11111 Chemistry 132 NT Instead of having “answers” on a math test, they should just call them...

11111

Chemistry 132 NT

Instead of having “answers” on a math test, they should just call them “ impressions”, and if you got a different “impression”, so what, can’t we all be brothers?

Jack Handey

33333

Acid-Base Equilibria

Chapter 16

Module 3

Sections 16.5, 16.6, and 16.7 Reaction of zinc metal

with hydrochloric acid.

44444

Review

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

55555

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 equilibrium above.

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

66666



If we were to add NaC2H3O2 to this solution, it would provide C2H3O2

- ions which are present on the right side of the equilibrium.


77777



The equilibrium composition would shift to the left and the degree of ionization of the acetic acid is decreased.


88888



This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect.


99999

A Problem To Consider

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

1010101010



The equilibrium constant expression is:

a2

23 K]OHCH[

]OCH][OH[

1111111111



Substituting into this equation gives:

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

1212121212



Assume that x is small compared with 0.018 and 0.025. Then

025.0)x025.0(

018.0)x018.0(

1313131313



The equilibrium equation becomes

4107.1)025.0()018.0(x

1414141414



Hence,

44 104.2018.0025.0

)107.1(x

1515151515



Note that x was much smaller than 0.018 or 0.025.

63.3)104.2log(pH 4

For comparison, the pH of 0.025 M formic acid alone (without the sodium formate) is 2.69.

1616161616



Note that x was much smaller than 0.018 or 0.025.

63.3)104.2log(pH 4

(see Example 16.10 and Problem 16.61)

1717171717

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.Buffers contain either a weak acid and its

conjugate base or a weak base and its conjugate acid.

Thus, a buffer contains both an acid species and a base species in equilibrium.

1818181818

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.Consider a buffer with equal molar amounts of

HA and its conjugate base A-.

When H3O+ is added to the buffer it reacts with the base A-.

)l(OH)aq(HA)aq(A)aq(OH 23

1919191919

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.Consider a buffer with equal molar amounts of

HA and its conjugate base A-.

When OH- is added to the buffer it reacts with the acid HA.

)aq(A)l(OH)aq(HA)aq(OH 2

2020202020

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.Two important characteristics of a buffer are its

buffer capacity and its pH.Buffer capacity depends on the amount of acid and conjugate base present in the solution.

The next example illustrates how to calculate the pH of a buffer.

2121212121

The pH of a Buffer

The solution described in Example 16.10 in your text is a buffer, because it is composed of a weak acid (0.10 M acetic acid) and its conjugate base (0.20 M acetate ion).

The example described how to calculate the pH of such a buffer (we obtained a pH of 5.07)

Given the recipe for making a buffer from volumes of solutions, we should then be able to calculate its pH. Let’s try one.

2222222222


Instructions for making up a buffer say to mix 60.0 mL of 0.100 M NH3 with 40.0 mL of 0.100 M NH4Cl. What is the pH of this buffer?

This buffer contains a base (NH3) and its conjugate acid (NH4

+) in equilibrium.

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

To do the equilibrium calculation, we need the starting concentrations in the solution obtained by mixing the NH3 and NH4Cl solutions.

2323232323



To do this, we must calculate the moles of NH3 and NH4Cl present and divide by the total volume of the solution.

This buffer contains a base (NH3) and its conjugate acid (NH4

+) in equilibrium.


2424242424



Note that the instructions say to add 60.0 mL (0.0600 L) of 0.100 M NH3, so…

In the same way, we find we have added 0.00400 mol NH4

+ (from NH4Cl).

soln. NH liters NH molarity NH moles 333

3NH mol 0.00600 L 0600.0 0.100 L

mol

2525252525



Total volume of buffer =

60.0 mL + 40.0 mL = 100.0 mL (0.100 L)Therefore, the concentration of base and conjugate acid are:

M 0600.0]NH[ L 100.0mol 00600.0

3

M 0400.0]NH[ L 100.0mol 00400.0

4

2626262626



Fill in the concentration table for the acid-base equilibrium (base ionization of NH3).


Starting 0.060 0.040 0Change -x +x +x

Equilibrium 0.060-x 0.040+x x

2727272727




b3

4 K]NH[

]OH][NH[

2828282828



Substituting into this equation gives:

5108.1)x060.0(

)x)(x040.0(

2929292929



Assuming that x is very small compared to 0.040 and 0.060, we can simplify

5108.1)060.0(

)x)(040.0(

3030303030



Therefore,

55040.0

060.0 107.2)108.1()(x

Thus, the hydroxide concentration is 2.7 x 10-5. The pH of the buffer is:

43.9)107.2log(00.14pOH00.14pH 5

(see Exercise 16.12 and Problem 16.65)

3131313131

Adding Acid or Base to a Buffer

Example 16.10 in your text describes a buffer composed of acetic acid and sodium acetate. This buffer controls the equilibrium below.


Adding a strong acid to this buffer would convert some of the C2H3O2

- (a base) back into HC2H3O2.

Adding a strong base to this buffer would convert more HC2H3O2 (an acid) into C2H3O2

-

(its conjugate base)

3232323232

Adding Acid or Base to a Buffer

Example 16.10 in your text describes a buffer composed of acetic acid and sodium acetate. This buffer controls the equilibrium below.


Let’s try an example where we add a strong acid to this buffer.

3333333333


Calculate the pH of 75 mL of the buffer described in Example 16.10 (0.10 M HC2H3O2 and 0.20 M NaC2H3O2) to which 9.5 mL of 0.10 M HCl is added.

When the hydronium ion (from the HCl) is added to the buffer, it reacts completely with the acetate ion.

)l(OH)aq(OHHC)aq(OHC)aq(OH 22322323

This initial reaction goes to completion until all of the H3O+ (from HCl) has been used up.

3434343434



We must first calculate the amounts of hydrogen ion, acetate ion and acetic acid present in the solution before reaction.

HCl mol 00095.0)HCl L 0095.0( HCl L 1HCl mol 10.0

since HCl is a strong acid.

H mol 00095.0

3535353535



We must first calculate the amounts of hydrogen ion, acetate ion and acetic acid present in the solution before reaction.

232HCl L 1OHC mol 20.0 OHC mol 015.0)soln L 075.0( 232

232HCl L 1OHHC mol 10.0 OHHC mol 0075.0)soln L 075.0( 232

3636363636



You can now assume that all of the hydronium ion added (0.00095 mol) reacts with acetate ion.

Therefore, 0.00095 mol of acetate ion is used up and 0.00095 mol of acetic acid is produced.

232 OHC mol )00095.0015.0(ion acetate of Moles 232 OHC mol 014.0

3737373737



You can now assume that all of the hydronium ion added (0.00095 mol) reacts with acetate ion.

Therefore, 0.00095 mol of acetate ion is used up and 0.00095 mol of acetic acid is produced.

232 OHHC mol 0085.0232 OHHC mol )00095.00075.0(acid acetic of Moles

3838383838



Now that we know the amounts of acetic acid and acetate ion remaining after the addition of the HCl, we can solve the equilibrium calculation.

Note that the total volume of solution (the buffer and the HCl) is 75 mL + 9.5 mL, or 85 mL (0.085 L) to two significant figures.

3939393939



M 10.0L 085.0mol 0085.0

]OHHC[ 232

Hence the concentrations after the acid addition are:

M 16.0L 085.0

mol 014.0]OHC[ 232

4040404040



Now we can construct the following table.

Starting 0.10 ~0 0.16Change -x +x +x

Equilibrium 0.10-x x 0.016+x

(aq)OHC(aq)OH 2323

4141414141




a232

2323 K]OHHC[

]OHC][OH[

4242424242



Substituting, you get:

5107.1)x10.0()x16.0(x

If we assume x is small, then:

16.0)x16.0( and 10.0)x10.0(

4343434343



And simplifying, we get

5107.1)10.0()16.0(x

or55 101.1

16.010.0

)107.1(x

4444444444



The H3O+ concentration is 1.1 x 10-5, so the pH is

96.4)101.1log(]OHlog[pH 53

The pH of the original buffer in Example 16.10 was 5.07.

4545454545




96.4)101.1log(]OHlog[pH 53

The addition of strong acid to the buffer dropped its pH by 0.11 pH units.

4646464646




96.4)101.1log(]OHlog[pH 53

This illustrates the stability of the buffer.

4747474747




96.4)101.1log(]OHlog[pH 53

Adding the same amount of acid to 75 mL of pure water results in a pH of 1.96.


4848484848

The Henderson-Hasselbalch Equation

How do you prepare a buffer of given pH?

A buffer must be prepared from a conjugate acid-base pair in which the Ka of the acid is approximately equal to the desired H3O+ concentration.To illustrate, consider a buffer of a weak acid HA and its conjugate base A-.

The acid ionization equilibrium is:

)aq(A)aq(OH )l(OH)aq(HA 32

4949494949



The acid ionization constant is:

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

]A[

]HA[K]O[H a3

]HA[]A][OH[

K 3a

5050505050



Taking the negative logarithm of both sides of the equation we obtain:

]A[

]HA[log)Klog(]Olog[H- a3

The previous equation can be rewritten

]HA[]A[

logKppH a

5151515151



More generally, you can write

This equation relates the pH of a buffer to the concentrations of the conjugate acid and base. It is known as the Henderson-Hasselbalch equation.

]acid[]base[

logKppH a

5252525252


A buffer is prepared by adding 45.0 mL of 0.15 M NaF to 35.0 mL of 0.10 M HF. What is the pH of this buffer? (The Ka for HF is 6.8 x 10-4)

The Henderson-Hasselbalch equation is ideal for solving buffer pH problems.

]acid[]base[

logKppH a

The negative log of the Ka, which was given.

The concentration of HF.

The concentration of F- (from NaF).

5353535353



First, let’s calculate the “new” molarities of HF and F-, once the solutions are mixed together.

HF mol 0035.0) (0.10L) (0.035 HF mol LHF mol

HF M 044.0volume total L 080.0

HF mol 0.0035 HF M new""

Total volume = 0.035 L + 0.045 L = 0.080 L

5454545454



First, let’s calculate the “new” molarities of HF and F-, once the solutions are mixed together.

-L

HF mol- F mol 0068.0) (0.15L) (0.045 F mol

--

- F M 084.0volume total L 080.0

F mol 0.0068 F M new""

5555555555



Now, we need the “pKa”.

17.3)108.6log(pK 4a

Now, we’re ready to assemble the Henderson-Hasselbalch equation.

5656565656



Substituting into the equation we get:

]acid[]base[

logKppH a

HF) M 044.0()F M 084.0(

log(3.17)pH

45.328.0(3.17)pH

Note that the pH of a buffer will always be close to its pKa. The log term simply “fine tunes” the buffer.

5757575757



So to prepare a buffer of a given pH (for example, pH 4.90) we need a conjugate acid-base pair with a pKa close to the desired pH.

The Ka for acetic acid is 1.7 x 10-5, and its pKa is 4.77.

You could get a buffer of pH 4.90 by increasing the ratio of [base]/[acid].

5858585858

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).

Such curves are used to gain insight into the titration process.

You can use titration curves to choose an appropriate indicator that will show when the titration is complete.

5959595959

Titration of a Strong Acid by a Strong Base

Figure 16.11 shows a curve for the titration of HCl with NaOH.

Note that the pH changes slowly until the titration approaches the equivalence point.

The equivalence point is the point in a titration when a stoichiometric amount of reactant has been added.

Curve for the titration of a strong acid by a strong base.

Figure 16.11

Some unneutralized acid. In this case simple stoichiometry tells you the concentration of remaining H+ from which you calculate pH.

Equivalence point. For a strong acid and strong base, this will always be pH 7.00

6161616161



At the equivalence point, the pH of the solution is 7.0 because it contains a salt, NaCl, that does not hydrolyze.

However, the pH changes rapidly from a pH of about 3 to a pH of about 11.

6262626262



To detect the equivalence point, you need an acid-base indicator that changes color within the pH range 3-11.

Phenolphthalein can be used because it changes color in the pH range 8.2-10. (see Figure 16.8)

6363636363


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.

Because the reactants are a strong acid and a strong base, the reaction is essentially complete.

)l(OH)l(OH)aq(OH)aq(OH 223

6464646464



We get the amounts of reactants by multiplying the volume of each (in liters) by their respective molarities.

mol 00250.0100.0L0250.0OH Mol Lmol

3

mol 00100.0100.0L0100.0OH Mol Lmol

6565656565



All of the OH- reacts, leaving an excess of H3O+

l0.00100)mo 00250.0(OH Excess 3

OH mol 0.00150 3

6666666666



You obtain the H3O+ concentration by dividing the mol H3O+ by the total volume of solution (0.0250 L + 0.0100 L=0.0350 L)

M 0429.0L 0.0350mol 00150.0

]OH[ 3

6767676767



Hence,

368.1)0429.0log(pH


6868686868

Titration of a Weak Acid by a Strong Base

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

The pH range of these titrations is shorter.

Before the equivalence point is reached, we have some un-neutralized acid and it’s conjugate base anion, that is, a buffer.

The pH in this “buffer” region can be determined using the Henderson-Hasselbalch equation.

6969696969

Titration of a Weak Acid by a Strong Base

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

The equivalence point will be on the basic side since the salt produced contains the anion of a weak acid.

Figure 16.12 shows the curve for the titration of nicotinic acid with NaOH.

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

Figure 16.12

Buffer region. Some un-neutralized acid and some conjugate base anion…you must use the Henderson Hasselbaclch equation to determine pH.

Equivalence point…only the conjugate base anion remains…must solve hydrolysis problem to determine pH.

7171717171


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.

At the equivalence point, equal molar amounts of acetic acid and sodium hydroxide react to give sodium acetate.

7272727272



First, calculate the concentration of the acetate ion.In this case, 25.0 mL of 0.10 M NaOH is needed to react with 25.0 mL of 0.10 M acetic acid.

7373737373



The molar amount of acetate ion formed equals the initial molar amount of acetic acid.

soln L 1e ionmol acetat 0.10

soln L 1025 3 e ionmol acetat 102.5 3-

7474747474



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

M 050.0L1050

mol102.5 ionconcentrat molar 3

3-

7575757575



You find the Kb for the acetate ion to be 5.9 x 10-10 and that the concentration of the hydroxide ion is 5.4 x 10-6. The pH is 8.73

The hydrolysis of the acetate ion follows the method given in an earlier module of this chapter.

7676767676

Titration of a Weak Base by a Strong Acid

The titration of a weak base with a strong acid is a reflection of our previous example.

Figure 16.13 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.

Curve for the titration of a weak base by a strong cid.

Figure 16.13

7878787878

A Little “Trick”

At the “half-way” point in the titration of a weak acid with a weak base, we can make a significant simplification.

At the halfway point in such a titration, exactly half of the original acid has been converted to its conjugate base anion.

]HA[]A[

logKppH a

This means that [A-] = [HA]

7979797979

A Little “Trick”

At the “half-way” point in the titration of a weak acid with a weak base, we can make a significant simplification.

At the halfway point in such a titration, exactly half of the original acid has been converted to its conjugate base anion.

)1log(KppH a

aKppH

The pH = pKa only at the halfway point. We’ll use this fact in lab to determine the Ka of an unknown acid.

8080808080

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

8181818181

Homework

Chapter 16 Homework: collected at the first exam.

Review Questions: none.Problems: 37, 41, 45, 49.

8282828282

11111 Chemistry 132 NT Instead of having “answers” on a math test, they should just call them...

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