1 PRECIPITATION REACTIONS Solubility of Salts Section 18.4.

Slides:



Advertisements
Similar presentations
Notes: Equilibrium: Le Châtelier’s Principle (18.1 & 18.2)
Advertisements

Solubility Equilibria AP Chemistry
Author: J R Reid Solubility KsKs Common Ion Effect.
Chapter 19 - Neutralization
Solubility Equilibrium
Monday, April 11 th : “A” Day Agenda  Homework questions/collect  Finish section 14.2: “Systems At Equilibrium”  Homework: Section 14.2 review, pg.
Solubility. Definition Q. How do you measure a compound’s solubility? A. The amount of that compound that will dissolve in a set volume of water. This.
CHEMISTRY 121/122 Solubility Equilibrium. What is a solution?  A solution is a mixture in which a solid has been dissolved into a liquid, usually water.
Lecture 62/2/05. Solubility vs. Solubility constant (K sp ) BaSO 4 (s) ↔ Ba 2+ (aq) + SO 4 2- (aq) K sp = [Ba 2+ ][SO 4 2- ] Ba 2+ (aq) + SO 4 2- (aq)
Lecture 71/31/07. Solubility vs. Solubility constant (K sp ) What is the difference? BaSO 4 (s) ⇄ Ba 2+ (aq) + SO 4 2- (aq) Ba 2+ (aq) + SO 4 2- (aq)
Lecture 82/07/05 Seminar: Monday 2/7 4:30 Room 006 Eric Howe “Untangling sickle-cell anemia and the discovery of heterozygote protection to address students’
Lecture 102/11/06. Solubility PbCl 2 (s) ⇄ Pb 2+ (aq) + 2Cl - (aq)
Lecture 92/05/07 Seminar Today. If HCl is added slowly to a solution that is 0.10 M Pb 2+ and 0.01 M Ag +. K sp (AgCl) = 1.6 x K sp (PbCl 2 ) =
Solubility. Solubility “Insoluble” salts are governed by equilibrium reactions, and are really sparingly soluble. There is a dynamic equilibrium between.
Lecture 61/30/06 Seminar TODAY at 4. Effect of a catalyst Increases the rate at which reaction gets to equilibrium  Doesn’t change the equilibrium concentrations.
The Solubility Product Principle. 2 Silver chloride, AgCl,is rather insoluble in water. Careful experiments show that if solid AgCl is placed in pure.
Chapter 17 SOLUBILITY EQUILIBRIA (Part II) 1Dr. Al-Saadi.
PRECIPITATION REACTIONS Chapter 19 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part.
Lecture 82/3/06. QUIZ 2 1. Ba(NO 3 ) 2 reacts with Na 2 SO 4 to form a precipitate. Write the molecular equation and the net ionic equation. 2. For the.
Lecture 72/1/06. Precipitation reactions What are they? Solubility?
1 Salt Solubility Chapter Solubility product constant K sp K sp Unitless Unitless CaF 2(s)  Ca 2+ (aq) + 2F - (aq) CaF 2(s)  Ca 2+ (aq) + 2F -
Solubility Product Constant
Solubility Equilibria
Solubility Product Constant
Solubility Equilibria. Write solubility product (K sp ) expressions from balanced chemical equations for salts with low solubility. Solve problems involving.
PRECIPITATION REACTIONS Chapter 17 Part 2 2 Insoluble Chlorides All salts formed in this experiment are said to be INSOLUBLE and form precipitates when.
A salt, BaSO4(s), is placed in water
Ksp and Solubility Equilibria
© 2009, Prentice-Hall, Inc. Solubility of Salts (Ksp) Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+
Section 4: Solubility Equilibrium. Objectives Explain what is meant by solubility product constants, and calculate their values. Calculate solubilities.
Solubility Equilibrium
Chapter 16 Lesson 1 Solubility and Complex Ion Equilibria.
Chemistry 102(01) Spring 2008 Instructor: Dr. Upali Siriwardane
Copyright Sautter SOLUBILITY EQUILIBRIUM Solubility refers to the ability of a substance to dissolve. In the study of solubility equilibrium we.
Solubility and Complex-ion Equilibria. 2 Solubility Equilibria Many natural processes depend on the precipitation or dissolving of a slightly soluble.
11111 Chemistry 132 NT The most difficult thing to understand is the income tax. Albert Einstein.
Solubility Equilibria
Chapter 18 Reaction Rates and Equilibrium 18.4 Solubility Equilibrium
Aqueous Equilibria Chapter 16 Solubility Equilibria.
Hannah Nirav Joe ↔. Big Ideas You will be able to Understand what K sp is Find K sp from a reaction.
Solubility Equilibrium Chapter 10. What happens to an insoluble salt when it is added to some water? What would you see? What does it mean to be “insoluble”?
Solutions and Equilibrium. Heterogeneous Equilibrium Reaction occurs in more than one phase K EQ = Products Reactants Solids and liquids are not included.
Le Chatelier’s Principle
Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Solubility Equilibrium TEXT REFERENCE Masterton and Hurley Chapter 16.1.
Solubility Equilibrium Solubility Product Constant Ionic compounds (salts) differ in their solubilities Most “insoluble” salts will actually dissolve.
1 PRECIPITATION REACTIONS Solubility of Salts Section 18.4.
John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend Chapter 18 Principles.
1 Solubility Equilibria Dissolution M m X x (s)  m M n+ (aq) + x X y- (aq) Precipitation m M n+ (aq) + x X y- (aq)  M m X x (s) For a dissolution process,
Solubility Lesson 6 Changing solubility/Common Ion Effect.
Chapter 16 Lesson 2 Solubility and Complex Ion Equilibria.
Complex Ion Equilibria and Solubility A complex ion can increase the solubility of a salt. Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) K f = [Ag(NH.
DO NOW: What is dissolution. What is precipitation
…concentrations are ________ constant …forward and reverse rates are ________ equal At Equilibrium…
SOLUBILITY I. Saturated Solution BaSO 4(s)  Ba 2+ (aq) + SO 4 2- (aq) Equilibrium expresses the degree of solubility of solid in water. Ksp = solubility.
CHE1102, Chapter 17 Learn, 1 Chapter 17 Solubility and Simultaneous Equilibria.
1 PRECIPITATION REACTIONS Solubility of Salts Section 18.4.
Solubility Equilibria.  Write a balanced chemical equation to represent equilibrium in a saturated solution.  Write a solubility product expression.
Solubility Constant (Ksp). © 2009, Prentice-Hall, Inc. Solubility of Salts (Ksp) Consider the equilibrium that exists in a saturated solution of BaSO.
SOLUBILITY – The maximum amount of solute that will dissolve in a specific amount of solvent EQUILIBRIA WITH SALTS SATURATED – A solution where the solid.
Solubility Equilibria Will it all dissolve, and if not, how much will?
Chapter 16 Solubility Equilibria. Saturated solutions of “insoluble” salts are another type of chemical equilibria. Ionic compounds that are termed “insoluble”
Chapter 17 Solubility and Simultaneous Equilibria
The Solubility Product Constant (Ksp)
Chem 30: Solubility The Common Ion Effect.
Solubility & Simultaneous Equilibria Part I: Ksp and Solubility
Unit 4 Solutions solubility constant.
Solubility & Simultaneous Equilibria Part I: Ksp and Solubility
Name: Nissana Khan Class: L6 Module 2- Kinetics and Equilibria
Solubility & Simultaneous Equilibria Part I: Ksp and Solubility
Presentation transcript:

1 PRECIPITATION REACTIONS Solubility of Salts Section 18.4

2 Metal Chloride Salts These products are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions.

3 Analysis of Silver Group The products are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) = Ag + (aq) + Cl - (aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED.

4 Analysis of Silver Group AgCl(s) = Ag + (aq) + Cl - (aq) When solution is SATURATED, expt. shows that [Ag + ] = 1.67 x M. This is equivalent to the SOLUBILITY of AgCl. What is [Cl - ]? [Cl - ] is equivalent to the AgCl solubility.

5 AgCl(s) = Ag + (aq) + Cl - (aq) Saturated solution has [Ag + ] = [Cl - ] = 1.67 x M Use this to calculate K c K c = [Ag + ] [Cl - ] = (1.67 x )(1.67 x ) = 2.79 x

6 K c = [Ag + ] [Cl - ] = 2.79 x Because this is the product of “solubilities”, we call it K sp = solubility product constant See Table 18.2 and Appendix J Solubility Product Constant

7

8 Lead(II) Chloride PbCl 2 (s) = Pb 2+ (aq) + 2 Cl - (aq) K sp = 1.9 x 10 -5

9 Solution Solubility = [Pb 2+ ] = 1.30 x M [I - ] = ? [I - ] = 2 x [Pb 2+ ] = 2.60 x M Solubility of Lead(II) Iodide Consider PbI 2 dissolving in water PbI 2 (s) = Pb 2+ (aq) + 2 I - (aq) Calculate K sp if solubility = M

10 Calculate K sp K sp = [Pb 2+ ] [I - ] 2 = X{2 X} 2 K sp = 4 X 3 = 4[Pb 2+ ] 3 Solubility of Lead(II) Iodide = 4 (solubility) 3 K sp = 4 (1.30 x ) 3 = 8.8 x 10 -9

11 Precipitating an Insoluble Salt Hg 2 Cl 2 (s) = Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x = [Hg 2 2+ ] [Cl - ] 2 If [Hg 2 2+ ] = M, what [Cl - ] is req’d to just begin the precipitation of Hg 2 Cl 2 ? That is, what is the maximum [Cl - ] that can be in solution with M Hg 2 2+ without forming Hg 2 Cl 2 ?

12 Precipitating an Insoluble Salt Hg 2 Cl 2 (s) = Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x = [Hg 2 2+ ] [Cl - ] 2 Recognize that. K sp = product of maximum ion concs. Precip. begins when product of ion concs. EXCEEDS the K sp.

13 K sp = 1.1 x = [Hg 2 2+ ] [Cl - ] 2 Solution [Cl - ] that can exist when [Hg 2 2+ ] = M, If this conc. of Cl - is just exceeded, Hg 2 Cl 2 begins to precipitate.

14 Hg 2 Cl 2 (s) = Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x Now use [Cl - ] = 1.0 M. What is the value of [Hg 2 2+ ] at this point? Solution [Hg 2 2+ ] = K sp / [Cl - ] 2 = K sp / (1.0) 2 = 1.1 x M The concentration of Hg 2 2+ has been reduced by !

15 The Common Ion Effect Adding an ion “common” to an equilibrium causes the equilibrium to shift back to reactant.

16 Common Ion Effect PbCl 2 (s) = Pb 2+ (aq) + 2 Cl - (aq) K sp = 1.9 x 10 -5

17 Calculate the solubility of BaSO 4 BaSO 4 (s) = Ba 2+ ( aq ) + SO 4 2- ( aq ) (a)In pure water and (b) in M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x The Common Ion Effect

18 K sp for BaSO 4 = 1.1 x BaSO 4 (s) = Ba 2+ (aq) + SO 4 2- (aq) Solution Solubility = [Ba 2+ ] = [SO 4 2- ] = x K sp = [Ba 2+ ] [SO 4 2- ] = x 2 x = (K sp ) 1/2 = 1.1 x M Solubility in pure water = 1.1 x M BaSO 4 in pure water

19 Solution Solubility in pure water = 1.1 x mol/L. Now starting with M Ba 2+. Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSO 4 be less than or greater than in pure water?___ BaSO 4 in in M Ba(NO 3 ) 2. BaSO 4 (s) = Ba 2+ (aq) + SO 4 2- (aq)

20 Solution [Ba 2+ ][SO 4 2- ] initial change equilib. The Common Ion Effect + y y yy

21 K sp = [Ba 2+ ] [SO 4 2- ] = ( y) (y) Because y < 1.1 x M (pure), y is about equal to Therefore, K sp = 1.1 x = (0.010)(y) y = 1.1 x M = solubility in presence of added Ba 2+ ion. The Common Ion Effect Solution

22 SUMMARY Solubility in pure water = x = 1.1 x M Solubility in presence of added Ba 2+ = 1.1 x M Le Chatelier’s Principle is followed! Add to the right: equilibrium goes to the left The Common Ion Effect BaSO 4 (s) = Ba 2+ (aq) + SO 4 2- (aq)

23 The Common Ion Effect

24 The Common Ion Effect