Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 120

What reagents would you employ to separate the following pairs of ions in solution: (a) \(\mathrm{Na}^{+}\) and \(\mathrm{Ba}^{2+},\) (b) \(\mathrm{K}^{+}\) and \(\mathrm{Pb}^{2+},\) (c) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+} ?\)

Short Answer

Expert verified
For Na+ and Ba2+, use a sulphate ion (SO4^2-) reagent. For K+ and Pb2+, use an Iodate ion (IO3-) reagent. For Zn2+ and Hg2+, use hydrochloric acid (HCL) reagent.

Step by step solution

01

Separate Na+ and Ba2+

Sulphate ion (SO4^2-) can be used as the reagent. Barium ion will react with the sulphate ion to form Barium Sulphate (BaSO4) which is insoluble in water and forms a precipitate while Sodium ion will remain in the solution as Sodium Sulphate (Na2SO4). Equation: Ba2+(aq) + SO4^2-(aq) -> BaSO4(s).
02

Separate K+ and Pb2+

Iodate ion (IO3-) can be used as a reagent for this separation. Lead ion will react with Iodate ion to form insoluble Lead(II) Iodate (Pb(IO3)2) while Potassium will remain in solution as Potassium Iodate (KIO3). Equation: Pb2+(aq) + 2IO3-(aq) -> Pb(IO3)2(s).
03

Separate Zn2+ and Hg2+

Hydrochloric acid (HCL) can be used as reagent for this separation. Mercury ion will react with Chloride ions (formed due to dissociation of HCL) to form insoluble Mercury(II) Chloride (HgCl2), while Zinc will remain in solution as Zinc Chloride (ZnCl2). Equation: Hg2+ (aq) + 2Cl- (aq) -> HgCl2 (s).

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

How many milliliters of \(1.0 \mathrm{M} \mathrm{NaOH}\) must be added to a \(200 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NaH}_{2} \mathrm{PO}_{4}\) to make a buffer solution with a pH of \(7.50 ?\)

Calculate the molar solubility of \(\mathrm{Fe}(\mathrm{OH})_{2}\) in a solution buffered at (a) \(\mathrm{pH} 8.00,\) (b) \(\mathrm{pH} 10.00\).

Consider the ionization of the following acid-base indicator: $$ \operatorname{HIn}(a q) \Longrightarrow \mathrm{H}^{+}(a q)+\operatorname{In}^{-}(a q) $$ The indicator changes color according to the ratios of the concentrations of the acid to its conjugate base as described inSection \(16.5 .\) Show that the \(\mathrm{pH}\) range over which the indicator changes from the acid color to the base color is \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} \pm 1,\) where \(K_{\mathrm{a}}\) is the ionization constant of the acid.

Calculate the concentrations of \(\mathrm{Cd}^{2+}, \mathrm{Cd}(\mathrm{CN})_{4}^{2-}\) and \(\mathrm{CN}^{-}\) at equilibrium when \(0.50 \mathrm{~g}\) of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) dissolves in \(5.0 \times 10^{2} \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{NaCN}\).

Explain how an acid-base indicator works in a titration. What are the criteria for choosing an indicator for a particular acid-base titration?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

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