1 acid-base equilibria. 2 solutions of a weak acid or base the simplest acid-base equilibria are...

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Acid-Base Equilibria

2

Solutions of a Weak Acid or Base

• The simplest acid-base equilibria are those in which a single acid or base solute reacts with water.

– In this chapter, we will first look at solutions of weak acids and bases.

– We must also consider solutions of salts, which can have acidic or basic properties as a result of the reactions of their ions with water.

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Acid-Ionization Equilibria

• Acid ionization (or acid dissociation) is the

• (See Animation: Acid Ionization Equilibrium)– When acetic acid is added to water it reacts as follows.

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• For a weak acid, the equilibrium concentrations of ions in solution are determined by the

– Consider the generic monoprotic acid, HA.

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• For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant).

– The corresponding equilibrium expression is:

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– Since the concentration of water remains relatively constant, we rearrange the equation to get:

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– Thus, Ka , the acid-ionization constant, equals the constant [H2O]Kc.

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– Table 17.1 lists acid-ionization constants for various weak acids.

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Experimental Determination of Ka

• The degree of ionization of a weak electrolyte is the fraction of molecules that react with water to give ions.

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A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C.

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StartingChange

Equilibrium

– Let x be the moles per liter of product formed.

)aq(Nic)aq(OH )l(OH)aq(HNic 32

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– The equilibrium-constant expression is:

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– Substituting the expressions for the equilibrium concentrations, we get

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– We can obtain the value of x from the given pH.

)pHlog(anti]OH[x 3

)39.3log(antix

00041.0101.4x 4

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– Substitute this value of x

– Note that

012.001159.0)00041.0012.0()x012.0( the concentration of unionized acid remains virtually unchanged.

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– Substitute this value of x

522

a 104.1)012.0()00041.0(

)x012.0(x

K

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– To obtain the degree of dissociation:

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Calculations With Ka

• Once you know the value of Ka, you can calculate the equilibrium concentrations of species HA, A-, and H3O+ for solutions of different molarities.

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• Note that in our previous example, the degree of dissociation was so small that “x” was negligible compared to the concentration of nicotinic acid.

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• How do you know when you can use this simplifying assumption?

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• How do you know when you can use this simplifying assumption?

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• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

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• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.– Note that

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26



StartingChange

Equilibrium

– These data are summarized below.

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– The equilibrium constant expression is

A Problem To Consider• What is the pH at 25°C of a solution

obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

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– If we substitute the equilibrium concentrations and the Ka into the equilibrium constant expression, we get



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– Rearranging the preceding equation to put it in the form ax2 + bx + c = 0, we get

– You can solve this equation exactly by using the quadratic formula.

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– Now substitute into the quadratic formula.



a2ac4bb

x2

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– Now substitute into the quadratic formula.



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– Taking the upper sign, we get



– Now we can calculate the pH.

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Polyprotic Acids

• Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

– Sulfuric acid, for example, can lose two protons in aqueous solution.

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Polyprotic Acids

– For a weak diprotic acid like carbonic acid, H2CO3, two simultaneous equilibria must be considered.


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– Each equilibrium has an associated acid-ionization constant.

Polyprotic Acids


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– Each equilibrium has an associated acid-ionization constant.

Polyprotic Acids


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Polyprotic Acids


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– However, reasonable assumptions can be made that simplify these calculations as we show in the next example.

Polyprotic Acids

– When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions.


Polyprotic Acids

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• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

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– The pH can be determined by simply solving the equilibrium problem posed by the first ionization.



2- ?


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– If we abbreviate the formula for ascorbic acid as H2Asc, then the first ionization is:



2- ?


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2- ?


(aq)AscH(aq)OH )l(OH)aq(AscH 322

StartingChange

Equilibrium

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2- ?


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– Substituting into the equilibrium expression



2- ?


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– Assuming that x is much smaller than 0.10, you get



2- ?


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– The hydronium ion concentration is 0.0028 M, so



2- ?


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– The ascorbate ion, Asc2-, which we will call y, is produced only in the second ionization of H2Asc.



2- ?


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– Assume the starting concentrations for HAsc- and H3O+ to be those from the first equilibrium.



2- ?


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2- ?


StartingChange

Equilibrium

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2- ?


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– Substituting into the equilibrium expression



2- ?


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– Assuming y is much smaller than 0.0028, the equation simplifies to



2- ?


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– Hence,



2- ?


122 106.1]Asc[y – The concentration of the ascorbate ion equals Ka2.

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Base-Ionization Equilibria

• Equilibria involving weak bases are treated similarly to those for weak acids.– Ammonia, for example, ionizes in water as

follows.

)aq(OH)aq(NH )l(OH)aq(NH 423

– The corresponding equilibrium constant is:

]OH][NH[]OH][NH[

K23

4c

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• Equilibria involving weak bases are treated similarly to those for weak acids.– Ammonia, for example, ionizes in water as

follows.

– The concentration of water is nearly constant.

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• Equilibria involving weak bases are treated similarly to those for weak acids.– In general, a weak base B with the base ionization

has a base ionization constant equal to

Table 17.2 lists ionization constants for some weak bases.

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• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.

1. Write the equation and make a table of concentrations.

2. Set up the equilibrium constant expression.

3. Solve for x = [OH-].

– As before, we will follow the three steps in solving an equilibrium.

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• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– Pyridine ionizes by picking up a proton from water

(as ammonia does).

StartingChange

Equilibrium

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• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– Note that

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• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– The equilibrium expression is

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• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– If we substitute the equilibrium

concentrations and the Kb into the equilibrium constant expression, we get

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• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– Using our simplifying assumption that the x

in the denominator is negligible, we get

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• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– Solving for x we get

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• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– Solving for pOH

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Acid-Base Properties of a Salt Solution

• One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.– Consider a solution of sodium cyanide, NaCN.

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• One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.


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• One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.


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• One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.– You can also see that OH- ion is a product, so you

would expect

– The reaction of the CN- ion with water is referred to as the hydrolysis of CN-.


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• The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

)aq(OH)aq(HCN )l(OH)aq(CN 2


71)aq(OH)aq(HCN )l(OH)aq(CN 2



72)aq(OH)aq(NH )l(OH)aq(NH 3324



73)aq(OH)aq(NH )l(OH)aq(NH 3324



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Predicting Whether a Salt is Acidic, Basic, or Neutral

• How can you predict whether a particular salt will be acidic, basic, or neutral?– The Brønsted-Lowry concept illustrates the

inverse relationship in the strengths of conjugate acid-base pairs.

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• How can you predict whether a particular salt will be acidic, basic, or neutral?

reaction no)l(OH)aq(Cl 2

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• How can you predict whether a particular salt will be acidic, basic, or neutral?

reaction no)l(OH)aq(Na 2

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• To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt.– Consider potassium acetate, KC2H3O2.

The potassium ion is the cation of a strong base (KOH) and does not hydrolyze.

reaction no)l(OH)aq(K 2

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Predicting Whether a Salt is Acidic, Basic, or Neutral• To predict the acidity or basicity of

a salt, you must examine the acidity or basicity of the ions composing the salt.– Consider potassium acetate, KC2H3O2.

OHOHCH )l(OH)aq(OHC 2322232

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• These rules apply to normal salts (those in which the anion has no acidic hydrogen)

1. A salt of a strong base and a strong acid.

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2. A salt of a strong base and a weak acid.

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3. A salt of a weak base and a strong acid.

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4. A salt of a weak base and a weak acid.

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The pH of a Salt Solution

• To calculate the pH of a salt solution would require the Ka of the acidic cation or the Kb of the basic anion. (see Figure 17.8)

– The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species.

– Thus the Kb for CN- is related to the Ka for HCN.

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• To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-.

)aq(CN)aq(OH )l(OH)aq(HCN 32


Ka

Kb

– When these two reactions are added you get the ionization of water.

)aq(OH)aq(OH )l(OH2 32 Kw

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Ka

Kb

– When two reactions are added, their equilibrium constants are multiplied.


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Ka

Kb

Therefore,


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• For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases.– The only difference is first obtaining the Ka

or Kb for the ion that hydrolyzes.

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• What is the pH of a 0.10 M NaCN solution at 25 °C? The Ka for HCN is 4.9 x 10-10.

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– The CN- ion is acting as a base, so first, we must calculate the Kb for CN-.

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– Let x = [OH-] = [HCN


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– This gives


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– Solving the equation, you find that


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The Common Ion Effect

• The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.– Consider a solution of acetic acid (HC2H3O2), in

which you have the following equilibrium.

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• The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.

(aq)OHC(aq)OH 2323 )l(OH)aq(OHHC 2232

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96




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• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.– Consider the equilibrium below.

(aq)OCH(aq)OH 23 )l(OH)aq(OHCH 22

Starting 0.025 0 0.018Change -x +x +x

Equilibrium

0.025-x x 0.018+x

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• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.– The equilibrium constant expression is:

a2

23 K]OHCH[

]OCH][OH[

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• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.– Substituting into this equation gives:

4107.1)x025.0()x018.0(x

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• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.

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• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.– The equilibrium equation becomes

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• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.– Hence,

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• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.

Buffers

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Buffers

• A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

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Buffers


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Buffers


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Buffers


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The Henderson-Hasselbalch Equation

• How do you prepare a buffer of given pH?

– To illustrate, consider a buffer of a weak acid HA and its conjugate base A-.

The acid ionization equilibrium is:

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– The acid ionization constant is:

– By rearranging, you get an equation for the H3O+ concentration.

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• How do you prepare a buffer of given pH?– Taking the negative logarithm of both sides of the

equation we obtain:

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– More generally, you can write

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• Calculate the pH of a 44ml solution of .202 M acetic acid and 15.5 ml of .185 M NaOH solution.

HC2H3O2 + NaOH NaC2H3O2 + H2O 44.0 ml 15.5 ml

.202 M .185 M


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• Calculate the pH of a 44ml solution of .202 M acetic acid and 15.5 ml of .185 M NaOH solution.

Calculating the new molarity in 59.5 ml

HC2H3O2 + H2O <->C2H3O2- + H3O+

1.01 x 10-1M 4.82 x 10-2M x

Titration Curves

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Acid-Ionization Titration Curves

• An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid).

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Titration of a Strong Acid by a Strong Base

Figure 17.12: Curve for the titration of a strong acid by a strong base.

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• Figure 17.12 shows a curve for the titration of HCl with NaOH.

– At the equivalence point, the pH of the solution is


121

• Figure 17.12 shows a curve for the titration of HCl with NaOH.

– To detect the equivalence point, you need


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• Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl.

NaOH + HCl NaCl + H2O

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NaOH + HCl NaCl + H2O– We get the amounts of reactants by multiplying

the volume of each (in liters) by their respective molarities.

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NaOH + HCl NaCl + H2O

125



126



– Hence,

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• The titration of a weak acid by a strong base gives a somewhat different curve.

Titration of a Weak Acid by a Strong Base

Figure 17.13: Curve for the titration of a weak acid by a strong base.

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• Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5.

HC2H3O2 + NaOH NaC2H3O2 + H2O– At the equivalence point,.

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HC2H3O2 + NaOH NaC2H3O2 + H2O

– First, calculate the concentration of the acetate ion.

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A Problem To Consider• Calculate the pH of the solution at the equivalence point when

25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5.

HC2H3O2 + NaOH NaC2H3O2 + H2O 2.5 x 10-3 mol

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– The total volume of the solution is 50.0 mL. Hence,

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HC2H3O2 + NaOH NaC2H3O2 + H2O

.025 M .025 M - .025 M - .025 M +.025 M 0 0 .025 M


134


NaC2H3O2 Na+ + C2H3O2-

.025 M .025 M .025 M


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• The Kb for the acetate ion is 5.9 x 10-10

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• The titration of a weak base with a strong acid is a reflection of our previous example.– Figure 17.14 shows the titration of NH3 with HCl.

– In this case, the pH declines slowly at first, then falls abruptly from about pH 7 to pH 3.

– Methyl red, which changes color from yellow at pH 6 to red at pH 4.8, is a possible indicator.

Titration of a Strong Acid by a Weak Base

Figure 17.14: Curve for the titration of a weak base by a strong acid.

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Just a Reminder…

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Building Equations with K values

• Weak acids and bases are assigned a Ka or Kb values based on the degree to which they ionize in water.

• Larger K values indicate a greater degree of ionization (strength).

• Ka and Kb, along with other K values that we will study later (Ksp, KD, Kf) are all

manipulated in the same manner.

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• When equations are added K values are multiplied.

• MnS ↔ Mn+2 + S-2 K= 5.1 x 10-15

• S-2 + H2O ↔HS- + OH- K= 1.0 x 10-19

• 2H+ + HS- OH- ↔ H2S +H2O K= 1.0 x 10-7

• MnS + 2H+ ↔ Mn+2+ H2K=


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• When equations are reversed the K values are reciprocated.

• • Al(OH)3 ↔ Al+3 + 3OH- K = 1.9 x 10-33

• 3OH- + Al+3 ↔ Al(OH)3 K = 1/1.9 x 10-33 =


142

• When equations are multiplied the K values are raised to the power.

• NH3 + H2O ↔ NH4++ OH- K = 1.8 x 10-5

• 2NH3 + 2H2O ↔ 2NH4++ 2OH- K = (1.8 x 10-5)2

=


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• When equations are divided the root of the K values are taken.

• 2HPO4-2 ↔ 2H+ + 2PO4

3- K = 1.3 x 10-25

• HPO4-2 ↔ H+ + PO4

3- K = √1.3 x 10-25 =


144

Operational Skills

• Determining Ka (or Kb) from the solution pH

• Calculating the concentration of a species in a weak acid solution using Ka

• Calculating the concentration of a species in a weak base solution using Kb

• Predicting whether a salt solution is acidic, basic, or neutral

• Obtaining Ka from Kb or Kb from Ka

• Calculating concentrations of species in a salt solution

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Operational Skills

• Calculating the common-ion effect on acid ionization

• Calculating the pH of a buffer from given volumes of solution

• Calculating the pH of a solution of a strong acid and a strong base

• Calculating the pH at the equivalence point in the titration of a weak cid with a strong base

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Animation: Acid Ionization Equilibrium

Return to slide 3

(Click here to open QuickTime animation)

147

Figure 17.3: Variation of percent ionization of a weak acid with concentration.

Return to slide 19

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Figure 17.8: A pH meter reading NH4Cl. Photo courtesy of American Color.

Return to slide 82

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Animation: Adding an Acid to a Buffer

Return to slide 104

(Click here to open QuickTime animation)

1 acid-base equilibria. 2 solutions of a weak acid or base the simplest acid-base equilibria are...

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