Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 45

Explain why \(\left[\mathrm{PO}_{4}^{3-}\right]\) in \(1.00 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) is not simply \(\frac{1}{3}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\) but much, much less than \(\frac{1}{3}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\)

Short Answer

Expert verified
The [PO4^3-] is not simply 1/3 [H3O+] because H3PO4 is a polyprotic acid and undergoes stepwise dissociation. Each additional step of dissociation becomes less likely, resulting in a much lower concentration of PO4^3-, the product of the third dissociation step, compared to the H3O+ concentration.

Step by step solution

01

Understanding phosphoric acid and its dissociation

Phosphoric acid, H3PO4, is a polyprotic acid. This means it has more than one hydrogen atom it can lose when it donates a proton. The dissociation of phosphoric acid is a three step process, each step representing the donation of one proton:\n1. H3PO4 -> H2PO4^- + H+2. H2PO4^- -> HPO4^2- + H+3. HPO4^2- -> PO4^3- + H+It should be noted that each step above is an equilibrium reaction and has its own equilibrium constant. Thus, the dissociation process is not complete and all species exist simultaneously in the solution.
02

Why is [PO4^3-] not simply 1/3 [H3O+] ?

It is tempting to believe that the concentration of PO43- would be one third the concentration of H3O+ since phosphoric acid 'donates' three protons. However, this is not the case. The reason lies in the stepwise dissociation process where each step becomes less likely as the phosphate ion becomes more negatively charged and therefore less likely to donate a proton. This results in a lesser concentration of PO43- compared to H3O+.
03

Understanding of pH and dissociation steps

This can be further understood by considering pH. For most acids, the second and third dissociations do not greatly contribute to [H3O+]. Therefore, even as [H3O+] might be high, [PO4^3-] (which is product of third dissociation step) is much less. Hence, [PO4^3-] is not simply 1/3 [H3O+], but much, much less than 1/3 [H3O+].

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.

Start your free trial

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

Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\mathrm{pH}\) in saturated \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\) which contains \(3.9 \mathrm{g} \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) per \(100 \mathrm{mL}\) of solution.

The equilibria \(\mathrm{OH}^{-}+\mathrm{HClO} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{ClO}^{-}\) and \(\mathrm{ClO}^{-}+\mathrm{HNO}_{2} \longrightarrow \mathrm{HClO}+\mathrm{NO}_{2}^{-}\) both lie to the right. Which of the following is a list of acids ranked in order of decreasing strength? (a) \(\mathrm{HClO}>\mathrm{HNO}_{2}>\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{ClO}^{-}>\mathrm{NO}_{2}^{-}>\mathrm{OH}^{-}\) (c) \(\mathrm{NO}_{2}^{-}>\mathrm{ClO}^{-}>\mathrm{OH}\) (d) \(\mathrm{HNO}_{2}>\mathrm{HClO}>\mathrm{H}_{2} \mathrm{O}\) (e) none of these

Predict which is the stronger acid: (a) \(\mathrm{HClO}_{2}\) or \(\mathrm{HClO}_{3} ;(\mathrm{b}) \mathrm{H}_{2} \mathrm{CO}_{3}\) or \(\mathrm{HNO}_{2} ;(\mathrm{c}) \mathrm{H}_{2} \mathrm{SiO}_{3}\) or \(\mathrm{H}_{3} \mathrm{PO}_{4}\) Explain.

Of the following, the amphiprotic ion is (a) \(\mathrm{HCO}_{3}^{-}\) (b) \(\mathrm{CO}_{3}^{2-} ;\) (c) \(\mathrm{NH}_{4}^{+} ;\) (d) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} ;\) (e) \(\mathrm{ClO}_{4}^{-}\).

\(3.00 \mathrm{mol}\) of calcium chlorite is dissolved in enough water to produce 2.50 L of solution. \(K_{\mathrm{a}}=2.9 \times 10^{-8}\) for \(\mathrm{HClO}\), and \(K_{\mathrm{a}}=1.1 \times 10^{-2}\) for \(\mathrm{HClO}_{2}\). Compute the pH of the solution.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation
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