Untitled Notes

The most useful point for writing the reactant quotient is when the reaction has reached equilibrium. Since it is a ratio of concentrations or pressures, this equilibrium constant is usually expressed as a number without units. The concentration of pure liquids that appear in the equilibrium expression are assumed to be 1, since their con centrations do not change.

The equilibrium constant's numerical value can show the extent of the reaction after equilibrium has been reached.

A reaction will reach equilibrium with the production of a certain amount of product. Not much will be formed if the equilibrium constant is small. Le Chatelier discovered that if a chemical system is stressed it will reestablish equilibrium by shifting the reactions involved. The final ratio will remain the same as the amounts of reac tants and products will change. Changes in concentration, pressure, and temperature are some of the ways in which the equilibrium may be stressed.

The use of a catalyst is often mentioned. The equilibrium amounts will not be affected by a catalyst because it affects both the forward and reverse reactions equally.

It will cause the reaction to reach equilibrium quicker.

If the equilibrium system is stressed by a change in concentration of one of the reactants or products, the equilibrium will react to remove that stress. The equilibrium will shift to produce more of it if the concentration is decreased. The concentration of chemical species on the other side of the reaction arrows will be decreased.

The equilibrium will shift if the concentration of a chemical species is increased.

The equilibrium shifts to the right if one increases the concentration of hydrogen gas. The concentration of ammonia will increase and the concentration of nitrogen gas will decrease. If the concentration of nitrogen gas was decreased, the equilibrium would shift to the left, the concentration of ammonia would decrease, and the concentration of hydrogen would increase.

If gases are involved, changes in pressure are not significant. The concentration of a gaseous species can be changed by changing the volume of the container, although this is really a change in concentration and can be treated as a con centration effect. There is an increase in the number of accidents on the inside walls of the container if the container becomes smaller. The equilibrium system will shift to reduce the pressure. The equilibrium can be shifted to the side of the equation that has less gas. The equilibrium will shift to the side containing more moles of gas if the container size is increased. Changing the pressure won't affect the equilibrium if the number of moles of gas is the same on both sides.

The equilibrium will shift to the right if the container is made smaller. The concentration of ammonia would increase.

Pressure effects are not important for gases.

The equilibrium constant is affected by temperature. It changes the amount of heat in the system and can be treated as a concentration effect. One needs to know which reaction is exothermic to treat it this way.

The equilibrium would shift to the left if the temperature of the reaction mixture were increased. The concentration of nitrogen and hydrogen gases would increase and the concentration of ammonia gas would decrease. In order to shift the reaction to the right, you would probably want to operate at a reduced temperature.

The equilibrium shifts to remove excess CO2.

The equilibrium changes to replace some of the CO2.

Solids don't shift equilibria unless they are completely removed.

When heated, right-endothermic reactions shift to the right.

The equilibrium will be shifted to the side with less gas if volume is decreased or pressure increased.

catalysts don't affect the position of an equilibrium.

The concept of acids and bases was introduced in the chapter.

Acids and bases are proton acceptors.

Consider two acids: HCl and CH3COOH.

There is no HCl left in the first reaction. The amount of reac tants and products left in solution is the second reaction.

The hydroxide and oxide ion are the two strongest bases to consider. The other bases are weak.

There is competition for an H+ in the theory.

CH3COOH(aq) NH3(aq) CH3COO (aq) NH4(aq) acetic acid donates a proton to ammonia in the equilibrium to form the acetate and ammonium ion. Ammonia and acetic acid can be formed in the reverse (right-to- left) reaction. The acid and base are acting on the ammonium ion. A conjugate acid-base pair is defined as acetic acid and the acetate ion.

The acid-base pair is also referred to as the ammonium ion. There is a competition for the H+ between the two groups.

The equilibrium will lie to the right.

In order to be able to do quantitative problems, we need to find good qualitative answers. A modification of the equilibrium constant is useful.

The acids are in the water. An equilibrium system can be established by the partial dissociation of weak acids.

The equilibrium constant expression does not show it.

The equilibrium molar concentration of the undissociated weak acid is not its initial concentration.

Before we look at the equilibrium behavior of weak bases, let's look at the behavior of water. In the initial discussion of acid-base equilibrium, we showed that water acted both as an acid and a base. The same nature is noted in pure water.

In the discussion of weak acids, we said that the H+ was A-. There are two sources of H+ in the system. The amount of H+ that is due to the water dissociation is very small and can be ignored.

A scale to easily represent the acidity of a solution was developed because the concentration of the H3O+ can vary greatly.

The solution for [H3O+] gives us 1.0 x 10.

The water's pH is.

The solution's pOH can be calculated. The definition of pOH is -log[OH-].

You can now calculate the pOH of the solution in any of the problems where H+ was calculated.

You can estimate the solution's pH by looking at its H+. The solution's pH would be 5 if it had an [H+] of 1 x 10-5. The value was determined from the value of the exponent.

Determine the pH of a solution made by adding 0.400 mol of strontium acetate to enough water to produce 2.000 L of solution using this relationship.

Ions that come from strong acids or bases may be ignored in this type of problem. Weak acids or bases will cause C 2H3O2 to undergo hydrolysis.

The acid-base properties of the ion present in the salt will affect the behavior of a salt. The solutions of the salt may be acidic, basic, or neutral. A generic term for a variety of reactions with water is hydra. The ion will change the pH.

An acid and a base will produce a salt. The base and anion of the acid will be contained in the salt. The anion from the acid is the conjugate base of the acid. The salt has a conjugate acid and a conjugate base. In principle, this is true. One or the other of these ion is not a true conjugate base or a conjugate acid. Even though the ion is not a true conjugate acid or base, we can still use it as if it were.

The conjugate acid of any strong acid is so weak that it will not undergo any sig nificant hydrolysis. Ions that do not undergo any significant hydrolysis will have no effect on the solution's pH. The solution will be neutral because of the pres ence of the conjugate bases. The cations from the strong bases will leave the solution neutral. Salts with a combination of cations and anions are neutral.

A basic solution can be produced from the conjugate base from any weak acid. The solution will be basic if the conjugate base of a weak acid is in a salt with the conjugate of a strong base. Basic salts are what this type is. Basic salts include the cation of a strong base and the anion of a weak acid.

The conjugate acid of a weak base is a strong acid that will be hydrolysis to make the solution acidic. The solution will be acidic if the conjugate acid is in a salt with the conjugate base of a strong acid. Salts of this type are acidic. The anion of a strong acid and the cation of a weak base are acidic salts.

There are salts that have the cation of a weak base with the anion of a weak acid. The acid-base character of these salts is not obvious because both ion undergo hydrolysis. The knowledge you need to score high can interfere with each other. The solution would be neutral if the two values are equal.

There is a strong base and a weak acid in sodium carbonate. Strong bases and weak acids are basic salts. The final answer must be basic, as a basic salt.

A weak acid and its conjugate base is the most common type of buffer. Any acid added to the solution will be mitigated by the weak acid and the weak base of the buffer.

The more concentrated these species are, the less acid and base can be neutralized.

The pH of a buffer can be calculated.

There are two ways to solve this problem.

Both chemistry students and chemists hope that the endpoints are close together.

equilibria will be established if the acid is a weak acid.

One may know the initial concentration of the weak acid, but may be interested in the changes during the titration.

A mixture of weak acid and conjugate base is formed. The buffer solution can be used in the calculations. Determine the moles of acid con sumed from the moles of conjugate base formed.

Taking into account the volume of titrant added, calculate the molar concentration of weak acid and conjugate base. Finally, apply your equations.

The weak acid has been converted to its conjugate base. The solution's pH needs to be calculated.

The excess strong base will determine the pH after the equivalence point.

A typical titration problem can be considered.

HNO2 is the only base present.

There is an acid and a base present and they are not conjugates. There is a balanced chemical equation and moles.

H2O Na+ + NO-2 could be written as NaNO2, but the separated ion are more useful.

NaOH is the limiting reagent because it is on the moles of acid and base.

Part of the problem is done.

HNO2 and NO2 are a conjugate acid-base pair.

This is a buffer problem since a CA/CB pair is present.

The CB/CA concentrations are simplified. Since both moles are in the same solution, the identical volumes are canceled.

Since both an acid and a base are present, this must be a problem again. There is a balanced chemical equation and moles.

NaOH is the limiting reagent because it is on the moles of acid and base.

Part of the problem is done.

HNO2 and NO2 are a conjugate acid-base pair.

Review the knowledge you need to score high since a CA/CB pair is present.

There is an acid and a base present and they are not conjugates. There is a balanced chemical equation and moles.

The problem and moles of acid and base are limiting reagents.

Part of the problem is done.

The NO2 solution is a conjugate base of a weak acid.

There is an acid and a base present and they are not conjugates. There is a balanced chemical equation and moles.

The limiting reagent is based on the amount of acid and base in the problem.

The strong base will be able to control the pH.

Part of the problem is done.

Since this is a solution of a strong base, it is a simple problem.

There is an acid and a base present and they are not conjugates. There is a balanced chemical equation and moles.

The limiting reagent is based on the amount of acid and base in the problem.

The strong base will control the pH.

Part of the problem is done.

Since this is a solution of a strong base, it is a simple problem.

Some salts are slightly soluble in water. When placed in water, these salts quickly reach their solubility limit and the ion establish an equilibrium system with the undissolved solid.

It is the result of the ionic concentrations raising the power of the coefficients in the balanced chemical equation. The concentration of a solid does not appear in the equilibrium constant expressions.

The concentration of the ion can be determined if the numerical value of the solubility product constant is known.

If a slightly soluble salt solution is at equilibrium and a solution containing one of the ion involved in the equilibrium is added, the solubility of the salt is decreased.

A solution of Na2SO4 could be added to the equilibrium system. Le Chatelier's principle states that the equilibrium will be disrupted by the additional sulfate ion. If you tried to remove PbSO4 from a solution of Na2SO4 instead of pure water, the solubility would be lower.

Knowing the value of the product constant can allow us to predict if two solutions containing the same ion component will form a precipitate.

The same treatment can be given to other types of equilibria.

An equilibrium constant is associated with the formation of complex ion.

Concentrations or pressures are usually zero for the products and a measured or calculated value for the reactants. The amount of at least one of the substances is determined once equilibrium has been restored. The amount of the other materials may be calculated based on the change in this one substance.

The pressure, mass, volume, and pH can be measured. Some experiments measure the color intensity with a spectrophotometer.

Don't make the mistake of measuring a change. Changes are always calculated.

Check the units and figures of your answer.

Use products over reactants when writing equilibrium constant expressions. Concentrations are raised to the power of the balanced chemical equation.

Pressure effects are only important for gases that are involved in the equilibrium in Le Chatelier problems.

When mixing two solu tions together, make sure you compensate for dilution.

To practice for the AP Chemistry exam, use these questions to review the content of this chapter. There are 28 multiple-choice questions that are similar to what you will see in the chemistry section of the exam. Section II of the exam has a long free-response question. You can make these questions even more authentic by following the instructions.

You can't use a calculator. The periodic table and equation sheet can be found at the back of the book.

A standard strong acid is used to make a solution of a weak base.

The best titration will be followed by a pH meter. Which is the best way to prepare a buffer with a pH of 8?

A change in volume will follow a constant temperature.

C(s) + H2O(g) CO(g) + H2(g) endothermic 6H5NH2) solution

An equilibrium mixture of reactants is put in a sealed container.

A 1.00 L flask is filled with 0.30 mole of CH4 and 0.40 mole of CO (A) HBrO 2 and allowed to come to equilibrium. There are 0.20 mole of CO in the flask.

When 2(g) was added to the flask, it allowed the mixture to return a solution with a pH > 7, which is a buffer to equilibrium at the same temperature.

A chemistry student adds ammonia to increase the volume solution. The temperature is decreased by decreasing the C.

Since you can estimate the answer, no actual and D can be eliminated. The calculation of a 0.1 molar solution is necessary.

A is eliminated by acetic acid, which is not a strong acid.

The solution must be acidic to be an acid-dissociation constant.

The answer must involve the H 2PO4 ion.

This is not a buffer because A is the salt of a strong acid and a weak base. This salt is acidic. B is a salt of a strong base and a weak acid; acid and a strong base are neutral.

C and D are salts of a weak acid and a strong base. No actual calculations are necessary because the two substances constitute a conjugate answer.

The acid is about 2.

There is a weak acid and a strong base in the salt. When an acid is added, the pH will not change.

The weak acid and weak base will not react and the pH will drop to give a neutral solution.

4 x 10-6.

The equilibrium shifts toward the side desired zinc ion concentration.

The amount of the products will be increased by two moles.

3 must be added to produce this concentration.

No actual calculations are necessary if you use the following table to estimate the answer.

The interaction of a ion with water starts the ion's hydrolysis.

The equilibrium values are entered into the equilib base of a weak acid.

2 will not go below its earlier equilibrium value since the ZnS(s) Zn2+(aq) + S2 excess was added.

The forward and reverse reactions are equal.

You need to temperature these equations.

The salts of weak acids are bases.

The H2SeO4 and H2CO3 won't be affected by volume or pressure changes if the ion is accepted as H 3PO4, HCO3 or both. B can behave as equilibrium.

The strongest acid in this group is expected to eliminate B and C.

There is acid and acidic salts in this equilibrium.

You have 20 minutes to answer the question. You can use the tables in the back of the book.

There are two products of the equilibrium.

You only need to show one form if resonance is possible.

You have to give yourself 1 point for this answer. If you put your values into the wrong equation, you can get 1 point.

You should give yourself 1 point for a correct calculation and 1 point for the correct conclusion.

The equilibrium must shift to the right since the concentration of cadmium increases.

When processes are heated, they shift to the right. This is in line with Le Chatelier's principle.

You have to give yourself 1 point for endothermic. If you mention Le Chatelier's principle, give yourself 1 point.

The formal charge for an atom is nonbonding electrons.

In case of a calculation error, the total is a check. The charge on the ion is equal to the total.

You get 1 point for a correct Lewis structure and 1 point for a set of formal charges.

There is a chance for 10 points. If a numerical answer has an incorrect number of significant figures, subtract 1 point.

A chemical equilibrium can be established when two different reactions occur in the same container at the same time and with the same rates of reaction.

The concentrations of the chemical species are not always equal.