Find study content
Learning Materials

Discover learning materials by subject, university or textbook.

Explanations

Textbooks

Exams
All Subjects

Anthropology

Archaeology

Architecture

Art and Design

Bengali

Biology

Business Studies

Greek

Chemistry

Chinese

Combined Science

Computer Science

Economics

Engineering

English

English Literature

Environmental Science

French

Geography

German

History

Hospitality and Tourism

Human Geography

Italian

Japanese

Law

Macroeconomics

Marketing

Math

Media Studies

Medicine

Microeconomics

Music

Nursing

Nutrition and Food Science

Physics

Polish

Politics

Psychology

Religious Studies

Sociology

Spanish

Sports Sciences

Translation

All Subjects

Biology

Business Studies

Chemistry

Combined Science

Computer Science

Economics

English

Environmental Science

Geography

History

Math

Physics

Psychology

Sociology

EXAM TYPES

GCSE

IGCSE

AS

A Level

International A Level

University Admissions Tests

GCSE SUBJECTS

GCSE 1

GCSE 2

GCSE 3

GCSE 4

GCSE 5

SOME-TEXT

IGCSE SUBJECTS

IGCSE 1

AS SUBJECTS

AS 1

A Level SUBJECTS

A Level 1

International A Level SUBJECTS

International A Level 1

University Admissions Tests SUBJECTS

University Admissions Tests 1
Features
Features

Discover all of these amazing features with a free account.

Flashcards

Vaia AI

Notes

Study Plans

Study Sets
What’s new?

Flashcards
Study your flashcards with three learning modes.

Study Sets
All of your learning materials stored in one place.

Notes
Create and edit notes or documents.

Study Plans
Organise your studies and prepare for exams.
Resources
Discover

All the hacks around your studies and career - in one place.

Find a job

Find a degree

Student Deals

Magazine

Mobile App
Featured

Magazine
Trusted advice for anyone who wants to ace their studies & career.

Job Board
The largest student job board with the most exciting opportunities.

Vaia Deals
Verified student deals from top brands.

Our App
Discover our mobile app to take your studies anywhere.

Go to App

Learning Materials

Features

Discover

Chapter 8: QCE (page 182)

Using activities, find $[{Ag}^{+}]$ in $0.060 MKSCN$ saturated with $AgSCN (s)$

Short Answer

Expert verified

The concentration of ${Ag}^{+}$ ion in 0.060MKSCN is $2.9 \times 10^{- 11} M$

Step by step solution

01

Define the formula for activity species.

Any species' activity value is calculated by multiplying its concentration by its activity coefficient.

For a species C, the activity is given by

$A_{c} = [C] γ_{c}$

where,

A - Activity

[c] - Concentration

- Activity coefficient.

02

Calculate the activity species.

Given

$0.060 {MKSCN}^{- 1}$ saturated with ${AgSCN}_{(s)}$

The ionic strength $(μ)$ is $0.060 M$ from KSCN,

The solubility product expression

$μ = 0.060 M K_{SP} = [{Ag}^{+}] γ_{{Ag}^{+}} [{SCN}^{-}] γ_{SCN} = 1.1 \times 10^{- 12} At and γ_{SCN} = 0.80 γ_{A g^{-}} = 0.79$

Substitute the values of concentration and activity coefficients into the solubility product expression

$K_{SP} = [{Ag}^{+}] (0.79) [0.060] (0.80) = 1.1 \times 10^{- 12} [{Ag}^{+}]$

The concentration of ${Ag}^{+}$ ion in $0.060 MKSCN is 2.9 \times 10^{- 11} M$

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

15.Activity coefficient of a neutral molecule. We use the approximation

that the activity coefficient $(γ)$ of neutral molecules is 1.00.A more accurate relation is $logγ = kμ$ , where $μ$ is ionic strengthand $k \approx 0.11$ for ${NH}_{3}$ and ${CO}_{2}$ and $k \approx 0.2$ for organic molecules.With activity coefficients for $HA, A^{-}$ ,and $H^{+}$ , predict the value ofthe quotient below for benzoic acid $^{(HA = C_{6} H_{5} {CO}_{2} H)}$ The observed quotient is $^{0.63 \pm 0.03}$ .

Concentration quotient $= \frac{\frac{[H^{+}] [A^{-}]}{[HA]} (at μ = 0)}{\frac{[H^{+}] [A^{-}]}{[HA]} (at μ = 0.1 M)}$

Ion pairing. As in Problem $8 - 30$ , find the concentration, ionic strength, and ion pair fraction in localid="1654942556135" $0.025 F {Na}_{2} {SO}_{4}$ . Assume that the size of the ${NaSO}^{-}_{4} = 500 pm$

Write the charge and mass balances for dissolving ${CaF}_{2}$ in water if the reactions are

role="math" localid="1654770961556" ${CaF}_{2} (s) ٟ {Ca}^{2 +} + 2 F {Ca}^{2 +} + H_{2} O ٟ {CaOH}^{+} + H^{+} {Ca}^{2 +} F ؚ {CaF}^{+} {CaF}_{2} (s) ؚ {CaF}_{2} (aq) F + H^{+} ؚ HF (aq) HF (aq) + F ؚ {HF}_{2}$

Assuming complete dissociation of the salts, calculate the ionic strength of

(a) $0.2 mM K {NO}_{3}$

(b) $0.2 mM {Cs}_{2} {CrO}_{4}$

(c) $0.2 m {MMgCl}_{2}$ plus $0.3 mM {AlCl}_{3}$

The equilibrium constant for dissolution of a non ionic compound, such as diethyl ether $({CH}_{3} {CH}_{2} {OCH}_{2} {CH}_{3})$ , in water can be written

$ether (I) ֏ ether (aq) K = [ether (aq)] γ_{ether}$

At low ionic strength, role="math" localid="1654837696819" $γ \approx 1$ for neutral compounds. At high ionic strength, most neutral molecules can be salted out of aqueous solution. That is, when a high concentration typically $(> 1 M)$ of a salt such as $NaCI$ is added to aqueous solution, neutral molecules usually becomes less soluble. Does the activity coefficient, $γ_{ether}$ increases or decreases at high ionic strength?

See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Physical 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