Compare the strengths of the following pairs of acids: (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SeO}_{4}\), (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{3} \mathrm{AsO}_{4}\).

Short Answer

Expert verified
H2SeO4 is a stronger acid than H2SO4. H3AsO4 is a stronger acid than H3PO4.

Step by step solution

01

Compare H2SO4 and H2SeO4

In this case, both acids have the same structure, but the central atom differs. Sulfur is less electronegative than Selenium. However, Selenium is larger in size and more polarizable than Sulfur. As a result, H2SeO4 can deform its electron cloud more easily to stabilize the charge after donating a proton. Thus, H2SeO4 is a stronger acid than H2SO4.
02

Compare H3PO4 and H3AsO4

Again, in these acids, the structure is similar but the central atom is different. Phosphorous is less electronegative and less polarizable than Arsenic. Considering these factors, H3AsO4 will be a stronger acid as it can more easily deform its electron cloud to stabilize charge after proton donation.

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!

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 \(0.400 M\) formic acid (HCOOH) solution freezes at \(-0.758^{\circ} \mathrm{C}\). Calculate the \(K_{\mathrm{a}}\) of the acid at that temperature. (Hint: Assume that molarity is equal to molality. Carry your calculations to three significant figures and round off to two for \(K_{\mathrm{a}}\).)

Calculate the \(\mathrm{pH}\) of a \(0.20 \mathrm{M} \mathrm{NaHCO}_{3}\) solution. (Hint: As an approximation, calculate hydrolysis and ionization separately first, followed by partial neutralization.)

HA and \(\mathrm{HB}\) are both weak acids although \(\mathrm{HB}\) is the stronger of the two. Will it take a larger volume of a \(\begin{array}{llll}0.10 & M & \text { NaOH } & \text { solution to } & \text { neutralize } & 50.0 & \text { mL }\end{array}\) of \(0.10 M\) HB than would be needed to neutralize \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) HA?

What is the \(\mathrm{pH}\) of \(250.0 \mathrm{~mL}\) of an aqueous solution containing \(0.616 \mathrm{~g}\) of the strong acid trifluoromethane sulfonic acid (CF \(_{3} \mathrm{SO}_{3} \mathrm{H}\) )?

You are given two beakers, one containing an aqueous solution of strong acid (HA) and the other an aqueous solution of weak acid (HB) of the same concentration. Describe how you would compare the strengths of these two acids by (a) measuring the \(\underline{p H},\) (b) measuring electrical conductance, (c) studying the rate of hydrogen gas evolution when these solutions are reacted with an active metal such as \(\mathrm{Mg}\) or \(\mathrm{Zn}\).

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