We continue our study of acid-base properties and reactions by considering the effect of common ion on the amount of acid in the solution.

The buffer solutions' pH remains largely unchanged after the addition of small amounts of acids and bases.

We conclude our study of acid-base chemistry by looking at acid-base titration. During any stage of titration, we learn to calculate the pH. Acid-base indicators are used to determine the end point of a titration. Heterogeneity is a type of equilibrium that deals with the solubility of substances. In order to express the solubility of these substances, we have to use a product. Different types of metal ion can be separated by their different products. We see how Le Chatelier's principle helps us understand the effects of common ion and pH. Lewis acid-base reaction, which is a type of ion formation, can enhance the solubility of an insoluble compound.

The solubility product principle is applied to qualitative analysis to identify ion in solution.

We will look at another type of equilibrium--that is between compounds and their ion in solution.

Weak acids and weak bases do not ionize completely. A weak acid solution has nonionized acid as well as H+ ion and the conjugate base. The system is an example of homogeneity because all of the species are dissolved.

In the second half of the chapter, we will look at another type of equilibrium, which involves the dissolution and precipitation of some substances. Heterogeneous equilibria refers to reactions in which components are in more than one phase.

We only talked about solutions containing a single solute in Chapter 15.

A weak acid or a weak base can be weakened by the presence of a common ion.

Adding CH3COO- ion from CH3COONa to a solution of CH3COOH will shift the equilibrium from right to left. The shift in equilibrium of the acetic acid is caused by the salt. The common ion is CH3COO because it is supplied by both CH3COOH and CH3COONa.

The common ion effect is an important factor in determining the pH of a solution and the amount of salt in it. The com mon ion effect is related to the pH of the solution. The common ion effect is a special case of Le Chatelier's principle.

A solution containing a weak acid, HA, and a salt of the weak acid, NaA, should be considered.

The conjugate base is A- and the acid is HA.

The Henderson-Hasselbalch equation is derived from the equilibrium constant expression. Regardless of the source of the conjugate base, it is valid.

We are usually given the starting concentrations of weak acid HA and its salt, such as NaA, in problems that involve the common ion effect. HA is a weak acid and the extent of the A- ion's hydrolysis is very small, so this is a valid approximation.

The presence of A- further suppresses the ionization of HA Student data and the presence of HA further suppresses the hydrolysis of A-.

The SmartBook can be used to view Equation.

The CH3COO- is the common ion here. The major species in solution are CH3CO OH, CH3COO-, Na+, H+, and H2O. The Na+ ion has no base or acid properties. If we know both CH3COO- and CH3COOH, we can calculate H+ at equilibrium.

When the common ion is present, the equilibrium shifts from right to left. The weak acid's ionization is decreased by this action. Fewer H+ ion are produced in (b) and the solution's pH is higher than in (a). You should always check the assumptions.

The common ion effect works in a solution with a weak base such as NH3 and a salt of the base.

The Henderson-Hasselbalch equation can be derived from this system.

The common ion NH4 suppresses the ioniza tion of NH3 in the solution containing both the base and the salt.

Chemicals and biological systems rely on buffer systems.

In a neutralization reaction, the acid and the base components of the buffer must not consume each other. The requirements are satisfied by an acid-base conjugate pair, for example, a weak acid and its conjugate base (supplied by a salt) or a weak base and its conjugate acid.

A simple buffer solution can be prepared by adding equal amounts of acetic acid and salt to water. The starting concentrations of the acid and conjugate base are the same as the equilib rium concentrations. A solution containing these two substances has the ability to destroy acid and base.

The color of the indicator is purple and yellow.

A buffer system can be represented as either salt-acid or conjugate base-acid.

The buffer system can be written as CH3COONa/ CH3COOH or simply CH3COO-/ CH3COOH.

buffer systems are different from acid-salt combinations that don't function as buffers.

Tell me your answer.

A buffer system needs a weak acid and its salt in order to function.

3PO4 is a weak acid and its conjugate base is a weak base.

This is a buffer system.

The conjugate base of 4 is an extremely weak base.

The H+ ion in solution will not combine with the ClO-4 ion. The system can't act as a buffer system.

This is a buffer system.

The buffer solution has an effect on the pH.

When the HCl is added, the volume of the solution will not change.

The conjugate base of a strong acid is the spectator ion in solution.

The conjugate base of the buffer, which is CH3COO-, reacts completely with the strong acid HCl.

Acid-Base Equilibria and Solubility Equilibria moles. When a substance is added, the volume of the solution may change. The number of moles will not be affected by a change in volume.

After neutralization of the acid, we can calculate the buffer's pH by dividing moles by 1.0 L of solution.

The addition of HCl decreases the pH by a small amount. The action of a buffer solution is consistent with this.

If 0.10 mol HCl were added to 1 L of water, we would see an increase in H+ ion concentration.

Suppose we want to make a buffer solution with a specific pH.

We work backwards to prepare a buffer solution.

Student data shows you may be close to the desired pH.

The equation is used to get the ratio. Access your eBook for additional quantities for the preparation of the buffer solution.

How would you prepare aphosphate buffer with a pH of 7.40?

The acid component must be equal to the conjugate base component in order for the buffer to function effectively.

The three stages of ionization are written because phosphoric acid is a triprotic acid.

2PO4 is close to the desired pH.

We could make a 1-L solution by dissolving 1.5 moles of Na2HPO4 and 1.0 mole of NaH2PO4 in water.

Carbonic acid (H2CO3), sodium hydrogen carbonate (NaHCO3), and sodium carbonate are provided to you.

The chemistry in action essay shows the importance of buffer systems in the human body.

The diagrams show solutions with a weak acid HA and/or a salt called NaA. The Na+ ion and water molecule are not included for clarity.

titrations involving a strong acid and a strong base, titrations involving a weak acid and a strong base, and titrations involving a strong acid and a weak base are some of the reactions we will consider. Titrations involving a weak acid and a weak base are complicated by the hydrolysis of the cat ion and the anion of the salt. It is not easy to determine the equivalency point in these cases. The titrations will not be dealt with here.

For convenience, we will only use three significant figures for volume and concentration. The acid's pH is given by -log (0. 100), or 1.00. The solution's pH increases slowly when NaOH is added. The curve rises almost vertically at the equivalence point, when the pH begins to rise steeply and equimolar amounts of acid and base have reacted.

The curve is referred to as a titration curve.

There are about 5 L of blood in an average adult. The blood that is circulating deep in the tissues carries oxygen and vitamins to keep the cells alive and removes carbon dioxide and other waste materials. Nature has provided an extremelyficient method for the delivery of oxygen and the removal of carbon dioxide using several buffer systems.

There are many compounds in the blood.

The main buffer systems in the erythrocyte are HCO-3 and H2CO3.

As a rough approximation, we can think of it as a monoprotic fusion of the erythrocyte and the blood plasma acid of the form HHb: to the lungs. This is the main way to remove carbon dioxide.

It diffuses into the blood capillaries and erythrocytes. The carbonic anhydrase converts it to carbonic acid. Two anions form HHbO2, which eventually splits into HHb and O2. Oxygen diffuses out of the erythrocytes and into the tissues because the partial pressure of O2 is higher in the erythrocytes. The erythrocytes diffuse out of them and carry the bicarbonate ion to the lungs. Oxygen diffuses from the lungs to the erythrocytes. They combine with HHb to form HHbO2. The erythrocytes from the plasma are able to form carbonic acid with the help of the HHbO2 protons. H2O and CO2 are converted to carbonic acid in the presence of carbonic anhydrase. The CO2 diffuses out of the erythrocytes and into the lungs.

The carbon dioxide diffuses to the lungs and is ex up by other cells.

The formation of the Hb- ion is due to the reaction when the blood returns to the lungs.

The entire cycle is repeated when the blood flows back to the body tissues.

Acid-Base Equilibria and Solubility Equilibria of the base can cause a large increase in the solution's pH. The pH increases with the addition of NaOH.

The solution's pH can be calculated at every stage of titration. There are three sample calculations.

The solution has a total volume of 35.0 mL.

The amount of HCl left after partial neutralization is either 2.50 x 10-3) or 1.50 x 10-3 mol.

This is a simple calculation that involves a neutralization reaction.

The total volume of the solution has increased.

The solution has 2.50 moles of HCl in it. The number of moles of NaOH left after neutralization is either 3.50 x 10-3) or 2.50 x 10-3).

The equivalence point is greater than 7 due to the hydrolysis of the salt.

The excess OH- ion formed. Access to the SmartBook is similar to the hydrolysis of CH3COONa.

At every stage of the titration, we can calculate the number of moles of base reacting with the acid, and the pH of the solution is determined by the excess acid or base left over. The neutralization is complete at the equivalence point and the solution's pH will be dependent on the amount of salt formed.

When two solutions are mixed, the solution volume increases. The number of moles will not change as the volume increases.

There is a buffer system made up of CH3COOH and CH3COO.

The acid and base concentrations are zero at the equivalence point.

The next step is to calculate the solution's pH, which is the result of the CH3COO- ion hydrolysis.

The two species that are responsible for making the solution basic are OH- and CH3COO-. Because OH- is a stronger base than CH3COO-, we can use only the concentration of OH- ion to calculate the solution's pH.

The pH at the equivalence point is lower than 7 because of salt hydrolysis.

NH4 and H2O are separated into 4Cl. The concentration of NH4 formed is determined first. The NH+4 ion hydrolysis causes the pH to be calculated. The conjugate base of a strong acid does not react with water. We don't pay attention to the ionization of water.

The number of moles of NH3 is equal to the number of moles of HCl.

The acid and base concentrations are zero at the equivalence point.

The solution's pH is determined by the NH+4 ion hydrolysis. Section 15.10 is where we follow the procedure.

The solution is acidic. We would expect this from the hydrolysis of the ion.

The number of moles of OH- ion added to a solution is equal to the number of moles of H+ ion originally. To determine the equivalence point in a titration, we need to know how much volume of a base to add from a buret to an acid in a flask. Adding a few drops of an acid-base indicator to the acid solution at the start of the titration is one way to achieve this goal. You will recall from Chapter 4 that an indicator is usually a weak organic acid or base that has different colors in its nonionized and ionized forms. The two forms are related to the solution in which the indicator is dissolved. The choice of indicator for a particular titration depends on whether the acid and base used in the titration is strong or weak. We can use the end point to determine the equivalence point if we choose the right indicator.

HIn is a weak monoprotic acid. To be an effective indicator, HIn and its conjugate base must have different colors. In a basic medium, the equilibrium shifts to the right and the color of the solution is due to the conju gate base.

Because the curve is steep, this choice ensures that the pH at the equivalence point will fall within the range that the indicator changes color. The indicator for the titration of NaOH and HCl is phenolphthalein. In acidic and neutral solutions, phenolphthalein is pink in basic solutions. The indicator is not red when the pH is less than 8.3, but it is when it is. The steepness of the pH curve near the equivalence point means that the addition of a very small quantity of NaOH brings about a large rise in the pH. The range over which phenolphthalein changes from pink to colorless is included in the steep portion of the pH profile. The indicator can be used to find the equivalence point of the titration when there is a cor respondence.

Plants have many acid-base indicators. By boiling chopped red cabbage in water we can get many different colors. There are a number of indicators used in acid-base titrations. The strength of the acid and base will affect the choice of an indicator.

The pH range for color change must overlap the steep portion of the titration curve in order for an indicator to be used. We can't use the color change to find the equivalence point.

The range over which the indicator changes from acid to base color is known as the pH range.

All of the indicators are suitable for use in the titration.

Refer to Table 16.1 to specify which indicator or indicators you would use for the following titrations: (a) HNO3 versus NaOH, (b) HNO2 versus KOH.

In industry, medicine, and everyday life, precipitation reactions are important. The preparation of essential industrial chemicals such as Na2CO3 is based on precipitation reactions. tooth de cay is caused by the dissolving of tooth enamel in an acidic medium. Barium sulfate, an insoluble compound that is opaque to X rays, is used to diagnose ailments of the digestive tract. Many foods, such as fudge, are produced by precipitation reactions in stalagmites and stalactites.

Section 4.2 introduces the general rules for predicting the solubility of ionic compounds in water. The solubility rules do not allow us to make predictions about how much of a compound will be dissolved in water. We start with what we know about chemical equilibrium to develop a quantitative approach.

A saturated solution of silver chloride is in contact with a solid silver chlo ride. It is assumed that the small amount of solid AgCl that is dissolved in water will break down into Ag+ and Cl- ion.

The product expression of each AgCl unit is very easy to write.

There are a number of salts of low solubility listed in Table 16.2.

Section 15.3 assumes ideal behavior for dissolved substances, but this assumption is not always valid. The neutral and charged ion pairs of BaF2 and BaF+ can be found in a solution of barium fluoride.

Many anions in the ionic compounds listed in Table 16.2 are conjugate bases of weak acids. Consider copper sulfide.

If the solution is saturated, or if the solution is supersaturated, then the dissolution of an ionic solid in a solution is possible.

The subscript 0 reminds us that the initial concentrations do not correspond to the equilibrium concentrations.

It is convenient to use both molars in the laboratory.

The CaSO4 is found to be 0.67 g/L.

CaSO4 can be found in water.

For every mole of CaSO4 that is dissolved, 1 mole of Ca2+ and 1 mole of SO2 are produced. 4.5 x 10-5 g/L is the solubility of lead chromate.

The same procedure can be used to calculate its molar solubility. The species present at equi librium are identified first. We have Ag+ here.

There are learning resources on this topic.

This approach is used in example 16.9

The data in Table 16.2 shows the solubility of copper in g/L.

It is a pesticide and can be used to treat seeds.

The concentration of OH- is twice that of Cu2+.

As shown in Examples 16.8 and 16.9, there is a correlation between the two products. We can calculate the other if we know one. The table shows the relationship between the two products.

A saturated solution is created by dissolving a substance in a certain amount of water.

Thebility product is an equilibrium constant.

When we mix two solutions or add a compound to a solution, we can predict whether a precipitate will form. Practical value is often achieved by this ability.

Predicting precipitation reactions is useful in medicine.

Oxalate ion (C 2- 2O4 ), derived from oxalic acid present in many vegetables, react with cal cium ion to form insoluble calcium oxalate, which can gradually build up in the kid neys. Proper adjustment of a patient's diet can help.

There are steps involved in predicting precipitation reactions.

The solution has Ba2+, K+, and SO2 in it. BaSO4 is the only precipitate that can form according to the rules listed in Table 4.2. We know the number of moles of the ion in the original solutions and the volume of the combined solution, so we can calculate Ba2+ and SO2 4. A sketch of the situation is helpful.

The total volume after combining the two solutions is 800 mL.

The diagrams show solutions of AgCl that do not affect the solubility of AgCl. If (a) is a saturated solution, the other solutions should be classified as saturated, saturated, or super saturated.

In chemical analysis, it is desirable to remove one type of ion from solution while leaving other types in solution. Adding sul fate ion to a solution containing both potassium and barium ion causes BaSO4 to be removed from the solution. K2SO4 will remain in solution. The BaSO4 can be separated from the solution.

When silver nitrate is added to the solution, AgI begins to form first, followed by AgBr and then AgCl.

If a solution containing more than two different types of ion can be formed, the procedure can be applied to it.

Solid AgNO3 is added to the solution without changing the volume.

The Ag+ ion combine with the other ion to form AgBr and AgCl. This is a fractional precipitation problem. To start precipitation, the concentration in the saturated solution must be greater than the Ag+ concentration.

There is a suspension of AgCl and AgBr.

The products of Ag3PO4 are 1.6 x 10-10 and 1.8 x 10-18. If Ag+ is added to a solution containing 0.10 mol Cl- and 0.10 mol PO3- 4 without changing the volume, you can calculate the concentration of Ag+ in the solution.

Just before AgCl begins to occur, 100 - 0.48 percent, or 99.52 percent, of Br- will have precipitated. The Cl- ion can be separated by this procedure.

Section 16.2 deals with the effect of a common ion on acid and base ionizations.

The relationship between the common ion effect and solubility will be looked at here.

Simple stoichiometry tells us that [Ag+] is the same as [Cl-].

Equality does not hold in all situations.

We could study a solution containing two dissolved substances that share the same ion.

The common ion effect is shown in the example.

Student data shows you may struggle with the effect of a common ion.

This is not a new problem. Ag+ is supplied by both AgNO3 and AgCl.

The NO3 is a spectator ion.

The practice exercise shows that AgCl can be found in pure water at a rate of 1.9 x 10-3 g/L. In the presence of AgNO3 the lower solubility is reasonable. Le Chatelier's principle can be used to predict the lower solubility. The equilibrium is shifted to the left when Ag+ ion is added.

The solubilities of substances depend on the solution's pH. Adding H+ ion causes the equilibrium to shift from left to right and increases the solubility of Mg(OH)2. In acidic solutions, insoluble bases tend to dis solve. Insoluble acids can be dissolved in basic solutions.

Let's first calculate the pH of a saturated Mg(OH)2 solution.

In a medium with a pH of less than 10.45, the solubility of Mg(OH)2 would increase.

The equilibrium condition is affected by [Mg2+] rising to main and more Mg(OH)2 dissolving.

The solubility of salts that contain a basic anion is influenced by the pH.

The equilibrium condition must be maintained as F- decreases.

The solubilities of salts that don't hydrolyze are unaffected. Examples of such anions are I-.

Write the reaction of the salt into its cation and anion.

The H+ ion and the cation both have positive charges. If the conjugate base of a weak acid is the anion, it will act as a proton acceptor.

The equilibrium will shift to the right to replace some of the S2 ion that was removed.

The weak acid HSO-4 is the conjugate base of the sulfate ion.

The equilibrium will shift to the right to replace some of the SO 2- 4 ion that were removed, which will increase the solubility of PbSO4.

The concentration of OH- is in a saturated solution. The concentration of NH3 that will supply this concentration is calculated. The precipitation of Fe(OH)2 will begin if the NH3 concentration is greater than the calculated value.

Ammonia reacts with water to form Fe.

The changes in concentrations are summarized.

There are many chemical and biological processes. The effect of complex ion formation on solubility will be considered here. We will discuss the chemistry of complex ion in more detail in Chapter 23.

The property allows them to act as Lewis acids in reactions with many molecule or ion that serve as electron donors or as Lewis bases.

The copper(II) ion is responsible for this color, as are many other sulfates.

The formation constants of a number of complex ion are listed in Table 16.4

Student data shows you may struggle with the effect of complex ion formation.

The concentration of Cu2+ will be very small. We can assume that all of the dissolved Cu2+ is Cu(NH 2+ 3)4 ion. There will be a very small amount of Cu2+ present.

The amount of NH3 used to form the complex ion is 4 x 0.20 mol.

The equilibrium for the process will be given by combining these two equilibria.

The species in solution are Ag+ and NH3.

The complex ion Ag(NH3)2 is produced by 3.

The formation constant for Ag(NH + 3)2 is large, so most of the silver is in the complexed form. In the absence of ammonia we have an equilibrium Ag+. As a result of complex ion formation, we can write Ag(NH + 3)2

eggshells are 40 percent calcium and weigh about 5 g. Most of the calcium in an eggshell is laid down in 16 hours. It is deposited at a rate of about 125 million units per hour. The hen can't consume calcium fast enough. The hen's long bones accumulate large reserves of cal cium for eggshell formation.

There are columns of calcite in the X-ray micrograph of an eggshell.

The blood is carried to the shell glands.

Chickens don't perspire and so must pant to cool their gland as more is provided by the dissociation of the protein selves. The CO2 from the chicken's body is expelled by panting.

The CO2H2CO3 equilibrium shown from the right metabolic byproduct will be shifted by the carbonate ion necessary for eggshell formation. When metabo is left, carbon dioxide is produced and the concentration of CO 3- 3 is lowered, which leads to the conversion of H2CO3 to carbonic acid. Chickens can drink in hot weather if they are given carbonated water.

The formation of the complex ion Ag(NH + 3)2 enhances the solubility of AgCl.

They are examples of Al(OH)3, Pb(OH)2, and Zn(OH)2. Other amphoteric hydroxides do the same thing.

In Section 4.6, we talked about the principle of gravimetric analysis, in which we know the amount of an ion in an unknown sample. The focus will be on the cations.

Adding precipitating reagents to an unknown solution is a general procedure for separating 20 ion.

The Ag+, Hg 2+ 2 and Pb2+ ion are insoluble in the solution. The chlorides in the other ion remain in solution.

The hydrogen sulfide is reacted with an unknown acidic solution after the chlorides have been removed. The concentration of the S2 in solution is insignificant.

Bi2S3, CdS, CuS, HgS, and SnS are listed in Table 16.

The solution is made basic with the addition of sodium hydroxide. The moresoluble sulfides are now out of solution. The hydroxides Al(OH)3 and Cr(OH)3 are less soluble than the sulfides, so the Al3+ and Cr3+ ion are more likely to form hydroxides. There are insoluble sulfides and hydroxides in the solution.

After the group 1, 2, and 3 cations have been removed from the basic solution, sodium carbonate is added to make BaCO3 and CaCO3. The precipitates are removed from the solution.

Left to right are the flame colors.

The metal ion gives a characteristic color when heated. The color emitted by Na+ ion is yellow, that of K+ ion is violet, and that of Cu2+ ion is green.

Figure 16.14 shows the separation scheme.

There are two points regarding qualitative analysis.

The cations in group 1 form insoluble sulfides. Second, the removal of cations at each step must be done as quickly as possible. If we don't add enough HCl to remove all the group 1 cations, they will become insoluble sulfides, interfering with further chemical analysis and leading to er roneous conclusions.

A solution has both Zn2+ and Pb2+ ion.

Determine the equilibrium concentrations of the solution. Discuss how to prepare a buffer. A buffer of a given pH can be made with the appropriate weak acid. The type of acid and base used will affect the titration's pH. Evaluate the pH throughout a titration between acids and bases. Examine a curve to find out what type of titration was done.

The acid and base species are used in a titration. Show how common ion affects the solubility of salts.

The ioniza 5 is suppressed by the common ion effect. The principle of Le Chatelier's can be explained.

A weak 6 is combined with a buffer solution. The presence of a common ion decreases the solubil acid and its weak conjugate base.

As the hydrogen ion con solution remains nearly constant, the pH of the basic anions increases as the solubility of slightly soluble salts containing added acid or base increases. Salts have a vital role in maintaining the pH of fluids.

The acid-base titra tion depends on the amount of salt formed in the 8. The solution is formed by a neutralization reaction. The complex point is greater than 7 and the pH at the equivalence point is less than 7.

The identification of cations 4 is the subject of a qualitative analysis. Weak organic acids or bases and anions in solution are acid-base indicators.

For a weak base, do the same.

Le Chatelier's principle can be used to explain how the 5.9 and 8.1 are. The common ion effect affects the pH of a solution.

The temperature is assumed to be 25 degrees.

The buffer system made up of each sphere has a pH of 0.1 mol.

The blood's pH is 7.40 Keep volume at 3 ]/[H2CO3].

The base is added to the acid in an Erlenmeyer 8.60 in order to make a buffer solution.

A 0.2688-g sample of a monoprotic acid neutralizes 2A, NaHA, and Na2A, where H2A is a weak diprotic acid.

The protons of the acid should be NH3 and HCl. Do you know the solution that correrated?

The concentra equivalency point is beyond the equivalence point.

Water and HCOOH have not been included for clarity.

The volume of base re quired was 18.4 mL.

HNO2 solution is measured with a KOH 16.33 to calculate the pH at the equivalence point.

The pH should be calculated at the equivalence point for the NaOH solution. The solution's pH read NaOH.

The diagrams show solutions at dif ferent stages in the titration of a weak acid HA with NaOH.

Is HCOOH versus NaOH greater than, less than, or equal to 7 at the equiv KOH? Is HNO3 versus CH3NH2 greater than, less than, or equal to 7 at the equiv KOH?

A student carried out a titration.

She saw a faint reddish- pink color at the equiva lence point. The solution gradually turned white after a few minutes.

The nonionized form is red and the ionized form is yellow.

A few drops when CuI starts to form.

BaSO4 can be used to distinguish between the three types of product.

AgNO3 is the same as pure water.

The solubility product of PbBr2 is small.

In a 2X3 of 1.00-L, the molar solubility of AgCl is 3.6 x 10-18 g/L.

The MOH is 9.68.

The molar solubility of Fe(OH)2 in a solu tion is calculated.

The product of Mg(OH)2 is 1.2 x 10.

The indicator methyl orange has concentrations of Cu2+ of 3.46.

To calculate the mass percent of iodide in a species at equilibrium, the 3 solution is needed: Al(OH)3 or sample.

The solution has a pH of 14.

CH3COONa should be mixed with 1.0 L of solution to get the original volume.

Both NaOH and AgOH are not.

The mixed solution had a higher pH than analysis.

It is an insoluble compound. A student adds HCl acid to a solution. What type of reaction is it?

Pb2+ is still in the solution.

Ba(OH)2 16.89 are white.

The acid is called caccialic acid. A 2.0-L kettle has 116 g of boiler scale solutions.

The solution is mixed.

In g/L, calculate the solubility of Ag2CO3.

What is Zn2+?

Adding an excess of HCl acid 3NH2 solution from a buret can be used to determine CH.

3NH2 solution has been added after 25.0 mL of CH. The 3NH2 solution has been added.

The NaIO3 solution is about 10 mol/L.

Identification of M. Pb(IO carbonate and identify it.

When a solution of 16.115 Acid-base reactions is added, they usually go to completion.

A student plotted the mass of the pre rium constant for each of the following cases: a cipitate versus the volume of the KI solution added strong acid reacting with a strong base; and a strong and obtained the following graph.

Strong acids and strong bases exist in solution.

The neutralization was added by the NaOH solu Volume of KI.

A patient drinks a suspension of solution of CH3COONa and reacts with a solution of 20 g BaSO4. We may be able to give a good estimate.

The stituents would change over time.

There is a chemical reason for the color change.

What reagents would you use to separate the NaP from the solution?

The NaP solution is formed when the salt is dissolved in water.

A solution with both compounds is saturated with NaF.

Plot distribution compounds are not used as buffer curves for CH3COOH.

The main reason for the change is a form of insoluble salts, or curds. There is a way to remove acid-base reaction. Adding washing with "polyphenols" which are weak acids and soda will tea the Ca2+ ion from hard water.

0.05 g solution is the maximum allowable concentration of Pb2+ lime for drinking water. Is this solution a guideline?

The building blocks of the human body are made of mino acids.

It is a weak monoprotic acid.

The gate base is one of the three possible forms of glycine, depending on the solution's pH.

NH of 4.5 to 5.0 is fully protonsated.

The gate acid-base pair shown in (a) is responsible for 7.0 and 12.0.

A sample of 0.96 L of HCl was found at 372mmHg and 22degC.

The solution is constant.

The maximum mass of each NaOH solution added when the pH of the following salts can be added is (a) 2.85 and (b) 3.15, and (c) 11.89.

2 without causing a pre cipitate to form.

Draw distribution curves for an Assume volume to remain constant.

Only two of the three compose the compound.

The present erated 2CO3, HCO3 and CO3 exerts in a small amount of concentrations.

The two curves are representative of the titration of an acid solution.

Which solution?

Use appropriate equations to account for the solubil of a weak diprotic acid.

The major species are present at low and high pHs.

Table 16.2 shows that silver bromide has a larger product than iron(II) hydroxide.

What is the ratio of neutral aspirin to deprotonated 2 groups?

The zinc(OH)2 will form.

Capacity will be created by AgBr.

When the pH FeS are more acidic in solution than in water, BaSO4 is accurate.

A lot of physical and chemical changes occur when an egg cooks.

The shell protects the inner components from the outside environment, but it has many tiny pores through which air can pass. The albumen is made of 88 percent water and 12 percent protein. The yolk has 50 percent water, 34 percent fat, 16 percent pro tein, and a small amount of iron.

They are made up of a group of acids. In solution, each long chain of a molecule folds in such a way that the parts of the molecule that are in contact with the solution are buried inside. The heat causes the molecule to unfold. acids and salt can be denatured. A semirigid opaque white solid is formed when the denatured parts of the proteins clump together. When the shell of a boiled egg is cracked, it can smell hydrogen sulfide, an unpleasant smelling gas that can sometimes be detected when the sulfur in them combines with hydrogen.

One of the hard-boiled eggs has been boiled for 12 minutes and the other has been over cooked. The outside of the overcooked yolk is green.

Hard-boiled eggs can crack in water. The best way to hard boil eggs is to place the eggs in cold water and then bring the water to a boil.

The cords that anchor the yolk to the shell are called chalazae.

If you want to hard boil eggs, place room-temperature eggs or cold eggs from the refrigerator in boiling water.

An experienced cook adds salt to the water in order to minimize the formation of streamers.

Write an equation for the formation of the green substance on the outer layer of the egg. If the egg is washed with cold water after it has been removed from the boiling water, the green yolk can be eliminated.

To distinguish a raw egg from a hard-boiled egg, spin the eggs.

A hard-boiled egg and a 12-minute egg.

The chapter begins with a discussion of the three laws of thermodynamics. We can see that the function of entropy is predicting the outcome of a reaction. The number of microstates associated with the system can be used to calculate the system's entropy. In practice, calorimetric methods are used to determine entropy and standard values are used for many substances.

The second law of thermodynamics states that the universe's entropy increases in a random process and stays the same in an equilibrium process. We learn how to make up for the change in the universe's entropy by calculating the change of a system and surroundings. The third law of thermodynamics allows us to determine the absolute value of a substance.

A new function called Gibbs free energy is needed to focus on the system. The change in free energy can be used to predict. The equilibrium constant of a reaction is related to the change in free energy. The chapter ends with a discussion of how living systems are applied to the elements. The principle of coupled reactions is important in many biological processes.

We can use informa tion gained from experiments on a system to draw conclusions about other parts of the system. Chapter 6 shows that it is possible to calculate the enthalpy of reaction from the standard enthalpies. The second law of thermodynamics is introduced in this chapter. The relationship between free energy and chemical equilibrium is discussed.

The first law of thermodynamics states that energy can be converted from one form to another, but it cannot be created or destroyed.

Chemical processes tend to favor one direction according to the second law of thermodynamics. The second law is an extension of the third law.

One of the main objectives in studying thermodynamics is to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions. Whether one is trying to understand the intricate biological processes in a cell, trying to synthesise compounds in a research laboratory, or manufacturing chemicals on an industrial scale, this knowledge is important.

A waterfall runs downhill but never up.

There is a lump of sugar in a cup of coffee, but it doesn't come back in its original form.

Water and ice melt at the same time.

The reverse of heat flows from a hotter object to a cooler one.

The expansion of a gas into a bulb is a process.

The gathering of all the molecule into one bulb is not a result of chance.

A piece of metal reacts with water to form something. Hydrogen gas does not react with sodium hydroxide to form water.

Iron exposed to water and oxygen does not spontaneously change back to it's original state.

Under the same conditions, processes that occur in one direction can not occur in the opposite direction.

We can explain why a ball rolls downhill and why a clock unwinds if we assume that there is a decrease in the energy of the system.

A large number of reactions are not planned.

The assumption that a system always decreases its en ergy fails. Even though the process is endothermic, ice can spontaneously melt above 0degC.

This process is also endothermic.

From a study of the examples mentioned and many more cases, we come to the conclusion that exothermicity does not favor the spontaneity of a reaction. It is possible for an exothermic reaction to be nonspontaneous, just as it is possible for an endothermic reaction to be nonspontaneous. We can't decide whether or not a chemical reaction will occur spontaneously because of changes in the system.

We need to introduce a new thermody namic quantity called entropy in order to predict the spontaneity of a process. The greater the dispersal, the more it is. Most processes have a change in entropy. A cup of hot water has a certain amount of en tropy due to the dispersal of energy among the various energy states of the water mole cules. The water loses heat if it's left on a table. There is an increase in the number of energy states in the air.

The system has a certain amount of power. The gas molecule now has access to the volume of both bulbs after opening the valve.

It is important to provide a proper definition of entropy before introducing the second law of thermodynamics. There is only one way to arrange the molecule in the left compartment, four ways to have three in the left compartment and one in the right compartment, and six ways to have two in each of the two compartments. Distribution III is the most probable because there are six ways to achieve it and distribution I is the least probable. The analysis shows that the probability of occurrence of a particular distribution depends on the number of ways in which the distribution can be achieved. It's not difficult to see that the number of molecules will be evenly distributed between the two compartments because of the large number of microstates in this distribution.

There are other ways to distribute the four molecule between the two compartments.

We can have all four molecule in the right compartment and three molecule in the right compartment and one molecule in the left compartment.

Like enthalpy, entropy is a state function. Take a look at a process in a system.

There is an increase in the system's entropy.

The increase in dispersal of energy resulted in an increase in the system's entropy. There is a correlation between the qualitative description of entropy in terms of dispersal of energy and the quantitative definition of entropy in terms of microstates given by Equation (17.1).

We will study several processes that lead to a change in the number of microstates of the system.

The number of microstates is small in a solid and the atoms are confined to fixed positions. As the atoms move away from the lat tice points, they can occupy many more positions. The number of microstates increases because there are more ways to arrange the particles. The vaporization process will lead to an in crease in the system. The increase will be greater than melting due to the fact that there are more microstates in the gas phase than in the liquid phase. The solution process can lead to an increase in entropy. The highly ordered structure of the solid and part of the ordered structure of water break down when a sugar crystal is dissolved in water. The solution has more microstates than the pure solute and pure solvent combined. Sider hydration causes water to become more ordered around the ion. The process reduces the number of solvent molecule microstates. For small, highly charged ion such as Al3+ and Fe3+, hydration can outweigh the increase in entropy due to mixing and dissociation so that the overall process can actually be negative. The system's entropy is increased by heating. The molecule can execute both rotational and vibrational motions. The energy associated with all types of motion increases as the temperature increases.

Increasing temperature causes more microstates to become avail able, thus increasing the system's entropy.

Diamonds have a smaller standard entropy than graphite. Consider methane and ethane. A more complex structure and more ways to execute motions increase its microstates. ethane has a higher standard entropy than methane.

There are more microstates associated with these atoms.

The hydrogen ion in solution is assigned a zero value of entropy because the individual ion cannot be studied. One can use this scale to determine the entropy of the chloride ion from the measurement on HCl, and so on. Some ion have positive and negative values. The signs are deterred by the amount of hydration. If the hydrogen ion has less hydration than the ion, then the ion's entropy has a negative value. Positive entropies hold the opposite.

Predict whether the change is greater or less than zero for each of the following processes.

We look at whether the number of microstates of the system increases or decreases.

The process is not spontaneously described.

Proceed according to example 17.2 if you access your eBook for additional.

There are learning resources on this topic.

The standard entropy changes can be calculated from the values in Appendix 2.

To calculate the standard entropy of a reaction, we look up the standard entropies of reactants and products in Appendix 2.

When 1 mole of CaCO3 becomes 1 mole of CaO

The subscript is omitted for simplicity.

It makes sense that gases have greater entropy than liquids.

Knowing the nature of reactants and products makes it possible to predict changes in entropy.

Predict the change of the system's entropy in the following reactions.

Predicting, not calculating, the sign of entropy is what we are asked to do. There are two factors that lead to an increase in entropy, a transition from a conjugate phase to the vapor phase and a reaction that produces more product than reactant molecule in the same phase. It's important to compare the product and reactant molecule's complexity.

All molecules are diatomic and have the same complexity.

The heat that is transferred to the surroundings enhances the motion of the molecule in the surroundings. There is an increase in the number of microstates. An endothermic process in the system absorbs heat from the surroundings and decreases the motion of the sur roundings.

The amount of heat absorbed depends on the temperature. The molecules are very energetic if the temperature is high. The effect of heat absorption from an exothermic process in the system will not have a big impact on the motion of the surrounding environment. If the temperature of the sur roundings is low, the addition of the same amount of heat will cause a bigger increase in motion.

Student data shows you may occur at an observable rate. The synthesis of ammonia is very slow.

There is a chance that a reaction will occur on this topic.

Chemical kinetics is the subject of reaction rates.

The larger the number of microstates a system possesses, the larger the system's entropy. At absolute zero, consider a perfect crystal substance.

The number of microstates increases as the temperature increases. The en tropy of any substance is greater than zero. The number of microstates would be greater than one if the crystal is impure or has defects.

Standard entropies are usually referred to as absolute entropies because they are measured at 1 atm. We can't have the absolute energy or enthalpy of a sub stance because the zero of energy or enthalpy is not defined. It has a zero en tropy value if it is a perfect crystalline substance. As it is heated, it increases in entropy. As the liquid state is formed, there is an increase in entropy. The en tropy of the liquid increases due to enhanced motion. The liquid-to-vapor transition causes a large increase in entropy at the boiling point. The temperature of the gas continues to rise.

Consider the gas-phase reaction of A2 and B2 to formAB3.

What happens in a particular system is usually what we are concerned with. If we only consider the system itself, we will not be able to determine whether a reaction will occur spontaneously.

There is a reaction in the forward direction.

The reaction is notpontaneous. There is a reaction in the direction of the op.

The system is in equilibrium. There is no change.

The table summarizes the standard states of pure substances as well as the solu solution 1 molar tions.

One of the founding fathers of thermodynamics was a modest and private individual who spent most of his professional life at Yale University. James Maxwell was so impressed with his works that he gave him a star rating, but because he pub lished most of his works in obscure journals, he never got a star rating. Today, very few people outside of chemistry and physics have ever heard of him.

As the engine they range from automobile engines to the giant steam turbine does work, some of the heat is given off to the surroundings, or that run generators to produce electricity.

A heat engine is shown in the figure.

The gas expands in the cylinder.

The apparatus will return to its original state. The up and down movement of the piston can be made to do chanical work by repeating this cycle.

When heat engines work, some heat must be given off to the surroundings.

At a power plant, superheated steam is used to drive a turbine. The heat sink has a temperature of 311 K.

The maximum efficiency of a steam turbine is only 40 percent because of a number of factors.

To calculate the standard free-energy change of a reaction, we have to look up the standard free energies of formation of reactants and products in Appendix 2.

The effects of the possibilities are summarized in Table 17.3.

Reaction happens at high temperatures.

Reaction happens at low temperatures.

Two specific applications of Equation will be considered.

Calcium oxide, also called quicklime, is an extremely valuable sub stance used in steelmaking, production of calcium metal, the paper industry, water treat ment, and pollution control.

To form CaCO3. The pressure of CO2 in equilibrium with CaCO3 and CaO increases with temperature. In the Source industrial preparation of quicklime, CO2 is removed from the kiln to shift the equilibrium from left to right, promoting the formation of calcium oxide.

The temperature at which the reaction begins to favor products is the most important information for the practical chemist. It is possible to make a reliable estimate of that temperature.

The pressure of CO2 is so low that it can't be measured.

Two points are worth making a calculation about.

The standard-state value is CO 1 below 1 atm.

The equilibrium pressure of CO2 is 1 atm.

We should first consider the ice-water equilibrium.

There is an increase in the entropy of 22.0 J/K * mol when 1 mole of ice is melted.

There are some very inter esting thermodynamic properties of a rubber band.

The following experiments can be performed with a rubber band that is at least half a cm wide. Press the rub ber band against your lips. You will feel warm. Next, change the process. Hold the rub ber band in place for a few seconds.

You will feel a cooling effect this time.

In the laboratory, we carry out either ice to water or water to ice transition.

The fusion and vaporization of benzene have a molar heat of 10.9 kJ/mol and 31.0 kJ/mol, respectively.

The boiling point and melting point of argon are -190degC and -188degC, respectively.

The closed flask shows the I2 at 45 degrees.

Not all reactants and products will be at their standard states during a chemical reaction.

The net reaction will go from left to right until a significant amount of product is formed.

The net reaction will go from right to left until a significant amount of reactant has been formed.

The equilibrium constant of a reaction is one of the most important equations in the field of thermodynamics, and Equation (17.14) is one of the most important equations for chemists.

As the reaction progresses and becomes zero at equilibrium, the free-energy change of the system varies.

There is a significant conversion of reactants to products.

At equilibrium, products are favored over reactants.

At equilibrium, products and reactants are equally favored.

At equilibrium, reactants are favored over products.

The examples show the use of Equations.

The small equilibrium constant is consistent with the fact that water does not spontaneously break down into hydrogen and oxygen gases at 25degC.

In Chapter 16 we talked about the solubility product.

The reactant and product are not at their standard state of 1 atm.

The net reaction proceeds from left to right.

Consider an industrial process. We might want to extract zinc from the sphalerite.

There are a wide variety of nonspontaneous reactions in biological systems.

In a living cell, this reaction doesn't happen in a single step, but in a series of steps. A release of 31 kJ/mol of free energy can be used to drive unfavorable reactions, such as the synthesis of proteins, under ap propriate conditions.

They are made from the same group of acids. The joining of individual amino acids is involved in the stepwise synthesis of a molecule. Consider the formation of the dipeptide from alanine and glycine.

The reaction doesn't favor the formation of product, and so only a small amount of the dipeptide would be formed at equilibrium.

Figure 17.10 shows the interconversions that act as energy storage and free-energy release.

The trends in standard entropy of atoms and molecules are compared.

Entropy is a measure of different ways. A system can be used for a chemical or physical process. The second law of thermodynamics can be used to predict the spontaneity of a verse.

calcu 6 is the standard entropy of a chemical reaction.

The law allows us to measure the free-energy change of the reaction.

There are constant temperature and pres 8. There are many biological reactions.

The comparison uses the same amount.

Tell the basis for your choice.

The reasons for your arrangement should be given.

Give reasons before 17.4.

The mol is 8.6 kJ/mol.

Couldn't put Humpty together again.

CO is 1.20 atm.

Give a description of the signs.

The following reaction is a result of stances as well as other liquids.

Explain why the ratio is smaller than 90 J/K.

Carbon monoxide and NO are pollutants in automobile exhaust.

These gases can be made to react with nitrogen and carbon dioxide.

The reactions were carried out under standard-state conditions.

The data in Appendix 2 to cal is needed for the result of the decomposition of barium carbonate.

Can a substance ever have a point inkelvins?

There is a reaction at 72degC.

The heating copper(II) oxide does not produce an ing point.

The coupled reaction is caused by the heat of the vaporization.

The internal engine of a 1200- kilogram car is designed to be 78.3 degrees C.

The removal of the car can be done on 1.0 gallon of fuel.

Consider the amount of magnesium carbonate that floats in the sea.

It begins to favor products.

Comment on the feasibility of copper mining will produce useful work.

Explain why the first law of thermodynamics makes it impossible to have such a machine.

An ocean liner can scoop up sulfur dioxide.

The first law of is transferred from a region of lower concentration to one of higher concentration does not apply to this process. This is a nonsponta from the ocean.

The machine can't exist because of the concen.

There is a need for large quantities of hydrogen.

The vapor pressure is calculated at 25degC.

The length of the thermodynamic values is observed to shrink slightly. Give the temperature.

The normal boiling point of liquid bromine is 512 kJ/mol and the standard molar enthalpy is 17.89.

The values are not related to temperature.

There is one species of temperature at which the process favors the de for which the standard entropy value is not listed in natured state.

The process of converting CC bonds to CC bonds in food forms of nicotinamide adenine dinucleotide is aided by the use of transition key compounds in the metabolism. The oxidation of NADH is the first step. H2 is adsorbed onto the surface of Ni metal.

H is equal to 0.010 atm.

An equilib equilibrium constant with temperature for the reaction rium process is shown in the diagram.

Comment on the correctness of the analogy that relates a student's dormitory room to an increase in entropy.

The number of kJ/mol and J/K *mol is the sum of the total number of kJ/mol and the total number of J/K *mol.

Table 17.2 contains A.

You can round your answers to two different processes.

Under standard conditions, the number of ways to divide is longer.

Nicotine is a compound found in tobacco.

The following chemical equation has the number 1 and 2 represented.

The direction is left to right.