Solubility Lesson 8 Titrations & Max Ion Concentration.

Slides:



Advertisements
Similar presentations
Solubility Equilibria
Advertisements

Precipitation Equilibria. Solubility Product Ionic compounds that we have learned are insoluble in water actually do dissolve a tiny amount. We can quantify.
Chapter 19 - Neutralization
Types of Chemical Reactions and Solution Stoichiometry.
Electrolytes Some solutes can dissociate into ions. Electric charge can be carried.
VI.Applications of Solubility 1.Chloride Ion Titration The concentration of chloride ion in a water sample is determined. Adding Ag + to the water sample.
The K sp of chromium (III) iodate in water is 5.0 x Estimate the molar solubility of the compound. Cr(IO 3 ) 3 (s)  Cr 3+ (aq) + 3 IO 3 - (aq)
Lecture 72/1/06. Precipitation reactions What are they? Solubility?
Solubility Product Constant
Solubility Product Constant
Solubility Equilibria. Write solubility product (K sp ) expressions from balanced chemical equations for salts with low solubility. Solve problems involving.
Sample Titration Problems
Molarity. Molarity Molarity (M) or molar concentration is a measure of the concentration of a solute in a solution. Unit for molar concentration is mol/L.
CHM 112 Summer 2007 M. Prushan Acid-Base Equilibria and Solubility Equilibria Chapter 16.
1. Which of the following formula equations matches the word equation below: Magnesium + hydrogen monochloride  Magnesium chloride + hydrogen a. Mg +
Chapter 16 Lesson 1 Solubility and Complex Ion Equilibria.
Acid-Base Equilibria and Solubility Equilibria Chapter 16 Dr. Ali Bumajdad.
Review of Chemistry 11. Ionic Compounds:Covalent Compounds: (Begins with a metal or NH 4 ) (Begins with a nonmetal) BaseSalt AcidNonacid NaOH (Metal +
Solubility Lesson 8 Review Notes. Adding a Crystal to a Saturated Solution Consider the saturated solution Cl - Ag +
Solutions: Molarity. A. Concentration – measure of the amount of solute that is dissolved in a given amount of solvent I. Concentration of Solutions Solutions:
Solubility Lesson 5 Trial Ion Product. We have learned that when two ionic solutions are mixed and if one product has low solubility, then there is a.
Drill: Determine the KQ/HQ ratio to make a buffer solution with a pH of 4.70 K a for HQ = 3.0 x
Which of the following solubility product expressions is incorrect?
Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Solubility Equilibrium TEXT REFERENCE Masterton and Hurley Chapter 16.1.
Solutions Concentration of Solutions. Objectives 1.Define molarity and calculate its value. 2.Define molality and calculate its value. 3.Apply the principles.
Chemistry 12 – Sorting out Solubility Problems This will help you with the following types of Solubility Problems: 1-ion concentrations in mixtures (no.
Solubility Equilibria 16.6 AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ]K sp is the solubility product constant MgF 2 (s) Mg 2+ (aq) + 2F - (aq)
Solubility Equilibrium Solubility Product Constant Ionic compounds (salts) differ in their solubilities Most “insoluble” salts will actually dissolve.
MEASURING CONCENTRATION OF IONS IN SOLUTION Molarity is ONE way to do this…we will learn others later in the year!!!
Chapter 11 Solutions Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings.
Ch. 13/14: Solutions Describing a Solution’s Composition.
Solubility Lesson 6 Changing solubility/Common Ion Effect.
Lesson 2 Ion Concentration. 1. What is the concentration of each ion in a M AlCl 3 solution? AlCl 3  Al 3+ +3Cl -
Steps for solving Stoichiometric Problems Involving Solution
Calculating Solubility Unit III: Solubility Equilibrium.
1 What weight of sulfur (FW = ) ore which should be taken so that the weight of BaSO 4 (FW = ) precipitate will be equal to half of the percentage.
Drill: Determine the KQ/HQ ratio to make a buffer solution with a pH of 4.70 K a for HQ = 3.0 x
Molarity moles of solute Liters of solution Unit for molarity mole mol LiterL = M.
Solubility Equilibria.  Write a balanced chemical equation to represent equilibrium in a saturated solution.  Write a solubility product expression.
If 36.2 mL of M CaCl 2 solution is added to 37.5 mL of M Na 2 CO 3, what mass of calcium carbonate, CaCO 3, will be precipitate? CaCl 2 (aq)
$200 $300 $400 $500 $100 $200 $300 $400 $500 $100 $200 $300 $400 $500 $100 $200 $300 $400 $500 $100 $200 $300 $400 $500 $100 Back to the Basics Solution.
16-2. Dilution Calculations and Molar Solubility A. MOLAR SOLUBILITY : Think of molar solubility as the molarity of a saturated solution!! Moles per Litre.
Drill: Determine the KQ/HQ ratio to make a buffer solution with a pH of 4.70 K a for HQ = 3.0 x
Solution Concentrations The measured amount of solute in a given amount of solvent or solution.
Solubility Lesson 3 Separating Ions.
Drill: Determine the KQ/HQ ratio to make a buffer solution with a pH of 4.70 Ka for HQ = 3.0 x 10-5.
To Precipitate or not 6-6.
Molarity calculations
Solubility and Complexation Equilibria
Solubility Constants: Ksp
Chapter 16: Solubility Equilibria
SCH4U:Solubility Equilibrium Lesson
Solubility Equilibria
Ion Concentration.
Formality (F): Wt.(g) /F.Wt. F = F.Wt = Formula weight V(L)
The Solubility Product Constant (Ksp)
Solubility Lesson 5 Trial Ion Product.
Solubility Lesson 8 Titrations & Max Ion Concentration.
Solubility Lesson 8 Review Notes.
Solubility Equilibria
Solubility Lesson 7 Changing solubility.
Ion Concentration.
Solubility Equilibria
Solubility Product Constant
Acid-Base Equilibria and Solubility Equilibria
Solubility Product Constant
Solubility & Simultaneous Equilibria Part I: Ksp and Solubility
Lesson 2 Ion Concentration.
Solubility & Simultaneous Equilibria Part I: Ksp and Solubility
Solubility & Simultaneous Equilibria Part I: Ksp and Solubility
Presentation transcript:

Solubility Lesson 8 Titrations & Max Ion Concentration

Review Questions 1.Mg(OH) 2 will have the greatest solubility in: A.NaOH B.Mg(NO 3 ) 2 C.H 2 O D.AgNO 3

Review Questions 1.Mg(OH) 2 will have the greatest solubility in: Mg(OH) 2 ⇌ Mg 2+ +2OH - A.NaOH B.Mg(NO 3 ) 2 C.H 2 O D.AgNO 3

Review Questions 1.Mg(OH) 2 will have the greatest solubility in: Mg(OH) 2 ⇌ Mg 2+ +2OH - A.NaOHNa + lowers solubility B.Mg(NO 3 ) 2 Mg 2+ lowers solubility C.H 2 ONo effect solubility D.AgNO 3 Ag + increases solubility by reacting with OH -

Review Questions 1.Mg(OH) 2 will have the greatest solubility in: Mg(OH) 2 ⇌ Mg 2+ +2OH - A.NaOHNa + lowers solubility B.Mg(NO 3 ) 2 Mg 2+ lowers solubility C.H 2 ONo effect solubility D.AgNO 3 Ag + increases solubility by reacting with OH -

Review Questions 2.Mg(OH) 2 will have the greatest solubility in: A.1.0 M NaNO 3 B.NaOH C.Sr(OH) 2

Review Questions 2.Mg(OH) 2 will have the greatest solubility in: Mg(OH) 2 ⇌ Mg 2+ +2OH - A.1.0 M NaNO 3 B.1.0 M NaOH C.Sr(OH) 2

Review Questions 2.Mg(OH) 2 will have the greatest solubility in: Mg(OH) 2 ⇌ Mg 2+ +2OH - A.1.0 M NaNO 3 No effect B.1.0 M NaOH1.0 M OH - lowers solubility C.Sr(OH) M OH - lowers solubility more

Review Questions 2.Mg(OH) 2 will have the greatest solubility in: Mg(OH) 2 ⇌ Mg 2+ +2OH - A.1.0 M NaNO 3 No effect B.1.0 M NaOH1.0 M OH - lowers solubility C.Sr(OH) M OH - lowers solubility more remember: Sr(OH) 2  Sr OH M1.0 M2.0 M

Review Questions 3.PbCl 2 will have the greatest solubility in: PbCl 2 ⇌ Pb 2+ +2Cl - A.1.0 M NaCl B.1.0 M MgCl 2 C.1.0 M AlCl 3 D.2.0 M CaCl 2

Review Questions 3.PbCl 2 will have the greatest solubility in: PbCl 2 ⇌ Pb 2+ +2Cl - A.1.0 M NaCl1.0 M Cl - B.1.0 M MgCl 2 C.1.0 M AlCl 3 D.2.0 M CaCl 2

Review Questions 3.PbCl 2 will have the greatest solubility in: PbCl 2 ⇌ Pb 2+ +2Cl - A.1.0 M NaCl1.0 M Cl - B.1.0 M MgCl M Cl - C.1.0 M AlCl 3 D.2.0 M CaCl 2

Review Questions 3.PbCl 2 will have the greatest solubility in: PbCl 2 ⇌ Pb 2+ +2Cl - A.1.0 M NaCl1.0 M Cl - B.1.0 M MgCl M Cl - C.1.0 M AlCl M Cl - D.2.0 M CaCl 2

Review Questions 3.PbCl 2 will have the greatest solubility in: PbCl 2 ⇌ Pb 2+ +2Cl - A.1.0 M NaCl1.0 M Cl - B.1.0 M MgCl M Cl - C.1.0 M AlCl M Cl - D.2.0 M CaCl M Cl -

1. In a titration 3.61 mL of M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl 2 solution. Calculate the [Pb 2+ ]. Pb 2+ +2I -  PbI 2(s) Equation L LData ? M M [Pb 2+ ]=

1. In a titration 3.61 mL of M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl 2 solution. Calculate the [Pb 2+ ]. Pb 2+ +2I -  PbI 2(s) Equation L LData ? M M [Pb 2+ ]= L I -

1. In a titration 3.61 mL of M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl 2 solution. Calculate the [Pb 2+ ]. Pb 2+ +2I -  PbI 2(s) Equation L LData ? M M [Pb 2+ ]= L I - x mol 1L

1. In a titration 3.61 mL of M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl 2 solution. Calculate the [Pb 2+ ]. Pb 2+ +2I -  PbI 2(s) Equation L LData ? M M [Pb 2+ ]= L I - x mol x 1 mol Pb 2+ 1L 2 mol I -

1. In a titration 3.61 mL of M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl 2 solution. Calculate the [Pb 2+ ]. Pb 2+ +2I -  PbI 2(s) Equation L LData ? M M [Pb 2+ ]= L I - x mol x 1 mol Pb 2+ 1L 2 mol I L

1. In a titration 3.61 mL of M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl 2 solution. Calculate the [Pb 2+ ]. Pb 2+ +2I -  PbI 2(s) Equation L LData ? M M [Pb 2+ ]= L I - x mol x 1 mol Pb 2+ 1L 2 mol I L = M

2. Determine the Ksp for PbCl 2 from the results of the last question. PbCl 2(s) ⇌ Pb 2+ +2Cl - ss2s Ksp=[Pb 2+ ][Cl - ] 2 Ksp=[s][2s] 2 Ksp=4s 3 Ksp=4( ) 3 Ksp=1.88 x 10 -7

Maximum Ion Concentration 1. The maximum concentration of [AgBrO 3 ] is lower in a solution of NaBrO 3 than it would be in pure water. This is because the solution already has BrO 3 - present. What is the maximum [Ag + ] possible in a 0.100M NaBrO 3 solution? M BrO 3 - [AgBrO 3 ] What molarity of [AgBrO 3 ] is possible before it precipitates? AgBrO 3(s) ⇌ Ag + +BrO M Ksp=[Ag + ][BrO 3 - ] 5.3 x =[Ag + ][0.100] [Ag + ]=[AgBrO 3 ]=5.3 x M

2. Calculate the maximum number of grams of AgNO 3 that will dissolve mL of 0.200M AlCl 3 without forming a precipitate. AgCl (s) ⇌ Ag + +Cl M Ksp=[Ag + ][Cl - ] 1.8 x =[Ag + ][0.600] [Ag + ]=3.0 x M

2. Calculate the maximum number of grams of AgNO 3 that will dissolve mL of 0.200M AlCl 3 without forming a precipitate. AgCl (s) ⇌ Ag + +Cl M Ksp=[Ag + ][Cl - ] 1.8 x =[Ag + ][0.600] [Ag + ]=3.0 x M L AgNO 3

2. Calculate the maximum number of grams of AgNO 3 that will dissolve mL of 0.200M AlCl 3 without forming a precipitate. AgCl (s) ⇌ Ag + +Cl M Ksp=[Ag + ][Cl - ] 1.8 x =[Ag + ][0.600] [Ag + ]=3.0 x M L AgNO 3 x 3.0 x moles 1 L

2. Calculate the maximum number of grams of AgNO 3 that will dissolve mL of 0.200M AlCl 3 without forming a precipitate. AgCl (s) ⇌ Ag + +Cl M Ksp=[Ag + ][Cl - ] 1.8 x =[Ag + ][0.600] [Ag + ]=3.0 x M L AgNO 3 x 3.0 x moles x g 1 L 1 mole

2. Calculate the maximum number of grams of AgNO 3 that will dissolve mL of 0.200M AlCl 3 without forming a precipitate. AgCl (s) ⇌ Ag + +Cl M Ksp=[Ag + ][Cl - ] 1.8 x =[Ag + ][0.600] [Ag + ]=3.0 x M L AgNO 3 x 3.0 x moles x g = 5.1 x g 1 L 1 mole