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

Which of the following is conjugate acid of \(\mathrm{SO}_{4}^{2-} ?\) (a) \(\mathrm{HSO}_{4}^{-}\) (b) \(\mathbf{H}^{+}\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{SO}_{4}^{2-}\)

Short Answer

Expert verified
\((\mathrm{HSO}_{4}^{-})\) is the conjugate acid of \((\mathrm{SO}_{4}^{2-})\).

Step by step solution

01

Understanding Conjugate Acid-Base Pairs

A conjugate acid is formed when a base gains a proton (H+). A conjugate base is formed when an acid loses a proton. In this context, the conjugate acid of sulfate \((\mathrm{SO}_{4}^{2-})\) will be the species that results from the base (sulfate ion) gaining a proton.
02

Identify the Conjugate Acid

In order to find the conjugate acid of \((\mathrm{SO}_{4}^{2-})\), we need to look for a species that is the same as the sulfate ion plus one proton (H+). As the sulfate ion has a -2 charge, adding one proton will result in a species with a -1 charge.
03

Review the Choices

Looking at the choices, \((\mathrm{HSO}_{4}^{-})\) is \((\mathrm{SO}_{4}^{2-})\) plus one proton, therefore \((\mathrm{HSO}_{4}^{-})\) is the conjugate acid of \((\mathrm{SO}_{4}^{2-})\). The other choices either do not involve adding a proton to the sulfate ion or are the same as the original ion.

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.

Acid-Base Reactions
Acid-base reactions are fundamental chemical processes where an acid and a base interact, commonly resulting in the production of a salt and water. At the heart of such reactions is the transfer of protons (H+). When an acid donates a proton, it becomes a conjugate base, and the substance accepting the proton is the conjugate acid. This forms a conjugate acid-base pair.

For example, when hydrogen chloride (HCl) reacts with water (H2O), HCl donates a proton to water, creating hydronium ions (H3O+) and chloride ions (Cl-). In this case, H2O functions as a base and HCl as an acid; hydronium is the conjugate acid of water, and chloride is the conjugate base of HCl.
Proton Transfer
Proton transfer is the movement of a proton from one species to another and is the cornerstone of acid-base chemistry. This transfer decides the direction of the reaction and the nature of the products formed.

In any acid-base reaction, a proton, which is simply a hydrogen ion (H+), is exchanged between molecules. The molecule that donates the proton is the acid, while the molecule that receives the proton is the base. For instance, in the exercise provided, the sulfate ion (SO42-) acts as a base that can accept a proton. Doing so forms the conjugate acid, bisulfate (HSO4-), which has one more proton than the original base.
Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products. It is important to note that while the concentrations remain constant at equilibrium, the reactions are still occurring, just at the same rate in both directions.

For an acid-base reaction, the equilibrium will involve both the forward reaction, where the acid donates a proton to the base, and the reverse reaction, where the conjugate acid, formed in the forward reaction, donates a proton back to the conjugate base. The strength of an acid or base, known as its equilibrium constant (Ka for acids and Kb for bases), will determine the position of the equilibrium and the proportions of conjugate acids and bases present at equilibrium.

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

What is the minimum concentration of \(\mathrm{SO}_{4}^{2-}\) required to precipitate \(\mathrm{BaSO}_{4}\) in a solution containing \(1 \times 10^{-4}\) mole of \(\mathrm{Ba}^{2+} ?\left(K_{i p}\right.\) for \(\left.\mathrm{BaSO}_{4}=4 \times 10^{-10}\right)\) (a) \(4 \times 10^{-10} \mathrm{M}\) (b) \(2 \times 10^{-10} \mathrm{M}\) (c) \(4 \times 10^{-6} \mathrm{M}\) (d) \(2 \times 10^{-3} \mathrm{M}\)

What is the percentage dissociation of \(0.1 \mathrm{M}\) solution of acetic acid? \(\left(K_{a}=10^{-5}\right)\) (a) \(10 \%\) (b) \(100 \%\) (c) \(1 \%\) (d) \(0.01 \%\)

If the equilibrium constant for the reaction, \(2 X Y \rightleftharpoons X_{2}+Y_{2}\) is 81 what is the value of equilibrium constant for the reaction $$ X Y \rightleftharpoons \frac{1}{2} X_{2}+\frac{1}{2} Y_{2} $$ (a) 81 (b) 9 (c) 6561 (d) \(40.5\)

What is \(\mathrm{pOH}\) of an aqueous solution with hydrogen ion concentration equal to \(3 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\) ? (a) \(9.47\) (b) \(4.52\) (c) \(12.69\) (d) \(11.69\)

Fill in the blanks in the given table with the appropriate choice.$$ \begin{array}{|c|c|c|} \hline \text { Species } & \text { Conjugate acid } & \text { Conjugate base } \\\ \hline \mathrm{HCO}_{3}^{-} & \rho & \mathrm{CO}_{3}^{2-} \\ \hline \mathrm{HSO}_{4}^{-} & \mathrm{H}_{2} \mathrm{SO}_{4} & q \\ \hline \mathrm{NH}_{3} & r & -s \\ \hline \mathrm{H}_{2} \mathrm{O} & t & \mathrm{OH}^{-} \\ \hline \end{array} $$ (a) \(\mathrm{H}_{2} \mathrm{CO}_{3} \quad \mathrm{SO}_{4}^{2-}\) \(\begin{array}{lll}\mathrm{NH}_{4}^{*} & \mathrm{NH}_{2}^{-} & \mathrm{H}_{3} \mathrm{O}^{*}\end{array}\) \(\begin{array}{lllll}\text { (b) } & \mathrm{HCO}_{3}^{-} & \mathrm{H}_{2} \mathrm{SO}_{3} & \mathrm{NH}_{2}^{*} & \mathrm{NH}_{4}^{*}\end{array}\) \(\mathrm{H}_{3} \mathrm{O}^{+}\) (c) \(\begin{array}{lllll}\mathrm{H}_{2} \mathrm{CO}_{3} & \mathrm{HSO}_{4}^{-} & \mathrm{NH}_{4}^{+} & \mathrm{NH}_{2}^{-} & \mathrm{H}_{2} \mathrm{O}\end{array}\) \(\begin{array}{lllll}\text { (d) } \mathrm{HCO}_{3}^{-} & \mathrm{H}_{2} \mathrm{SO}_{4} & \mathrm{NH}_{2}^{+} & \mathrm{NH}_{2}^{-} & \mathrm{OH}^{-}\end{array}\)

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