Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 15

A diprotic acid, \(\mathrm{H}_{2} \mathrm{~A}\), has the following ionization constants: \(K_{\mathrm{a}_{1}}=1.1 \times 10^{-3}\) and \(K_{\mathrm{a}_{2}}=2.5 \times 10^{-6} .\) To make up a buffer solution of \(\mathrm{pH} 5.80\), which combination would you choose: \(\mathrm{NaHA} / \mathrm{H}_{2} \mathrm{~A}\) or \(\mathrm{Na}_{2} \mathrm{~A} / \mathrm{NaHA} ?\)

Short Answer

Expert verified
To make a buffer solution of pH 5.80, the combination to choose would be \(\mathrm{Na}_{2}\mathrm{A} / \mathrm{NaHA}\).

Step by step solution

01

Identify the acid and base pairs

For a diprotic acid, the first ionization step is \(\mathrm{H}_{2}\mathrm{A} \rightarrow \mathrm{HA}^- + \mathrm{H}^+\), forming the pair \(\mathrm{H}_{2}\mathrm{A} / \mathrm{HA}^- \). And the second step is \(\mathrm{HA}^- \rightarrow \mathrm{A}^{2-} + \mathrm{H}^+\), forming the pair \(\mathrm{HA}^- / \mathrm{A}^{2-}\). Therefore, \(\mathrm{NaHA} / \mathrm{H}_{2} \mathrm{A}\) gives us the pair for the first dissociation, and \(\mathrm{Na}_{2} \mathrm{A} / \mathrm{NaHA}\) for the second.
02

Identify appropriate buffer range

Each ${K}_a$ relates to a particular ionization step. At ${K}_{a1}$, the buffer pair would create a pH range within one pH unit of ${pK}_{a1}$. Same for ${K}_{a2}$, within one pH unit of ${pK}_{a2}$. ${pK}_{a} = -log{K}_{a}$, so we calculate ${pK}_{a1}$ and ${pK}_{a2}$ to see in which buffer range the pH 5.80 falls.
03

Calculate $pK_{a1}$ and $pK_{a2}$

${pK}_{a1} = -log(1.1 \times 10^{-3}) = 3.00$ and ${pK}_{a2} = -log(2.5 \times 10^{-6}) = 5.60$
04

Compare $pK_{a1}$ and $pK_{a2}$ to given pH

Given pH is 5.80. Comparing this with ${pK}_{a1} = 3.00$ and ${pK}_{a2} = 5.60$, it is clear that the pH 5.80 is within one pH unit of ${pK}_{a2}$ (between 4.60 and 6.60), which corresponds to the second ionization, so the buffer pair is \(\mathrm{Na}_{2}\mathrm{A} / \mathrm{NaHA}\)

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

Which of these ionic compounds will be more soluble in acid solution than in water: (a) \(\mathrm{BaSO}_{4}\) (b) \(\mathrm{PbCl}_{2},\) (c) \(\mathrm{Fe}(\mathrm{OH})_{3}\) (d) \(\mathrm{CaCO}_{3}\) ? Explain.

Calculate \(x\), the number of molecules of water in oxalic acid hydrate, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot x \mathrm{H}_{2} \mathrm{O},\) from the following data: \(5.00 \mathrm{~g}\) of the compound is made up to exactly \(250 \mathrm{~mL}\) solution and \(25.0 \mathrm{~mL}\) of this solution requires \(15.9 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) solution for neutralization.

The \(\mathrm{p} K_{\mathrm{a}}\) of butyric acid (HBut) is 4.7 . Calculate \(K_{\mathrm{b}}\) for the butyrate ion (But \(^{-}\) ).

What are the criteria for choosing an indicator for a particular acid-base titration?

In a group 1 analysis, a student obtained a precipitate containing both \(\mathrm{AgCl}\) and \(\mathrm{PbCl}_{2}\). Suggest one reagent that would allow her to separate \(\operatorname{AgCl}(s)\) from \(\mathrm{PbCl}_{2}(s)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

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