The document discusses aqueous equilibria involving slightly soluble ionic compounds and complex ions. It defines key concepts such as solubility product constants (Ksp), ion-product expressions (Qsp), and how Ksp relates to the solubility of an ionic compound. It provides examples of calculating Ksp from experimental solubility data and calculating solubility from known Ksp values. It also discusses how common ions and pH can impact the solubility of ionic compounds. Sample problems demonstrate applying these concepts to calculate solubilities, Ksp values, and predict solubility changes upon addition of strong acids or common ions.
The document provides information on solubility constants (Ksp) and equilibria involving complex ions and slightly soluble ionic compounds in aqueous solutions. It discusses solubility product principles, writing ion-product expressions, determining Ksp values from solubility data and vice versa, the effect of common ions on solubility, and how pH affects the solubility of salts containing anions of weak acids. Sample problems demonstrate calculating solubility, Ksp, and the impact of adding acids or salts containing common ions. Key concepts covered include solubility constants, solubility product expressions, and factors that influence the solubility of sparingly soluble salts.
REDOX POTENTIAL MEASUREMENT AND ITS APPLICATION IN SOIL FERTILITYKARTHIKEYANB30
What is mean redox reaction, How to measure redox condition in soil, Application of redox potential in nutrient management, Kinetics of pH, EC, Nutrients in soil
This document discusses ionic equilibrium and electrolytes. It defines electrolytes as substances that conduct electricity when dissolved in water by dissociating into ions. Electrolytes are divided into strong and weak based on their degree of ionization. Strong electrolytes almost completely ionize while weak electrolytes ionize only partially. The document discusses Arrhenius theory of electrolytic dissociation and factors that affect the degree of ionization like concentration, temperature, and presence of a common ion. It also defines concepts like isohydric solutions and dissociation constants.
PHYSICAL CHEMISTRY 1.2- ACIDS,BASES AND SALTSshahzadebaujiti
This document discusses acids, bases, and pH. It defines acids and bases according to Arrhenius, Bronsted-Lowry, and Lewis theories. The key points are:
1) Arrhenius defined acids as producing H+ ions in water and bases as producing OH- ions. Bronsted-Lowry expanded this to include proton donors and acceptors in any solvent.
2) Lewis defined acids and bases in terms of electron pair acceptance and donation.
3) Conjugate acid-base pairs are related - the conjugate base of an acid is its corresponding base, and vice versa.
4) pH measures hydrogen ion concentration on a logarithmic scale from 0-14.
Discusses the chemical of slightly soluble compounds. Ksp and factors affecting solubility are included as well as solved problems.
**More good stuff available at:
www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f
1. The solubility product constant (Ksp) is the product of the molar concentrations of ions in a saturated solution.
2. Ksp values can be used to calculate solubilities and predict if precipitates will form when solutions are combined.
3. To predict precipitation, the ion product is calculated and compared to the Ksp - if the ion product is greater than Ksp, precipitation will occur until equilibrium is re-established.
This document discusses weak bases and how they react with water to form the conjugate acid and hydroxide ions. It defines the base dissociation constant Kb and explains how it refers to the equilibrium of a base reacting with water. It provides examples of calculating the concentration of hydroxide ions produced from a weak base solution and calculating Kb or Ka values for conjugate acid-base pairs using known constants. The document also discusses how the properties of salt solutions are determined by the constituent ions and how buffers resist changes in pH upon addition of acids or bases.
The document discusses several key topics regarding acids and bases:
1. Group 1 and 2 hydroxides such as LiOH, NaOH, and Ca(OH)2 are strong bases. NaOH and KOH are common laboratory reagents while alkaline earth hydroxides have low solubility.
2. Calculating pH involves determining the hydroxide ion concentration from solubility equations or acid/base equilibria. Examples show calculating the pH of NaOH and NH3 solutions.
3. Many bases like NH3 produce hydroxide ions when dissolved in water by reacting with water. The acid dissociation constant Kb describes this reaction.
4. Polyprotic acids dissociate
The document provides information on solubility constants (Ksp) and equilibria involving complex ions and slightly soluble ionic compounds in aqueous solutions. It discusses solubility product principles, writing ion-product expressions, determining Ksp values from solubility data and vice versa, the effect of common ions on solubility, and how pH affects the solubility of salts containing anions of weak acids. Sample problems demonstrate calculating solubility, Ksp, and the impact of adding acids or salts containing common ions. Key concepts covered include solubility constants, solubility product expressions, and factors that influence the solubility of sparingly soluble salts.
REDOX POTENTIAL MEASUREMENT AND ITS APPLICATION IN SOIL FERTILITYKARTHIKEYANB30
What is mean redox reaction, How to measure redox condition in soil, Application of redox potential in nutrient management, Kinetics of pH, EC, Nutrients in soil
This document discusses ionic equilibrium and electrolytes. It defines electrolytes as substances that conduct electricity when dissolved in water by dissociating into ions. Electrolytes are divided into strong and weak based on their degree of ionization. Strong electrolytes almost completely ionize while weak electrolytes ionize only partially. The document discusses Arrhenius theory of electrolytic dissociation and factors that affect the degree of ionization like concentration, temperature, and presence of a common ion. It also defines concepts like isohydric solutions and dissociation constants.
PHYSICAL CHEMISTRY 1.2- ACIDS,BASES AND SALTSshahzadebaujiti
This document discusses acids, bases, and pH. It defines acids and bases according to Arrhenius, Bronsted-Lowry, and Lewis theories. The key points are:
1) Arrhenius defined acids as producing H+ ions in water and bases as producing OH- ions. Bronsted-Lowry expanded this to include proton donors and acceptors in any solvent.
2) Lewis defined acids and bases in terms of electron pair acceptance and donation.
3) Conjugate acid-base pairs are related - the conjugate base of an acid is its corresponding base, and vice versa.
4) pH measures hydrogen ion concentration on a logarithmic scale from 0-14.
Discusses the chemical of slightly soluble compounds. Ksp and factors affecting solubility are included as well as solved problems.
**More good stuff available at:
www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f
1. The solubility product constant (Ksp) is the product of the molar concentrations of ions in a saturated solution.
2. Ksp values can be used to calculate solubilities and predict if precipitates will form when solutions are combined.
3. To predict precipitation, the ion product is calculated and compared to the Ksp - if the ion product is greater than Ksp, precipitation will occur until equilibrium is re-established.
This document discusses weak bases and how they react with water to form the conjugate acid and hydroxide ions. It defines the base dissociation constant Kb and explains how it refers to the equilibrium of a base reacting with water. It provides examples of calculating the concentration of hydroxide ions produced from a weak base solution and calculating Kb or Ka values for conjugate acid-base pairs using known constants. The document also discusses how the properties of salt solutions are determined by the constituent ions and how buffers resist changes in pH upon addition of acids or bases.
The document discusses several key topics regarding acids and bases:
1. Group 1 and 2 hydroxides such as LiOH, NaOH, and Ca(OH)2 are strong bases. NaOH and KOH are common laboratory reagents while alkaline earth hydroxides have low solubility.
2. Calculating pH involves determining the hydroxide ion concentration from solubility equations or acid/base equilibria. Examples show calculating the pH of NaOH and NH3 solutions.
3. Many bases like NH3 produce hydroxide ions when dissolved in water by reacting with water. The acid dissociation constant Kb describes this reaction.
4. Polyprotic acids dissociate
Chapter 18.3 : Equilibria of Acids, Bases, and SaltsChris Foltz
The document discusses acid-base equilibria, including acid ionization constants, the ionization of acetic acid as a weak acid, and buffer solutions. It explains that buffer solutions contain a weak acid and salt of the weak acid and can resist changes in pH when small amounts of acid or base are added. The document also covers the ionization of water, hydrolysis of salts, and how hydrolysis can affect the pH of salt solutions depending on whether the salt contains ions from strong vs. weak acids and bases.
Salt hydrolysis can produce acidic, basic, or neutral solutions depending on the salt. Salts are classified based on whether they contain strong acids/bases or weak acids/bases. Salts of strong acids and bases do not undergo hydrolysis and produce neutral solutions. Salts of weak acids and strong bases produce basic solutions, while salts of strong acids and weak bases produce acidic solutions. Salts of weak acids and bases may produce neutral, acidic, or basic solutions. Solubility products define the solubility of sparingly soluble salts in solution. The common ion effect suppresses the dissociation of weak electrolytes in the presence of strong electrolytes with a common ion. Solubility products and common ion effects can be applied
3rd Lecture on Ionic Equilibria | Chemistry Part I | 12th StdAnsari Usama
Buffer solutions resist changes in pH when small amounts of acid or base are added. There are two types: acidic buffers containing a weak acid and its salt, and basic buffers containing a weak base and its salt. The pH of a buffer solution is related to the ratio of the salt and acid or base concentrations. Buffer solutions have important applications in biological systems, agriculture, medicine, and analytical chemistry. They maintain pH despite additions of H+ or OH-.
- Water is a polar solvent due to its molecular structure, with oxygen having a partial negative charge and hydrogen having partial positive charges. This allows it to dissolve ionic compounds by interacting with and separating the ions.
- Ionic compounds dissolve to varying degrees in water depending on how strongly the ions are attracted to each other versus water molecules. Solubility can be measured in g/L.
- Acids donate H+ ions in water and are classified as strong or weak based on how completely they ionize. Their strength affects pH calculations. Bases accept H+ and similarly ionize more or less completely.
Concept of solubility euilibria deals with the extent of solubility of different salts or compounds in water or other solvents and also tells about different factors which control the solubility of different salts.
This document discusses acid-base equilibria, including weak acids, weak bases, and water. It explains how weak acids and bases form equilibrium reactions in water, and defines the equilibrium constants KA and KB. It also describes how water forms an equilibrium with a constant of KW. Additionally, it introduces conjugate acids and bases and how acid-base reactions can produce new materials with acidic or basic properties. Finally, it provides an example of calculating the pH of a sodium acetate solution through considering its hydrolysis equilibrium.
2nd Lecture on Ionic Equilibria | Chemistry Part I | 12th StdAnsari Usama
1) This document discusses ionic equilibria, specifically the autoionization of water and the pH scale. It defines key terms like ionic product of water (Kw), pH, and pOH.
2) Salts can undergo hydrolysis when dissolved in water. Salts of strong acids and bases do not hydrolyze and are neutral. Salts of strong acids and weak bases hydrolyze to produce H3O+ ions, making the solution acidic. Salts of weak acids and strong bases hydrolyze to produce OH- ions, making the solution basic.
3) For salts of weak acids and bases, the solution can be acidic, basic, or neutral depending on whether the Ka
This document discusses various concepts related to gravimetric analysis methods. It covers three key points:
1) Gravimetric analysis involves selectively precipitating the analyte of interest and weighing the precipitate to determine the amount of analyte. Factors like solubility products (Ksp), common ion effects, and pH can impact precipitation.
2) Key steps in gravimetric analysis are discussed, including filtering, drying, and transferring precipitates. Equipment like filters, crucibles, and drying ovens are also mentioned.
3) Solubility is impacted by various equilibrium concepts like Ksp values, common ion effects, salt effects, pH, complexation, and temperature. These concepts are illustrated through
Acid base Theories
Role of the solvents
Acid base dissociation constant,
Relative strength of acids and bases
Distribution of acid base species with pH
Buffer solution
Henderson Hasselbalch equation,
Indicators, Mixed indicators
Different type of titrations (Neutralization curves)
Polyprotic systems,
Phosphoric acid system,
Polyamine and amino acid systems.
Titration of sodium carbonate
Lavoisier definition
Liebig definition
Arrhenius Acids and Bases
Bronsted-Lowry Acid and Base
Lewis Acid and Base
Solvent-system Concept
Lux-Flood Concept
Pearson’s Concept
Historically, the first of the scientific concepts of acids and bases was provided by the French chemist Antoine Lavoisier, circa 1776.
Lavoisier's knowledge of strong acids was mainly restricted to oxyacids, which tend to contain central atoms in high oxidation states surrounded by oxygen, such as HNO3 and H2SO4, and he was not aware of the true composition of the hydrohalic acids, HCl, HBr, and HI. From his limited knowledge,
He defined acids in terms of their content of oxygen, and he named oxygen from Greek words meaning "acid-former"
1. The document discusses ionic equilibria, including acids and bases, and how to identify strong and weak acids/bases. It also discusses calculating pH and pOH values.
2. It explains how to determine if a salt is neutral, acidic, or basic based on whether the cation or anion comes from a strong or weak acid/base.
3. For sparingly soluble salts, it discusses using Ksp expressions and concentrations to determine if a precipitate will form from mixing solutions of ions.
This document provides an overview of Chapter 22 from a chemistry textbook, which covers topics related to ionic equilibria including:
- pH, Ka, pKa and Kw values and their use in calculations involving strong and weak acids and bases.
- Acid-base titration curves and how they differ for strong-strong, strong-weak, weak-strong, and weak-weak acid-base titrations.
- How acid-base indicators work and their use in determining the endpoint of a titration.
It also lists learning outcomes for understanding these concepts and performing related calculations.
This document contains homework questions asking a student to:
1) Write balanced chemical equations for reactions of alkali metals and alkaline earth metals with water.
2) Identify the formula for lime.
3) Write the balanced equation for forming slaked lime.
4) Write the balanced equation for forming lime from calcium carbonate.
It also includes a bonus question to write and explain the equation for bubbling carbon dioxide through limewater.
This document discusses salt hydrolysis, which is the reaction of salt ions with water that can result in an acidic, basic, or neutral solution. Salts formed from a weak acid and strong base produce a basic solution, as the conjugate base of the weak acid will accept protons from water. Salts formed from a weak base and strong acid produce an acidic solution, as the conjugate acid of the weak base will donate protons to water. Salts formed from a strong acid and strong base produce a neutral solution. Examples of sodium ethanoate and ammonium chloride are provided to illustrate basic and acidic salt hydrolysis.
This chapter of the general chemistry textbook discusses solubility and complex ion equilibria. It covers topics such as the solubility product constant Ksp, the common ion effect, limitations of Ksp, criteria for precipitation, fractional precipitation, effects of pH on solubility, and equilibria involving complex ions. It also describes the process of qualitative cation analysis using selective precipitation of cations in different solubility groups.
This document contains answers to questions about chemical changes from a student book. It includes answers about:
- Reactions of metals like lithium and zinc with acids producing salts and hydrogen gas.
- Displacement reactions where more reactive metals displace less reactive ones from solutions.
- Extracting metals like zinc through chemical reduction of metal oxides.
- Producing salts by reacting metals with acids or insoluble bases with acids.
- Measuring pH and differences between strong and weak acids.
- Calculations involving moles of gases.
Acid soil formation and classification of acid soil in indiaKARTHIKEYANB30
Genesis of soil acidity,acid soil forming factors, pedogenic process influence the acid soil, acid soil classification, amelioration of soil acidity-chemistry of liming, equivalent acidity,neutralizing value or calcium carbonate equivalent
Learning objectives
Introduction
Preparation of a standard solution used for redox titration
Oxidizing and reducing agents used in volumetric analysis
N/10 potassium permanganate preparation
N/10 potassium dichromate preparation
N/10 Iodine solution preparation
Examples of redox titrations
Conclusion
References
The solubility product is a kind of equilibrium constant and its value depend...RidhaTOUATI1
The solubility product is a kind of equilibrium constant and its value depends on temperature. Ksp usually increases with an increase in temperature due to increased solubility. Solubility is defined as a property of a substance called solute to get dissolved in a solvent in order to form a solution
The document discusses chemical equilibrium and solubility. It provides examples of:
1) A chemical equation at equilibrium for the reaction of hydrogen and oxygen gases.
2) How the solubility product constant (Ksp) is defined and relates to the solubility of ionic compounds in water.
3) How adding a common ion to a saturated solution can decrease the solubility through Le Chatelier's principle.
This document discusses different types of chemical reactions including decomposition, synthesis, combustion, double replacement, and single replacement reactions. It provides examples of each type of reaction and explains the key features that define them. Double replacement reactions are highlighted, where a metal replaces a metal in a compound and a nonmetal replaces a nonmetal, forming a precipitate if one of the products is insoluble. Guidelines are provided for writing molecular, total ionic, and net ionic equations for double replacement reactions.
Solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature (in grams or moles)
•
The molar solubility (mol/L) is the number of moles of solute that will dissolve in 1L of a saturated solution.
•
The molarity of the dissolved solute in a saturated solution.
•
Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.
•
A saturated solution contains the maximum amount of solute possible at a given temperature in equilibrium with an undissolved excess of the substance.
This document summarizes Chapter 19 from a general chemistry textbook. The chapter covers solubility product constants, solubility equilibria, common ion effects, limitations of Ksp, precipitation criteria, fractional precipitation, solubility and pH, and equilibria involving complex ions. It provides examples and explanations of these concepts. Qualitative cation analysis is also discussed, describing how cations can be selectively precipitated into groups for identification purposes.
Chapter 18.3 : Equilibria of Acids, Bases, and SaltsChris Foltz
The document discusses acid-base equilibria, including acid ionization constants, the ionization of acetic acid as a weak acid, and buffer solutions. It explains that buffer solutions contain a weak acid and salt of the weak acid and can resist changes in pH when small amounts of acid or base are added. The document also covers the ionization of water, hydrolysis of salts, and how hydrolysis can affect the pH of salt solutions depending on whether the salt contains ions from strong vs. weak acids and bases.
Salt hydrolysis can produce acidic, basic, or neutral solutions depending on the salt. Salts are classified based on whether they contain strong acids/bases or weak acids/bases. Salts of strong acids and bases do not undergo hydrolysis and produce neutral solutions. Salts of weak acids and strong bases produce basic solutions, while salts of strong acids and weak bases produce acidic solutions. Salts of weak acids and bases may produce neutral, acidic, or basic solutions. Solubility products define the solubility of sparingly soluble salts in solution. The common ion effect suppresses the dissociation of weak electrolytes in the presence of strong electrolytes with a common ion. Solubility products and common ion effects can be applied
3rd Lecture on Ionic Equilibria | Chemistry Part I | 12th StdAnsari Usama
Buffer solutions resist changes in pH when small amounts of acid or base are added. There are two types: acidic buffers containing a weak acid and its salt, and basic buffers containing a weak base and its salt. The pH of a buffer solution is related to the ratio of the salt and acid or base concentrations. Buffer solutions have important applications in biological systems, agriculture, medicine, and analytical chemistry. They maintain pH despite additions of H+ or OH-.
- Water is a polar solvent due to its molecular structure, with oxygen having a partial negative charge and hydrogen having partial positive charges. This allows it to dissolve ionic compounds by interacting with and separating the ions.
- Ionic compounds dissolve to varying degrees in water depending on how strongly the ions are attracted to each other versus water molecules. Solubility can be measured in g/L.
- Acids donate H+ ions in water and are classified as strong or weak based on how completely they ionize. Their strength affects pH calculations. Bases accept H+ and similarly ionize more or less completely.
Concept of solubility euilibria deals with the extent of solubility of different salts or compounds in water or other solvents and also tells about different factors which control the solubility of different salts.
This document discusses acid-base equilibria, including weak acids, weak bases, and water. It explains how weak acids and bases form equilibrium reactions in water, and defines the equilibrium constants KA and KB. It also describes how water forms an equilibrium with a constant of KW. Additionally, it introduces conjugate acids and bases and how acid-base reactions can produce new materials with acidic or basic properties. Finally, it provides an example of calculating the pH of a sodium acetate solution through considering its hydrolysis equilibrium.
2nd Lecture on Ionic Equilibria | Chemistry Part I | 12th StdAnsari Usama
1) This document discusses ionic equilibria, specifically the autoionization of water and the pH scale. It defines key terms like ionic product of water (Kw), pH, and pOH.
2) Salts can undergo hydrolysis when dissolved in water. Salts of strong acids and bases do not hydrolyze and are neutral. Salts of strong acids and weak bases hydrolyze to produce H3O+ ions, making the solution acidic. Salts of weak acids and strong bases hydrolyze to produce OH- ions, making the solution basic.
3) For salts of weak acids and bases, the solution can be acidic, basic, or neutral depending on whether the Ka
This document discusses various concepts related to gravimetric analysis methods. It covers three key points:
1) Gravimetric analysis involves selectively precipitating the analyte of interest and weighing the precipitate to determine the amount of analyte. Factors like solubility products (Ksp), common ion effects, and pH can impact precipitation.
2) Key steps in gravimetric analysis are discussed, including filtering, drying, and transferring precipitates. Equipment like filters, crucibles, and drying ovens are also mentioned.
3) Solubility is impacted by various equilibrium concepts like Ksp values, common ion effects, salt effects, pH, complexation, and temperature. These concepts are illustrated through
Acid base Theories
Role of the solvents
Acid base dissociation constant,
Relative strength of acids and bases
Distribution of acid base species with pH
Buffer solution
Henderson Hasselbalch equation,
Indicators, Mixed indicators
Different type of titrations (Neutralization curves)
Polyprotic systems,
Phosphoric acid system,
Polyamine and amino acid systems.
Titration of sodium carbonate
Lavoisier definition
Liebig definition
Arrhenius Acids and Bases
Bronsted-Lowry Acid and Base
Lewis Acid and Base
Solvent-system Concept
Lux-Flood Concept
Pearson’s Concept
Historically, the first of the scientific concepts of acids and bases was provided by the French chemist Antoine Lavoisier, circa 1776.
Lavoisier's knowledge of strong acids was mainly restricted to oxyacids, which tend to contain central atoms in high oxidation states surrounded by oxygen, such as HNO3 and H2SO4, and he was not aware of the true composition of the hydrohalic acids, HCl, HBr, and HI. From his limited knowledge,
He defined acids in terms of their content of oxygen, and he named oxygen from Greek words meaning "acid-former"
1. The document discusses ionic equilibria, including acids and bases, and how to identify strong and weak acids/bases. It also discusses calculating pH and pOH values.
2. It explains how to determine if a salt is neutral, acidic, or basic based on whether the cation or anion comes from a strong or weak acid/base.
3. For sparingly soluble salts, it discusses using Ksp expressions and concentrations to determine if a precipitate will form from mixing solutions of ions.
This document provides an overview of Chapter 22 from a chemistry textbook, which covers topics related to ionic equilibria including:
- pH, Ka, pKa and Kw values and their use in calculations involving strong and weak acids and bases.
- Acid-base titration curves and how they differ for strong-strong, strong-weak, weak-strong, and weak-weak acid-base titrations.
- How acid-base indicators work and their use in determining the endpoint of a titration.
It also lists learning outcomes for understanding these concepts and performing related calculations.
This document contains homework questions asking a student to:
1) Write balanced chemical equations for reactions of alkali metals and alkaline earth metals with water.
2) Identify the formula for lime.
3) Write the balanced equation for forming slaked lime.
4) Write the balanced equation for forming lime from calcium carbonate.
It also includes a bonus question to write and explain the equation for bubbling carbon dioxide through limewater.
This document discusses salt hydrolysis, which is the reaction of salt ions with water that can result in an acidic, basic, or neutral solution. Salts formed from a weak acid and strong base produce a basic solution, as the conjugate base of the weak acid will accept protons from water. Salts formed from a weak base and strong acid produce an acidic solution, as the conjugate acid of the weak base will donate protons to water. Salts formed from a strong acid and strong base produce a neutral solution. Examples of sodium ethanoate and ammonium chloride are provided to illustrate basic and acidic salt hydrolysis.
This chapter of the general chemistry textbook discusses solubility and complex ion equilibria. It covers topics such as the solubility product constant Ksp, the common ion effect, limitations of Ksp, criteria for precipitation, fractional precipitation, effects of pH on solubility, and equilibria involving complex ions. It also describes the process of qualitative cation analysis using selective precipitation of cations in different solubility groups.
This document contains answers to questions about chemical changes from a student book. It includes answers about:
- Reactions of metals like lithium and zinc with acids producing salts and hydrogen gas.
- Displacement reactions where more reactive metals displace less reactive ones from solutions.
- Extracting metals like zinc through chemical reduction of metal oxides.
- Producing salts by reacting metals with acids or insoluble bases with acids.
- Measuring pH and differences between strong and weak acids.
- Calculations involving moles of gases.
Acid soil formation and classification of acid soil in indiaKARTHIKEYANB30
Genesis of soil acidity,acid soil forming factors, pedogenic process influence the acid soil, acid soil classification, amelioration of soil acidity-chemistry of liming, equivalent acidity,neutralizing value or calcium carbonate equivalent
Learning objectives
Introduction
Preparation of a standard solution used for redox titration
Oxidizing and reducing agents used in volumetric analysis
N/10 potassium permanganate preparation
N/10 potassium dichromate preparation
N/10 Iodine solution preparation
Examples of redox titrations
Conclusion
References
The solubility product is a kind of equilibrium constant and its value depend...RidhaTOUATI1
The solubility product is a kind of equilibrium constant and its value depends on temperature. Ksp usually increases with an increase in temperature due to increased solubility. Solubility is defined as a property of a substance called solute to get dissolved in a solvent in order to form a solution
The document discusses chemical equilibrium and solubility. It provides examples of:
1) A chemical equation at equilibrium for the reaction of hydrogen and oxygen gases.
2) How the solubility product constant (Ksp) is defined and relates to the solubility of ionic compounds in water.
3) How adding a common ion to a saturated solution can decrease the solubility through Le Chatelier's principle.
This document discusses different types of chemical reactions including decomposition, synthesis, combustion, double replacement, and single replacement reactions. It provides examples of each type of reaction and explains the key features that define them. Double replacement reactions are highlighted, where a metal replaces a metal in a compound and a nonmetal replaces a nonmetal, forming a precipitate if one of the products is insoluble. Guidelines are provided for writing molecular, total ionic, and net ionic equations for double replacement reactions.
Solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature (in grams or moles)
•
The molar solubility (mol/L) is the number of moles of solute that will dissolve in 1L of a saturated solution.
•
The molarity of the dissolved solute in a saturated solution.
•
Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.
•
A saturated solution contains the maximum amount of solute possible at a given temperature in equilibrium with an undissolved excess of the substance.
This document summarizes Chapter 19 from a general chemistry textbook. The chapter covers solubility product constants, solubility equilibria, common ion effects, limitations of Ksp, precipitation criteria, fractional precipitation, solubility and pH, and equilibria involving complex ions. It provides examples and explanations of these concepts. Qualitative cation analysis is also discussed, describing how cations can be selectively precipitated into groups for identification purposes.
There are five main types of chemical reactions: synthesis, decomposition, single replacement, double replacement, and combustion. A chemical reaction occurs when atoms bond together to form new compounds or when compounds separate to form other compounds. Chemical reactions can be identified by certain characteristics like gas produced, light produced, temperature change, color change, or precipitate formed.
This document provides an overview of chemical equations and reactions. It discusses:
- Chemical equations, reactants, products, and how atoms rearrange during reactions.
- Balancing chemical equations by ensuring equal numbers of each atom on both sides.
- Information that can be obtained from a balanced chemical equation, such as moles of substances.
- Four main types of chemical reactions: combination, decomposition, displacement, and double displacement. Examples of each type are provided along with general reaction equations.
The document discusses solubility equilibria and solubility product constants (Ksp). It defines Ksp as the product of ion concentrations raised to their coefficients in the solubility reaction equation. It provides examples of using Ksp to calculate solubility from solubility data and vice versa. Additionally, it discusses factors that affect solubility such as common ion effect, pH, and complexation.
Solubility and precipitation equilibrium .pptxFatmaBITAM
This course of analytical chemistry is distinated to students of chemistry, Pharmacy and biology. It is concerning the equilibrium of solubility and precipitation of salts. The constants of this equilibrium and effecting factors are given with some examples demonstrations
1. Solubility is defined as the amount of a substance that dissolves in water completely to give free ions, and can be expressed in units of g/L or mol/dm3. Molar solubility refers to the amount of solute in moles that dissolves in a given volume of solvent.
2. The solubility product constant (Ksp) represents the equilibrium between ions in solution and undissolved solid for sparingly soluble salts. Ksp is the product of the concentrations of ions raised to their stoichiometric coefficients at the point of saturation.
3. The molar solubility of a salt can be calculated from its Ksp value by setting up an ionic
Chemical reactions and equations activity based question 10thBharathbabu68
The document contains questions and answers related to chemical reactions and equations. Some key points:
- Hydrogen gas is evolved when zinc reacts with dilute sulfuric acid. Copper sulfate crystals change color from blue to white on heating due to loss of water of crystallization.
- When iron is added to copper sulfate solution, a displacement reaction occurs forming a brown coating of copper on the iron. Barium sulfate precipitate forms when sodium sulfate solution is added to barium chloride.
- Zinc hydroxide precipitate forms when sodium hydroxide is added to zinc sulfate solution. Lead nitrate decomposes on heating with a crackling sound, producing nitrogen dioxide, oxygen and lead oxide.
This document discusses acid-base equilibria and solubility equilibria. It covers the common ion effect and how it impacts equilibrium, buffer solutions, acid-base titrations and indicators, solubility equilibria including Ksp expressions and calculations, and the effects of pH and common ions on solubility. It also briefly mentions complex ion equilibria.
The document summarizes key concepts regarding ionic equilibrium and solubility products. It discusses four types of salt hydrolysis and how they result in acidic, basic, or neutral solutions. It then defines solubility product and solubility product expressions for different types of salts. The solubility product principle is introduced, stating that precipitation will occur when the ionic product exceeds the solubility product constant.
1. The document discusses chemical equations, including word equations, balanced equations, and how to interpret and balance chemical equations.
2. It explains the key parts of a chemical equation including reactants, products, phases (solid, liquid, gas), and how to write ionic equations.
3. The document also covers different types of chemical reactions like synthesis, decomposition, single replacement, and double replacement reactions and how to predict products.
A document discusses various topics relating to chemistry solutions including:
1) The definition of solutions, solvents, and solutes. A solution is a homogeneous mixture of substances where the solute is the smaller component dissolved in the solvent.
2) Properties of aqueous solutions including that electrolytes can conduct electricity while nonelectrolytes cannot. Strong electrolytes dissociate completely while weak electrolytes only partially dissociate.
3) Reactions involving solutions such as precipitation reactions, acid-base reactions, and redox reactions. Precipitation occurs when an insoluble solid forms. Acid-base reactions involve acids and bases reacting to form water and a salt. Redox reactions involve the transfer of electrons
This document outlines the key concepts and learning objectives for Chapter 17, which covers solubility equilibria and complex-ion equilibria. The chapter will examine how to determine solubility product constants (Ksp) and use them to calculate solubility. It will also explore how the common ion effect and pH can impact solubility. The chapter will then discuss the formation of complex ions and how they relate to solubility and precipitation. It concludes by looking at applications to qualitative metal ion analysis.
This document discusses solubility equilibria and the formation of precipitates. It defines key terms like solubility product constant (Ksp), explains how to calculate Ksp values from molar solubility and vice versa, and shows examples of using Ksp to determine whether precipitates will form when solutions are mixed. The key points are that solubility is dependent on equilibrium, saturated solutions have concentrations where Ksp = Q, and precipitates form when mixing produces Q > Ksp (supersaturation).
This document discusses several key topics regarding bases:
1. The hydroxides of Group 1 and 2 elements are strong bases, with NaOH and KOH being common laboratory reagents. The alkaline earth hydroxides have low solubility.
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3. 19-3
Ionic Equilibria in Aqueous Systems
19.3 Equilibria of Slightly Soluble Ionic Compounds
19.4 Equilibria Involving Complex Ions
4. 19-4
Goals & Objectives
• See the following Learning Objectives on
page 870.
• Understand These Concepts:
• 19.9-17
• Master these Skills:
• 19.7-11, 13-16.
5. 19-5
Equilibria of Slightly Soluble Ionic Compounds
Any “insoluble” ionic compound is actually slightly
soluble in aqueous solution.
We assume that the very small amount of such a compound that
dissolves will dissociate completely.
For a slightly soluble ionic compound in water, equilibrium
exists between solid solute and aqueous ions.
PbF2(s) Pb2+(aq) + 2F-(aq)
Qc = [Pb2+][F-]2
[PbF2]
Qsp = Qc[PbF2] = [Pb2+][F-]2
6. 19-6
Solubility Product Principle
• Although all compounds dissolve in
water to some extent, some compounds
are classified as “insoluble”. The
solubility of these insoluble compounds
can be described in terms of their
solubility product constant, Ksp.
• Memorize the Solubility generalization
handout.
7. 19-7
Solubility Product Principle
• Consider AgCl dissolved in water
• AgCl(s) = Ag+ + Cl-
• Ksp = [Ag+][Cl-]
• Consider Ag2S dissolved in water
• Ag2S(s) = 2Ag+ + S2-
• Ksp = [Ag+]2[S2-]
9. 19-9
Qsp and Ksp
Qsp is called the ion-product expression for a slightly
soluble ionic compound.
For any slightly soluble compound MpXq, which consists
of ions Mn+ and Xz-,
Qsp = [Mn+]p[Xz-]q
When the solution is saturated, the system is at equilibrium,
and Qsp = Ksp, the solubility product constant.
The Ksp value of a salt indicates how far the dissolution
proceeds at equilibrium (saturation).
10. 19-10
Metal Sulfides
Metal sulfides behave differently from most other slightly
soluble ionic compounds, since the S2- ion is strongly
basic.
We can think of the dissolution of a metal sulfide as a
two-step process:
MnS(s) Mn2+(aq) + S2-(aq)
S2-(aq) + H2O(l) → HS-(aq) + OH-(aq)
MnS(s) + H2O(l) Mn2+(aq) + HS-(aq) + OH-(aq)
Ksp = [Mn2+][HS-][OH-]
11. 19-11
Sample Problem 19.5 Writing Ion-Product Expressions
SOLUTION:
PROBLEM: Write the ion-product expression at equilibrium for each
compound:
(a) magnesium carbonate (b) iron(II) hydroxide
(c) calcium phosphate (d) silver sulfide
PLAN: We write an equation for a saturated solution of each
compound, and then write the ion-product expression at
equilibrium, Ksp. Note the sulfide in part (d).
(a) MgCO3(s) Mg2+(aq) + CO3
2-(aq) Ksp = [Mg2+][CO3
2-]
(b) Fe(OH)2(s) Fe2+(aq) + 2OH-(aq) Ksp = [Fe2+][OH-]2
(c) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO4
3-(aq) Ksp = [Ca2+]3[PO4
3-]2
13. 19-13
Determination of Ksp
• One liter of saturated silver chloride
solution contains 0.00192g of dissolved
AgCl at 25oC. Calculate the molar
solubility of , and Ksp for, AgCl.
• The molar solubility of CaF2 is 2.14x10-4
at 25oC. Determine Ksp for CaF2.
17. 19-17
Sample Problem 19.6 Determining Ksp from Solubility
PROBLEM: (a) Lead(II) sulfate (PbSO4) is a key component in lead-
acid car batteries. Its solubility in water at 25°C is
4.25x10-3 g/100 mL solution. What is the Ksp of
PbSO4?
(b) When lead(II) fluoride (PbF2) is shaken with pure
water at 25°C, the solubility is found to be 0.64 g/L.
Calculate the Ksp of PbF2.
PLAN: We write the dissolution equation and the ion-product
expression for each compound. This tells us the number of
moles of each ion formed. We use the molar mass to convert
the solubility of the compound to molar solubility (molarity),
then use it to find the molarity of each ion, which we can
substitute into the Ksp expression.
18. 19-18
Sample Problem 19.6
Ksp = [Pb2+][SO4
2-]
Ksp = [Pb2+][SO4
2-] = (1.40x10-4)2
SOLUTION:
= 1.96x10-8
(a) PbSO4(s) Pb2+(aq) + SO4
2-(aq)
Converting from g/mL to mol/L:
4.25x10-3g PbSO4
100 mL soln
x 1000 mL
1 L
x 1 mol PbSO4
303.3 g PbSO4
= 1.40x10-4 M PbSO4
Each mol of PbSO4 produces 1 mol of Pb2+ and 1 mol of SO4
2-, so
[Pb2+] = [SO4
2-] = 1.40x10-4 M
19. 19-19
Sample Problem 19.6
Ksp = [Pb2+][F-]2
Ksp = [Pb2+][F-]2 = (2.6x10-3)(5.2x10-3)2 = 7.0x10-8
(b) PbF2(s) Pb2+(aq) + F-(aq)
Converting from g/L to mol/L:
0.64 g PbF2
1 L soln
x 1 mol PbF2
245.2 g PbF2
= 2.6x10-3 M PbF2
Each mol of PbF2 produces 1 mol of Pb2+ and 2 mol of F-, so
[Pb2+] = 2.6x10-3 M and [F-] = 2(2.6x10-3) = 5.2x10-3 M
20. 19-20
Sample Problem 19.7 Determining Solubility from Ksp
PROBLEM: Calcium hydroxide (slaked lime) is a major component of
mortar, plaster, and cement, and solutions of Ca(OH)2
are used in industry as a strong, inexpensive base.
Calculate the molar solubility of Ca(OH)2 in water if the
Ksp is 6.5x10-6.
PLAN: We write the dissolution equation and the expression for Ksp.
We know the value of Ksp, so we set up a reaction table that
expresses [Ca2+] and [OH-] in terms of S, the molar solubility.
We then substitute these expressions into the Ksp expression
and solve for S.
SOLUTION:
Ksp = [Ca2+][OH-]2 = 6.5x10-6Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)
21. 19-21
Sample Problem 19.7
-Initial 0 0
Change - +S + 2S
Ksp = [Ca2+][OH-]2 = (S)(2S)2 = 4S3 = 6.5x10-6
= 1.2x10-2 M
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration (M)
Equilibrium - S 2S
S = =
22. 19-22
Uses of Ksp
• Determine the molar solubility of BaSO4
in pure water at 25oC. Ksp=1.1x10-10
• Calculate the molar solubility and pH of
a saturated solution of Mg(OH)2 at 25oC.
Ksp= 1.5x10-11
• Determine the molar solubility of BaSO4
in 0.010M sodium sulfate solution at
25oC. Ksp= 1.1x10-10
26. 19-26
Table 19.3 Relationship Between Ksp and Solubility at 25°C
No. of Ions Formula Cation/Anion Ksp Solubility (M)
2 MgCO3 1/1 3.5x10-8 1.9x10-4
2 PbSO4 1/1 1.6x10-8 1.3x10-4
2 BaCrO4 1/1 2.1x10-10 1.4x10-5
3 Ca(OH)2 1/2 6.5x10-6 1.2x10-2
3 BaF2 1/2 1.5x10-6 7.2x10-3
3 CaF2 1/2 3.2x10-11 2.0x10-4
3 Ag2CrO4 2/1 2.6x10-12 8.7x10-5
The higher the Ksp value, the greater the solubility, as long as we
compare compounds that have the same total number of ions in
their formulas.
27. 19-27
Figure 19.12 The effect of a common ion on solubility.
PbCrO4(s) Pb2+(aq) + CrO4
2-(aq)
If Na2CrO4 solution is added to a saturated solution of PbCrO4, it
provides the common ion CrO4
2-, causing the equilibrium to shift to
the left. Solubility decreases and solid PbCrO4 precipitates.
28. 19-28
Sample Problem 19.8 Calculating the Effect of a Common Ion
on Solubility
PROBLEM: In Sample Problem 19.7, we calculated the solubility of
Ca(OH)2 in water. What is its solubility in 0.10 M
Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.
PLAN: The addition of Ca2+, an ion common to both solutions, should
lower the solubility of Ca(OH)2. We write the equation and Ksp
expression for the dissolution and set up a reaction table in
terms of S, the molar solubility of Ca(OH)2. We make the
assumption that S is small relative to [Ca2+]init because Ksp is
low. We can then solve for S and check the assumption.
SOLUTION:
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = [Ca2+][OH-]2
29. 19-29
Sample Problem 19.8
Change - +S + 2S
Ksp = [Ca2+][OH-]2 = 6.5x10-6 ≈ (0.10)(2S)2 = (0.10)(4S2)
= 4.0x10-3 M
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration (M)
Equilibrium - 0.10 + S 2S
[Ca2+]init = 0.10 M because Ca(NO3)2 is a soluble salt, and dissociates
completely in solution.
4S2 ≈
6.5x10-6
0.10
so S ≈
Checking the assumption: 4.0x10-3 M
0.10 M
x 100 = 4.0% < 5%
-Initial 0.10 0
30. 19-30
Effect of pH on Solubility
Changes in pH affects the solubility of many slightly
soluble ionic compounds.
The addition of H3O+ will increase the solubility of a salt
that contains the anion of a weak acid.
CaCO3(s) Ca2+(aq) + CO3
2-(aq)
CO3
2-(aq) + H3O+(aq) → HCO3
-(aq) + H2O(l)
HCO3
-(aq) + H3O+(aq) → [H2CO3(aq)] + H2O(l) → CO2(g) + 2H2O(l)
The net effect of adding H3O+ to CaCO3 is the removal
of CO3
2- ions, which causes an equilibrium shift to the
right. More CaCO3 will dissolve.
31. 19-31
Sample Problem 19.9 Predicting the Effect on Solubility of
Adding Strong Acid
PROBLEM: Write balanced equations to explain whether addition of
H3O+ from a strong acid affects the solubility of each ionic
compound:
PLAN: We write the balanced dissolution equation for each compound
and note the anion. The anion of a weak acid reacts with H3O+,
causing an increase in solubility.
SOLUTION:
(a) lead(II) bromide (b) copper(II) hydroxide (c) iron(II) sulfide
(a) PbBr2(s) Pb2+(aq) + 2Br-(aq)
Br- is the anion of HBr, a strong acid, so it does not react with H3O+. The
addition of strong acid has no effect on its solubility.
32. 19-32
Sample Problem 19.9
(b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)
OH- is the anion of H2O, a very weak acid, and is in fact a strong base. It
will react with H3O+:
The addition of strong acid will cause an increase in solubility.
OH-(aq) + H3O+(aq) → 2H2O(l)
(c) FeS(s) Fe2+(aq) + S2-(aq)
S2- is the anion of HS-, a weak acid, and is a strong base. It will react
completely with water to form HS- and OH-. Both these ions will react
with added H3O+:
The addition of strong acid will cause an increase in solubility.
HS-(aq) + H3O+(aq) → H2S(aq) + H2O(l)
OH-(aq) + H3O+(aq) → 2H2O(l)
33. 19-33
Figure 19.14 Limestone cave in Nerja, Málaga, Spain.
Limestone is mostly CaCO3 (Ksp = 3.3x10-9).
CO2(g) CO2(aq)
CO2(aq) + 2H2O(l) H3O+(aq) + HCO3
-(aq)
Ground water rich in CO2 trickles over
CaCO3, causing it to dissolve. This
gradually carves out a cave.
CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3
-(aq)
Water containing HCO3
- and Ca2+ ions
drips from the cave ceiling. The air has
a lower P than the soil, causing CO2
to come out of solution. A shift in
equilibrium results in the precipitation
of CaCO3 to form stalagmites and
stalactites.
CO2
34. 19-34
Predicting the Formation of a Precipitate
For a saturated solution of a slightly soluble ionic salt,
Qsp = Ksp.
When two solutions containing the ions of slightly soluble
salts are mixed,
If Qsp = Ksp,
the solution is saturated and no change will occur.
If Qsp > Ksp,
a precipitate will form until the remaining solution is saturated.
If Qsp =< Ksp,
no precipitate will form because the solution is unsaturated.
35. 19-35
Sample Problem 19.10 Predicting Whether a Precipitate Will
Form
PROBLEM: A common laboratory method for preparing a precipitate is
to mix solutions containing the component ions. Does a
precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is mixed
with 0.200 L of 0.060 M NaF?
PLAN: First we need to decide which slightly soluble salt could form,
look up its Ksp value in Appendix C, and write the dissolution
equation and Ksp expression. We find the initial ion
concentrations from the given volumes and molarities of the
two solutions, calculate the value for Qsp and compare it to Ksp.
SOLUTION:
The ions present are Ca2+, NO3
-, Na+, and F-. All Na+ and NO3
- salts are
soluble, so the only possible precipitate is CaF2 (Ksp = 3.2x10-11).
CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = [Ca2+][F-]2
36. 19-36
Sample Problem 19.10
Ca(NO3)2 and NaF are soluble, and dissociate completely in solution.
We need to calculate [Ca2+] and [F-] in the final solution.
Amount (mol) of Ca2+ = 0.030 M Ca2+ x 0.100 L = 0.030 mol Ca2+.
[Ca2+]init = 0.030 mol Ca2+
0.100 L + 0.200 L
= 0.10 M Ca2+
Amount (mol) of F- = 0.060 M F- x 0.200 L = 0.012 mol F-.
[F-]init = 0.012 mol F-
0.100 L + 0.200 L
= 0.040 M F-
Qsp = [Ca2+]init[F-]2
init = (0.10)(0.040)2 = 1.6x10-4
Since Qsp > Ksp, CaF2 will precipitate until Qsp = 3.2x10-11.
37. 19-37
Uses of Ksp
• Predict whether a precipitate of PbSO4
will form in a solution having [Pb2+] =
0.050M and [SO4
2-] = 0.0050M. Ksp =
1.8x10-8
40. 19-40
Selective Precipitation
Selective precipitation is used to separate a solution
containing a mixture of ions.
A precipitating ion is added to the solution until the Qsp
of the more soluble compound is almost equal to its Ksp.
The less soluble compound will precipitate in as large a
quantity as possible, leaving behind the ion of the more
soluble compound.
41. 19-41
Sample Problem 19.12 Separating Ions by Selective Precipitation
PROBLEM: A solution consists of 0.20 M MgCl2 and 0.10 M CuCl2.
Calculate the [OH-] that would separate the metal ions as
their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of
Cu(OH)2 is 2.2x10-20.
PLAN: Both compounds have 1/2 ratios of cation/anion, so we can
compare their solubilities by comparing their Ksp values.
Mg(OH)2 is 1010 times more soluble than Cu(OH)2, so
Cu(OH)2 will precipitate first. We write the dissolution
equations and Ksp expressions. Using the given cation
concentrations, we solve for the [OH-] that gives a saturated
solution of Mg(OH)2. Then we calculate the [Cu2+] remaining
to see if the separation was successful.
42. 19-42
Sample Problem 19.12
SOLUTION:
Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = [Mg2+][OH-]2 = 6.3x10-10
Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = [Cu2+][OH-]2 = 2.2x10-20
[OH-] = = = 5.6x10-5 M
This is the maximum [OH-] that will not precipitate Mg2+ ion.
Calculating the [Cu2+] remaining in solution with this [OH-]
Ksp
[OH-]2
[Cu2+] = =
2.2x10-20
(5.6x10-5)2
= 7.0x10-12 M
Since the initial [Cu2+] is 0.10 M, virtually all the Cu2+ ion is precipitated.
43. 19-43
Figure B19.1 Formation of acidic precipitation.
Chemical Connections
Since pH affects the solubility of many slightly soluble ionic compounds, acid
rain has far-reaching effects on many aspects of our environment.
44. 19-44
Complex Ion Equilibria
• Many complex ions are known to exist.
The majority consist of a metal ion with
several anions or molecules coordinated
to it. The class of compounds is called
coordination compounds.
45. 19-45
Figure 19.15 Cr(NH3)6
3+, a typical complex ion.
A complex ion consists of a central metal ion covalently bonded to
two or more anions or molecules, called ligands.
46. 19-46
Figure 19.16 The stepwise exchange of NH3 for H2O in M(H2O)4
2+.
The overall formation constant is given by
Kf =
[M(NH3)4
2+]
[M(H2O)4
2+][NH3]4
49. 19-49
Sample Problem 19.13 Calculating the Concentration of a
Complex Ion
PROBLEM: An industrial chemist converts Zn(H2O)4
2+ to the more
stable Zn(NH3)4
2+ by mixing 50.0 L of 0.0020 M
Zn(H2O)4
2+ and 25.0 L of 0.15 M NH3. What is the final
[Zn(H2O)4
2+] at equilibrium? Kf of Zn(NH3)4
2+ is 7.8x108.
PLAN: We write the reaction equation and the Kf expression, and use
a reaction table to calculate equilibrium concentrations. To set
up the table, we must first find [Zn(H2O)4
2+]init and [NH3]init
using the given volumes and molarities. With a large excess
of NH3 and a high Kf, we assume that almost all the
Zn(H2O)4
2+ is converted to Zn(NH3)4
2+.
SOLUTION:
Zn(H2O)4
2+(aq) + 4NH3(aq) Zn(NH3)4
2+(aq) + 4H2O(l)
Kf =
[Zn(NH3)4
2+]
[Zn(H2O)4
2+][NH3]4
50. 19-50
Sample Problem 19.13
[Zn(H2O)4
2+]initial = = 1.3x10-3 M50.0 L x 0.0020 M
50.0 L + 25.0 L
[NH3]initial = = 5.0x10-2 M
25.0 L x 0.15 M
50.0 L + 25.0 L
Initial 1.3x10-3 5.0x10-2 0 -
Change ~(-1.3x10-3) ~(-5.2x10-3) ~(+1.3x10-3) -
Zn(H2O)4
2+(aq) + 4NH3(aq) Zn(NH3)4
2+(aq) + 4H2O(l)Concentration (M)
Equilibrium x 4.5x10-2 1.3x10-3 -
4 mol of NH3 is needed per mol of Zn(H2O4)2+, so
[NH3]reacted = 4(1.3x10-3 M) = 5.2x10-3 M and
[Zn(NH3)4
2+] ≈ 1.3x10-3 M
54. 19-54
Complex Ion Equilibria
Determine the concentration of silver ions in
a solution that is 0.10M in [Ag(NH3)2]+. Kd =
6.3x10-8
Determine the [Ag+] in a solution that is
0.10M in Ag(NH3)2NO3 and 0.10M in NH3.
58. 19-58
Net Ionic Equations
• Rules for writing:
– A. List predominant species
• All soluble salts, strong acids and strong bases are
written as their component ions. All others are
written as the molecule.
59. 19-59
Net Ionic Equations
• Rules (continued):
– B. Combine ions of opposite charge and
look for one or more of the following:
• 1. formation of a weak acid
• 2. formation of a weak base
• 3. formation of water
• 4. formation of an insoluble substance
• 5. formation of a complex ion
60. 19-60
Net Ionic Equations
• Rules (continued):
– C. If no reaction in B. above, then look at
each of the following in the order given as a
source of secondary species:
• 1. solubility equilibria
• 2. complex equilibria
• 3. weak acid equilibria
• 4. weak base equilibria
• 5. hydrolysis equilibria (only if necessary)
61. 19-61
Net Ionic Equations
• Rules (continued):
– D. If no reaction in C, then try secondary
species reacting with secondary species
again leaving hydrolysis to last.
62. 19-62
Writing Net Ionic Equations
• Give net ionic equations for each of the
following reactions occurring in aqueous
solution. Also indicate the form of K in
terms of other constants such as Ka, Kb,
Kw, Ksp or Kd.
– 1. AgNO3 + NaCl
– 2. BaCl2 + Na2SO4
– 3. FeCl3 + NaOH