Chapter 16 Ionic Equilibria: Acids and Bases.

Chapter 16 Ionic Equilibria: Acids and Bases

Chapter Goals A Review of Strong Electrolytes
The Autoionization of Water The pH and pOH Scales Ionization Constants for Weak Monoprotic Acids and Bases Polyprotic Acids Solvolysis Salts of Strong Bases and Strong Acids

Chapter Goals Salts of Strong Bases and Weak Acids
Salts of Weak Bases and Strong Acids Salts of Weak Bases and Weak Acids Salts That Contain Small, Highly Charged Cations

A Review of Strong Electrolytes
This chapter details the equilibria of weak acids and bases. We must distinguish weak acids and bases from strong electrolytes. Weak acids and bases ionize or dissociate partially, much less than 100%. In this chapter we will see that it is often less than 10%! Strong electrolytes ionize or dissociate completely. Strong electrolytes approach 100% dissociation in aqueous solutions.

There are three classes of strong electrolytes. Strong Water Soluble Acids Remember the list of strong acids from Chapter 4.

Strong Water Soluble Bases The entire list of these bases was also introduced in Chapter 4.

Most Water Soluble Salts The solubility guidelines from Chapter 4 will help you remember these salts.

The calculation of ion concentrations in solutions of strong electrolytes is easy. Example 18-1: Calculate the concentrations of ions in M nitric acid, HNO3.

Example 18-2: Calculate the concentrations of ions in M strontium hydroxide, Sr(OH)2, solution. You do it!

The Autoionization of Water
Pure water ionizes very slightly. The concentration of the ionized water is less than one-millionth molar at room temperature.

We can write the autoionization of water as a dissociation reaction similar to those previously done in this chapter. Because the activity of pure water is 1, the equilibrium constant for this reaction is:

Experimental measurements have determined that the concentration of each ion is 1.0 x 10-7 M at 25oC. Note that this is at 25oC, not every temperature! We can determine the value of Kc from this information.

This particular equilibrium constant is called the ion-product for water and given the symbol Kw. Kw is one of the recurring expressions for the remainder of this chapter and Chapters 19 and 20.

Example 18-3: Calculate the concentrations of H3O+ and OH- in M HCl.

Use the [H3O+] and Kw to determine the [OH-]. You do it!

The increase in [H3O+] from HCl shifts the equilibrium and decreases the [OH-]. Remember from Chapter 17, increasing the product concentration, [H3O+], causes the equilibrium to shift to the reactant side. This will decrease the [OH-] because it is a product!

Now that we know the [H3O+] we can calculate the [OH-]. You do it!

The pH and pOH scales A convenient way to express the acidity and basicity of a solution is the pH and pOH scales. The pH of an aqueous solution is defined as:

The pH and pOH scales In general, a lower case p before a symbol is read as the ‘negative logarithm of’ the symbol. Thus we can write the following notations.

The pH and pOH scales If either the [H3O+] or [OH-] is known, the pH and pOH can be calculated. Example 18-4: Calculate the pH of a solution in which the [H3O+] =0.030 M.

The pH and pOH scales Example 18-5: The pH of a solution is What is the concentration of H3O+? You do it!

The pH and pOH scales A convenient relationship between pH and pOH may be derived for all dilute aqueous solutions at 250C. Taking the logarithm of both sides of this equation gives:

The pH and pOH scales Multiplying both sides of this equation by -1 gives: Which can be rearranged to this form:

The pH and pOH scales Remember these two expressions!!
They are key to the next three chapters!

The pH and pOH scales The usual range for the pH scale is 0 to 14.
And for pOH the scale is also 0 to 14 but inverted from pH. pH = 0 has a pOH = 14 and pH = 14 has a pOH = 0.

The pH and pOH scales

The pH and pOH scales Example 18-6: Calculate the [H3O+], pH, [OH-], and pOH for a M HNO3 solution. Is HNO3 a weak or strong acid? What is the [H3O+] ?

The pH and pOH scales Example 18-6: Calculate the [H3O+], pH, [OH-], and pOH for a M HNO3 solution.

The pH and pOH scales [H3O+] [OH-] pH pOH 1.0 M 1.0 x 10-14 M 0.00
To help develop familiarity with the pH and pOH scale we can look at a series of solutions in which [H3O+] varies between 1.0 M and 1.0 x M. [H3O+] [OH-] pH pOH 1.0 M 1.0 x M 0.00 14.00 1.0 x 10-3 M 1.0 x M 3.00 11.00 1.0 x 10-7 M 7.00 2.0 x M 5.0 x 10-3 M 11.70 2.30

The pH and pOH scales Example 18-7: Calculate the number of H3O+ and OH- ions in one liter of pure water at 250C. You do it!

Ionization Constants for Weak Monoprotic Acids and Bases
Let’s look at the dissolution of acetic acid, a weak acid, in water as an example. The equation for the ionization of acetic acid is: The equilibrium constant for this ionization is expressed as:

The water concentration in dilute aqueous solutions is very high. 1 L of water contains 55.5 moles of water. Thus in dilute aqueous solutions:

The water concentration is many orders of magnitude greater than the ion concentrations. Thus the water concentration is essentially that of pure water. Recall that the activity of pure water is 1.

We can define a new equilibrium constant for weak acid equilibria that uses the previous definition. This equilibrium constant is called the acid ionization constant. The symbol for the ionization constant is Ka.

In simplified form the dissociation equation and acid ionization expression are written as:

The ionization constant values for several acids are given below. Which acid is the strongest? Acid Formula Ka value Acetic CH3COOH 1.8 x 10-5 Nitrous HNO2 4.5 x 10-4 Hydrofluoric HF 7.2 x 10-4 Hypochlorous HClO 3.5 x 10-8 Hydrocyanic HCN 4.0 x 10-10

From the above table we see that the order of increasing acid strength for these weak acids is: The order of increasing base strength of the anions (conjugate bases) of these acids is:

Example 18-8: Write the equation for the ionization of the weak acid HCN and the expression for its ionization constant.

Example 18-9: In a 0.12 M solution of a weak monoprotic acid, HY, the acid is 5.0% ionized. Calculate the ionization constant for the weak acid. You do it!

Since the weak acid is 5.0% ionized, it is also 95% unionized. Calculate the concentration of all species in solution.

Use the concentrations that were just determined in the ionization constant expression to get the value of Ka.

Example 18-10: The pH of a 0.10 M solution of a weak monoprotic acid, HA, is found to be What is the value for its ionization constant? pH = 2.97 so [H+]= 10-pH

Use the [H3O+] and the ionization reaction to determine concentrations of all species.

Calculate the ionization constant from this information.

Example 18-11: Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH, solution. It is always a good idea to write down the ionization reaction and the ionization constant expression.

Next, combine the basic chemical concepts with some algebra to solve the problem.

Next we combine the basic chemical concepts with some algebra to solve the problem

Substitute these algebraic quantities into the ionization expression.

Solve the algebraic equation, using a simplifying assumption that is appropriate for all weak acid and base ionizations.

Complete the algebra and solve for the concentrations of the species.

Note that the properly applied simplifying assumption gives the same result as solving the quadratic equation does.

Let us now calculate the percent ionization for the 0.15 M acetic acid. From Example 18-11, we know the concentration of CH3COOH that ionizes in this solution. The percent ionization of acetic acid is

Example 18-12: Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. Ka= 4.0 x for HCN You do it!

The percent ionization of 0.15 M HCN solution is calculated as in the previous example.

Let’s look at the percent ionization of two weak acids as a function of their ionization constants. Examples and will suffice. Solution Ka [H+] pH % ionization 0.15 M acetic acid 1.8 x 10-5 1.6 x 10-3 2.80 1.1 0.15 M HCN 4.0 x 10-10 7.7 x 10-6 5.11 0.0051 Note that the [H+] in 0.15 M acetic acid is 210 times greater than for 0.15 M HCN.

All of the calculations and understanding we have at present can be applied to weak acids and weak bases! One example of a weak base ionization is ammonia ionizing in water.

All of the calculations and understanding we have at present can be applied to weak acids and weak bases! Example 18-13: Calculate the concentrations of the various species in 0.15 M aqueous ammonia.

The percent ionization for weak bases is calculated exactly as for weak acids.

Example 18-14: The pH of an aqueous ammonia solution is Calculate the molarity (original concentration) of the aqueous ammonia solution. You do it!

Use the ionization equation and some algebra to get the equilibrium concentration.

Substitute these values into the ionization constant expression.

Examination of the last equation suggests that our simplifying assumption can be applied. In other words (x-2.3x10-3)  x. Making this assumption simplifies the calculation.

Polyprotic Acids Many weak acids contain two or more acidic hydrogens.
Examples include H3PO4 and H3AsO4. The calculation of equilibria for polyprotic acids is done in a stepwise fashion. There is an ionization constant for each step. Consider arsenic acid, H3AsO4, which has three ionization constants. Ka1 = 2.5 x 10-4 Ka2 = 5.6 x 10-8 Ka3 = 3.0 x 10-13

Polyprotic Acids The first ionization step for arsenic acid is:

Polyprotic Acids The second ionization step for arsenic acid is:

Polyprotic Acids The third ionization step for arsenic acid is:

Polyprotic Acids Notice that the ionization constants vary in the following fashion: This is a general relationship. For weak polyprotic acids the Ka1 is always > Ka2, etc.

Polyprotic Acids Example 18-15: Calculate the concentration of all species in M arsenic acid, H3AsO4, solution. Write the first ionization step and represent the concentrations. Approach this problem exactly as previously done.

Polyprotic Acids Substitute the algebraic quantities into the expression for Ka1.

Polyprotic Acids Use the quadratic equation to solve for x, and obtain both values of x.

Polyprotic Acids Next, write the equation for the second step ionization and represent the concentrations.

Polyprotic Acids Substitute the algebraic expressions into the second step ionization expression.

Polyprotic Acids

Polyprotic Acids Finally, repeat the entire procedure for the third ionization step.

Polyprotic Acids Substitute the algebraic representations into the third ionization expression.

Polyprotic Acids Use Kw to calculate the [OH-] in the M H3AsO4 solution.

Polyprotic Acids A comparison of the various species in M H3AsO4 solution follows. Species Concentration H3AsO4 0.095 M H+ M H2AsO4- HAsO42- 5.6 x 10-8 M AsO43- 3.4 x M OH- 2.0 x M

Solvolysis This reaction process is the most difficult concept in this chapter. Solvolysis is the reaction of a substance with the solvent in which it is dissolved. Hydrolysis refers to the reaction of a substance with water or its ions. Combination of the anion of a weak acid with H3O+ ions from water to form nonionized weak acid molecules.

Solvolysis Hydrolysis refers to the reaction of a substance with water or its ions. Hydrolysis is solvolysis in aqueous solutions. The combination of a weak acid’s anion with H3O+ ions, from water, to form nonionized weak acid molecules is a form of hydrolysis.

Solvolysis The reaction of the anion of a weak monoprotic acid with water is commonly represented as:

Solvolysis Recall that at 25oC in neutral solutions:
[H3O+] = 1.0 x 10-7 M = [OH-] in basic solutions: [H3O+] < 1.0 x 10-7 M and [OH-] > 1.0 x 10-7 M in acidic solutions: [OH-] < 1.0 x 10-7 M and [H3O+] > 1.0 x 10-7 M

Solvolysis Remember from BrØnsted-Lowry acid-base theory:
The conjugate base of a strong acid is a very weak base. The conjugate base of a weak acid is a stronger base. Hydrochloric acid, a typical strong acid, is essentially completely ionized in dilute aqueous solutions.

Solvolysis The conjugate base of HCl, the Cl- ion, is a very weak base. The chloride ion is such a weak base that it will not react with the hydronium ion. This fact is true for all strong acids and their anions.

Solvolysis HF, a weak acid, is only slightly ionized in dilute aqueous solutions. Its conjugate base, the F- ion, is a much stronger base than the Cl- ion. The F- ions combine with H3O+ ions to form nonionized HF. Two competing equilibria are established.

Solvolysis Dilute aqueous solutions of salts that contain no free acid or base come in four types: Salts of Strong Bases and Strong Acids Salts of Strong Bases and Weak Acids Salts of Weak Bases and Strong Acids Salts of Weak Bases and Weak Acids

Salts of Strong Bases and Weak Acids
Salts made from strong acids and strong soluble bases form neutral aqueous solutions. An example is potassium nitrate, KNO3, made from nitric acid and potassium hydroxide.

Salts made from strong soluble bases and weak acids hydrolyze to form basic solutions. Anions of weak acids (strong conjugate bases) react with water to form hydroxide ions. An example is sodium hypochlorite, NaClO, made from sodium hydroxide and hypochlorous acid.

We can combine these last two equations into one single equation that represents the total reaction.

The equilibrium constant for this reaction, called the hydrolysis constant, is written as:

Algebraic manipulation of the previous expression give us a very useful form of the expression. Multiply the expression by one written as [H+]/ [H+]. H+/H+ = 1

Which can be rewritten as:

Which can be used to calculate the hydrolysis constant for the hypochlorite ion:

This same method can be applied to the anion of any weak monoprotic acid.

Example 18-16: Calculate the hydrolysis constants for the following anions of weak acids. The fluoride ion, F-, the anion of hydrofluoric acid, HF. For HF, Ka=7.2 x 10-4.

The cyanide ion, CN-, the anion of hydrocyanic acid, HCN. For HCN, Ka = 4.0 x You do it!

Example 18-17: Calculate [OH-], pH and percent hydrolysis for the hypochlorite ion in 0.10 M sodium hypochlorite, NaClO, solution. “Clorox”, “Purex”, etc., are 5% sodium hypochlorite solutions.

Set up the equation for the hydrolysis and the algebraic representations of the equilibrium concentrations.

Substitute the algebraic expressions into the hydrolysis constant expression.

The percent hydrolysis for the hypochlorite ion may be represented as:

If a similar calculation is performed for 0.10 M NaF solution and the results from 0.10 M sodium fluoride and 0.10 M sodium hypochlorite compared, the following table can be constructed. Solution Ka Kb [OH-] (M) pH % hydrolysis NaF 7.2 x 10-4 1.4 x 10-11 1.2 x 10-6 8.08 0.0012 NaClO 3.5 x 10-8 2.9 x 10-7 1.7 x 10-4 10.23 0.17

Salts of Weak Bases and Strong Acids
Salts made from weak bases and strong acids form acidic aqueous solutions. An example is ammonium bromide, NH4Br, made from ammonia and hydrobromic acid.

The reaction may be more simply represented as:

Or even more simply as: The hydrolysis constant expression for this process is:

Multiplication of the hydrolysis constant expression by [OH-]/ [OH-] gives:

Which we recognize as:

In its simplest form for this hydrolysis:

Example 18-18: Calculate [H+], pH, and percent hydrolysis for the ammonium ion in 0.10 M ammonium bromide, NH4Br, solution. Write down the hydrolysis reaction and set up the table as we have done before:

Substitute the algebraic expressions into the hydrolysis constant.

Complete the algebra and determine the concentrations and pH.

The percent hydrolysis of the ammonium ion in 0.10 M NH4Br solution is:

Salts of Weak Bases and Weak Acids
Salts made from weak acids and weak bases can form neutral, acidic or basic aqueous solutions. The pH of the solution depends on the relative values of the ionization constant of the weak acids and bases. Salts of weak bases and weak acids for which parent Kbase =Kacid make neutral solutions. An example is ammonium acetate, NH4CH3COO, made from aqueous ammonia, NH3,and acetic acid, CH3COOH. Ka for acetic acid = Kb for ammonia = 1.8 x 10-5.

The ammonium ion hydrolyzes to produce H+ ions. Its hydrolysis constant is:

The acetate ion hydrolyzes to produce OH- ions. Its hydrolysis constant is:

Because the hydrolysis constants for both ions are equal, their aqueous solutions are neutral. Equal numbers of H+ and OH- ions are produced.

Salts of weak bases and weak acids for which parent Kbase > Kacid make basic solutions. An example is ammonium hypochlorite, NH4ClO, made from aqueous ammonia, NH3,and hypochlorous acid, HClO. Kb for NH3 = 1.8 x 10-5 > Ka for HClO = 3.5x10-8

The ammonium ion hydrolyzes to produce H+ ions. Its hydrolysis constant is:

The hypochlorite ion hydrolyzes to produce OH- ions. Its hydrolysis constant is: Because the Kb for ClO- ions is three orders of magnitude larger than the Ka for NH4+ ions, OH- ions are produced in excess making the solution basic.

Salts of weak bases and weak acids for which parent Kbase < Kacid make acidic solutions. An example is trimethylammonium fluoride,(CH3)3NHF, made from trimethylamine, (CH3)3N,and hydrofluoric acid acid, HF. Kb for (CH3)3N = 7.4 x 10-5 < Ka for HF = 7.2 x 10-4

Both the cation, (CH3)3NH+, and the anion, F-, hydrolyze.

The trimethylammonium ion hydrolyzes to produce H+ ions. Its hydrolysis constant is:

The fluoride ion hydrolyzes to produce OH- ions. Its hydrolysis constant is: Because the Ka for (CH3)3NH+ ions is one order of magnitude larger than the Kb for F- ions, H+ ions are produced in excess making the solution acidic.

Summary of the major points of hydrolysis up to now. The reactions of anions of weak monoprotic acids (from a salt) with water to form free molecular acids and OH-.

The reactions of anions of weak monoprotic acids (from a salt) with water to form free molecular acids and OH-.

Aqueous solutions of salts of strong acids and strong bases are neutral. Aqueous solutions of salts of strong bases and weak acids are basic. Aqueous solutions of salts of weak bases and strong acids are acidic. Aqueous solutions of salts of weak bases and weak acids can be neutral, basic or acidic. The values of Ka and Kb determine the pH.

Hydrolysis of Small Highly-Charged Cations
Cations of insoluble bases (metal hydroxides) become hydrated in solution. An example is a solution of Be(NO3)3. Be2+ ions are thought to be tetrahydrated and sp3 hybridized.

In condensed form it is represented as: or, even more simply as:

The hydrolysis constant expression for [Be(OH2)4]2+ and its value are: or, more simply

Example 18-19: Calculate the pH and percent hydrolysis in 0.10 M aqueous Be(NO3)2 solution. The equation for the hydrolysis reaction and representations of concentrations of various species are:

Algebraic substitution of the expressions into the hydrolysis constant:

Calculate the percent hydrolysis of Be2+.

This table is a comparison of 0.10 M Be(NO3 )2 solution and 0.10 M CH3COOH solution. Solution [H3O+] pH % hydrolysis or % ionization 0.10 M Be(NO3)2 1.0 x 10-3 M 3.00 1.0% 0.10 M CH3COOH 1.3 x 10-3 M 2.89 1.3% Notice that the Be solution is almost as acidic as the acetic acid solution.

Synthesis Question Rain water is slightly acidic because it absorbs carbon dioxide from the atmosphere as it falls from the clouds. (Acid rain is even more acidic because it absorbs acidic anhydride pollutants like NO2 and SO3 as it falls to earth.) If the pH of a stream is 6.5 and all of the acidity comes from CO2, how many CO2 molecules did a drop of rain having a diameter of 6.0 mm absorb in its fall to earth?

Synthesis Question

Group Question A common food preservative in citrus flavored drinks is sodium benzoate, the sodium salt of benzoic acid. How does this chemical compound behave in solution so that it preserves the flavor of citrus drinks?

End of Chapter 18 Weak aqueous acid-base mixtures are called buffers. They are the subject of Chapter 19.

Chapter 16 Ionic Equilibria: Acids and Bases.

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Chapter 16 Ionic Equilibria: Acids and Bases.

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