{4}^{2-}+2 \mathrm{Mn}^{2+}+3 \mathrm{H}_{2} \mathrm{O}\) (b) \(2 \mathrm{NO}_{2}(\mathrm{g})+7 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+} \longrightarrow\) \(2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+2 \mathrm{H}_{2} \mathrm{O}\)

A piece of marble (assume it is pure \(\mathrm{CaCO}_{3}\) ) reacts with \(2.00 \mathrm{L}\) of \(2.52 \mathrm{M} \mathrm{HCl}\). After dissolution of the marble, a \(10.00 \mathrm{mL}\) sample of the resulting solution is withdrawn, added to some water, and titrated with 24.87 mL of 0.9987 M NaOH. What must have been the mass of the piece of marble? Comment on the precision of this method; that is, how many significant figures are justified in the result?

Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

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