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Overview of Acid-Base Equilibria Questions -- KEY

The following questions are designed to give you an overview of the topics in Acids and Bases

and Equilibria in Acid-Base Solutions. Although not comprehensive, these few questions do a

good job of covering the majority of the concepts and calculations related to this subject area.

Please do your best to answer the questions completely. Keys will be posted.

Acid-Base Equilibria

1. a. A typical weak acid, HA, is prepared in such a way as to have an initial concentration

in water of 0.100 M. At equilibrium the solution has pH = 2.87, what is Kafor this acid?

The molarity of the conjugate base and the molarity of the hydronium ion are known since the

pH is given (pH = 2.87 gives [A-] = [H

+] = 10

-pH= 1.35x10

-3M.

The reaction information is summarized as . . .

HA(aq) + H2O(l) H3O+

(aq) + A-(aq)

initial 0.10 M ---- ~ 0 M 0 M

change - 1.35x10-3

M ---- + 1.35x10-3

M + 1.35x10-3

M

equilibrium0.10 - 1.35x10

-3

M---- 1.35x10

-3M 1.35x10

-3M

As you can see, all the concentrations are known and it is a simple matter to solve for Ka.

[ ]

( )( )( )

5

3

333

108.11035.110.0

1035.11035.1

+

=

== xx

xx

HA

AOH

Ka

As you can plainly see the reaction only proceeded to a small extent in the forward

direction in order to reach equilibrium . . . so this is a WEAK acid!

b. The conjugate base of the above acid exists as a sodium salt, NaA. What is Kbfor this

conjugate base.

Remember for conjugate acid base pairs (and only for pairs): Kw= Ka*KbSo Kb= Kw/Ka= (1.00x10

-14)/( 1.8x10

-5) = 5.6x10

-10

c. For a 0.100 M solution of the conjugate base in part b, what is the [OH-] atequilibrium AND what is the pH?

NOTE: in this problem I referred to the base as B-. . . this is an inconsequential change.

This is a straightforward equilibrium problem. The changes in concentrations and the final

equilibrium concentrations will be expressed in terms of a single variable . . .

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B-(aq) + H2O(l) OH

-(aq)

+ HB

(aq)

initial 0.10 M ---- ~ 0 M 0 M

change - x M ---- + x M + x Mequilibrium 0.10 - x M ---- x M x M

The variable expressions of the concentrations can be plugged into the equilibrium constant

expression and this expression can be solved. This should all look very familiar by now!

[ ][ ]

( )( )( )

10.0

106.5

10.0

210 x

x

x

xx

B

HBOHKb

==

Giving x = [OH-] = [HB] = 7.5x10

-6M and the following equilibrium conditions

B-(aq) + H2O(l) OH

-(aq) + HB(aq)

equilibrium ~0.10 M ---- 7.5x10-6

M 7.5x10-6

M

And pOH = 5.12 along with pH = 14.00 - pOH = 8.88

d. For the same pair of weak acid and conjugate base in the questions above, what is the

pH of a buffer in which [HA] = [A-] = 0.25 M?

For a buffer system,[ ]HAA

pKpH a

+= log

with pKa= -logKa= 4.74 When [HA] = [A-], then pH = pKa= 4.74

e. For the same pair of weak acid and conjugate base to form a buffer with pH = 5.00

when [HA] = 0.25, what does [A-] have to be?

In a buffer system,

[ ]HA

ApKpH a

+= log

with pKa= -logKa= 4.74 and [HA] = 0.25 M

Then for pH = 5.00, we find

5.00 - 4.74 = log[A-]- log[0.25] which then solves to -0.342 = log [A

-] so . . .

[A-] = 0.455M

This makes good sense, b/c for pH > pKa, [A-] should be > [HA]

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f. When 4.00 L of the buffer in part e (the pH = 5.00 buffer) has 0.100 mole of HCl (g)

added, what is the new pH? HINT: should it be higher or lower than 5.00? Why?

The simplest solution is,[ ][ ]

93.4100.025.0*4

100.045.0*4log74.4 5 =

+

+=pH

Note: moles of A-

were used instead of [A-

] and moles of HA instead of [HA]. This is perfectlyacceptable since both species exist in the same 4.00 L volume. Also note that the answer

makes good sense b/c an addition of acid should decrease the pH

g. When 0.100 mole of HCl (g)is addedto 4.0 L ofpure water (pH = 7.00), what is the

new pH? What are [H+] and [OH-] and pOH? How does the pH change of the water

compare to that of the above buffer?

The added HCl creates a [H+] = 0.100 mole/4.0 L = 0.025 M

This gives: pH = 1.60, pOH = 12.40, and [OH-] = 4.0x10

-13M (check [H

+][OH

-] = Kw) OK

Note: water is not a good buffer!!

2. a. 0.100 M NaOH(aq)is used to titrate 25.00 mL of 0.200 M HCl(aq). Calculate the pH

during this titration at the following volumes of added NaOH: 0.00 mL, 25.00 mL, 49.00mL, the equivalence point, 51.00 mL, and 60.00 mL.

The balanced reaction: HCl (aq) + NaOH (aq) NaCl (aq) + H2O(l)

Note: M = mole per liter may also be mmole per mL . . . which for problems like this one can save a great deal of effort

initial mmole HCl = (0.200 mmole/mL)(25.00 mL) = 5.00 mmole HCl

mmole NaOH added = (0.100 mmole/mL)(volume added in mL) = 3rd

column

mL NaOH need to reach equiv. pt. = 5.00 mmole needed / 0.100 M = 50.0 mL NaOH soln

NOTE: at the equivalence point only NaCl is present in the solution and pH = 7.00

mmole excess H+= mmole initial HCl mmole added NaOH = 4

thcolumn

mmole excess OH-= mmole added NaOH mmole initial HCl = 5

thcolumn

[H+] = mmole excess H

+/ total volume = 6

thcolumn before equiv. pt.

[OH-] = mmole excess H

+/ total volume = 6

thcolumn after equiv. pt.

pH = -log [H+], note after the equivlance pt. I used pH = 14.00 - pOH

mLNaOH

mmole H+initial

mmoleOH- added

mmole H+excess

mmole OH-excess

[H+], M pH

0.00 5.00 0.00 5.00 - 2.00E-01 0.70

25.00 5.00 2.50 2.50 - 5.00E-02 1.30

49.00 5.00 4.90 0.10 - 1.35E-03 2.87

50.00 5.00 5.00 0.00 0.00 1.00E-07 7.00

[OH-], M

51.00 5.00 5.10 - 0.10 1.32E-03 11.12

60.00 5.00 6.00 - 1.00 1.18E-02 12.07

Note: see sketch of titration curve below . . . due to limitations of Excel, the shape to the curve is a little odd

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titration curve of HCl with NaOH

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0.00 20.00 40.00 60.00

mL 0.100 M NaOH

pH

b. 0.100 M NaOH(aq)is used to titrate 25.00 mL of 0.200 M acetic acid(aq).

[for acetic acid, Ka= 1.8 x 10-5] Calculate the pH during this titration at the following

volumes of added NaOH: 0.00 mL, 25.00 mL, 49.00 mL, the equivalence point, 51.00

mL, and 60.00 mL.

The balanced reaction: HOAc (aq) + NaOH (aq) NaOAc (aq) + H2O(l)

Notes: OAc-represents the acetate ion and HOAc represents acetic acid

M = mole per liter may also be mmole per mL . . . which for problems like this one can save a great deal of effort

initial mmole HOAc = (0.200 mmole/mL)(25.00 mL) = 5.00 mmole HOAc

mmole NaOH added = (0.100 mmole/mL)(volume added in mL) = 3rd

column

mL NaOH need to reach equiv. pt. = 5.00 mmole needed / 0.100 M = 50.0 mL NaOH soln

pH at 0.00 mL is found using an ICE table for 0.20 M acetic acid [Ka= 1.8x10-5

]

[H+] = square root of Ka*0.20 = 1.9x10

-3M, pH = -log[H

+] = 2.72

mLNaOH

mmoleHOAc

initial

mmoleOH-

added

mmoleHOAc

excess

mmoleOAc-

formed

[H+], M pH

0.00 5.00 0.00 5.00 - 1.90E-03 2.72

25.00 5.00 2.50 2.50 2.50 1.82E-05 4.74

49.00 5.00 4.90 0.10 4.90 1.35E-03 6.43

[OH-], M

50.00 5.00 5.00 0.00 5.00 6.11E-06 8.79

51.00 5.00 5.10 - 5.00 1.32E-03 11.12

60.00 5.00 6.00 - 5.00 1.18E-02 12.07

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at 25.00 mL pH = pKa because [HOAc] =[OAc-] since one half of the acid has been converted

into the conjugate base and

pH = pKa+ log [base]/[acid], where base means the basic buffer species and acid is the acidic

buffer species

at 49.00 mL added NaOH the remaining HOAc = 0.100 mmole and the OAc-= 4.90 mmole

sincepH = pKa+ log [base]/[acid] we find pH = 4.74 + log (4.90/0.10) = 6.43

Note that we can still use this equation even though we are outside of the buffer range because

our assumptions about the mmole of basic and acidic buffer species are valid

At the equivalence pt. we need to solve an ICE table for the acetate ion with Kb= 5.6x10-10

since

Kb= Kw/Kawe find [OH-] = square root of 5.6x10

-10*(5.00 mmole/75.00 mL) = 6.1x10

-6M

Using what should now be standard practice we find pH = 14.00 pOH = 8.79

Beyond the equivalence point the pH is controlled by the excess NaOH added and thecalculations are the same as found in part (a) at 51.00 and 60.00 mL added NaOH.

Note: see sketch of titration curve below . . . due to limitations of Excel, the shape to the curve is a little odd, adding more datapoint could improve the shape.

titration of HOAc with NaOH

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0.00 10.00 20.00 30.00 40.00 50.00 60.00

mL 0.100 M NaOH

pH

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c. Make rough sketches of the above titrations. Note any significant differences and/or

similarities. [ex. look at initial pH, pH at equiv. pt., vol. of NaOH at equiv. pt., etc.]

See (a) and (b) above

similarities: since the stoichiometry of both titrations are 1:1 and the concentration and volumes

used are the same, the equivalence point volume of NaOH is the same. Also, since theconcentration of NaOH and volumes are the same, the pH beyond the equivalence pt. is

the same for both curvesdifferences: HCl is a strong acid and acetic acid is a weak acid, therefore, the pH of the HCl

curve is lower at the start of the titration. Also, the acetic acid titration curve passes

through a buffer region [centered on pH = pKaat the equivalence point]

d. Explain how you might select an indicator for the above titrations? HINT: pKIn.

Use the Henderson-Hasselbalch equation to understand that we want to choose an indicator that

will change color over a pH range of about 2 pH units [for the same reason that buffers

are effective over about a 2 pH unit range]. Then select an acid-base indicator with a pKathat falls on the steep part of the titration curve . . . ideally pKa of the indicator will

perfectly match the pH at the equivalence point. When this is not true, there can be a

difference between the added titrant volume at which the color change of the indicator isobserved (known as the end-point) vs. the volume at which the equivalence point is

reached. Our goal is to choose an indicator that will minimize this end-point error.

See the table or figure of indicators in your text:

For the HCl titration we might use chlorphenol red pKa ~ 6, bromthymol blue pKa ~ 6.8, orphenol red pKa ~ 7.2 . . . but since the equivalence pt. region is VERY steep for a strong

acid-strong base titration, we can even get away with using phenolphthalein pKa ~8.8and only have a small end-pt error

For the HOAc titration we will want to avoid the chlorphenol red, bromthymol blue, and phenol

red options listed above. Sticking with phenolphthalein for this titration is an excellentidea [phenolphthalein is a very common choice of indicator for a weak acid-strong base

titration indicator since the pH at the equivalence point will be controlled by a weak base

in these tirations the equivalence point pH will be > 7 but also < 11]

Note, that for the next curve [weak base-strong acid titration] we will select an indicator that

changes color at a pH < 7 but > 3 since a weak acid will be present at the equivalence

point.

e. Based on what you have learned from the above, make a sketch of the titration curve

of 25.00 mL 0.200 M ammonia [Kb= 1.8 x 10-5] with 0.100 M HCl. You neednt make

calculations, but do understand how you would do so if asked.

See the next page: no calcs. discussed, but a table of values is included.

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mL HCl mmoleNH3

initial

mmoleH+

added

mmoleNH3

excess

mmoleNH4+

formed

[OH-], M pH

0.00 5.00 0.00 5.00 - 1.90E-03 11.28

25.00 5.00 2.50 2.50 2.50 5.50E-10 9.26

49.00 5.00 4.90 0.10 4.90 1.35E-03 7.57

[H+], M

50.00 5.00 5.00 0.00 5.00 6.11E-06 5.21

51.00 5.00 5.10 - 5.00 1.32E-03 2.88

60.00 5.00 6.00 - 5.00 1.18E-02 1.93

Note: see sketch of titration curve below . . . due to limitations of Excel, the shape to the curve is a little odd, adding more datapoint could improve the shape.

titration of NH3 with HCl

0.00

2.00

4.00

6.00

8.00

10.00

12.00

0.00 10.00 20.00 30.00 40.00 50.00 60.00

mL 0.100 M HCl

pH

3. Calculate the pH of the following two solutions: (i) 0.100 M H3PO4and (ii) 0.100 M

Na3PO4. Also, find the concentrations of the following species in each of the two

solutions: H3PO4, H2PO4-, HPO4-2, and PO4-3. HINT: make sure you choose the

correct equilibrium constants at the correct times.

For H3PO4 Acidicspecies

Relationship BasicConjugate

For Na3PO4

Ka1= 6.9 x 10-3

H3PO4 Kw= Ka1*Kb3 H2PO41-

Kb3= 6.9 x 10-3

Ka2= 6.2 x 10-8

H2PO41-

Kw= Ka2*Kb2 HPO42-

Kb2= 6.2 x 10-8

Ka3= 4.5 x 10-13

HPO42-

Kw= Ka3*Kb1 PO43-

Kb1= 4.5 x 10-

13

Note: the best acid has the least good conjugate base

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(i) First phosphoric acid, 0.100 M

H3PO4(aq) + H2O(l) H3O+

(aq) + H2PO41-

(aq)

initial 0.10 M ---- ~ 0 M 0 M

change - x M ---- + x M + x M

equilibrium 0.10 - x M ---- x M x M

[ ] ( )3

-1423

1 106.610.0

]][[POH +

=

== xx

xx

HA

OHKa

Since this Kais reasonably large we cannot simplify by the assumption thatxis small . . . oh wellat least this is not on an exam! So we solve the quadratic equation and take the one root

that is reasonable giving x = [H3O+] = 2.3 x 10

-2M = [H2PO4

1-] and pH = 1.64

and [H3PO4] = 0.077 M

Note: this is 23% ionization so we certainly cannot avoid the quadratic!

Next we carry through these findings into the second equilibria . . .

H2PO41-

(aq) + H2O(l) H3O+

(aq) + HPO42-

(aq)

initial 2.3 x 10-2

M ---- 2.3 x 10-2

M 0 M

change -y M ---- + y M +y M

equilibrium 2.3 x 10-2

- y M ---- 2.3 x 10-2

+ y M y M

[ ]

8

2

2

-1

42

-243

2 102.6]103.2[

]][103.2[

POH

HPO

+

=

+== x

yx

yyxOHKa

This time we can assume thatyis small because this Kais small! When we do this we find thaty = [HPO4

2-] = Ka2= 6.2 x 10

-8M and our assumption is valid and the pH = 1.64 is still correct

Next we carry through findings from steps one and two into the third equilibria . . .

HPO41-

(aq) + H2O(l) H3O+

(aq) + PO43-

(aq)

initial 6.2 x 10-8

M ---- 2.3 x 10-2

M 0 M

change -z M ---- + z M +z M

equilibrium 6.2 x 10-8

-z M ---- 2.3 x 10-2

+ z M z M

[ ]13

8

2

-24

-343

3 108.4]102.6[

]][103.2[

HPO

PO

+

=

+== x

zx

zzxOHKa

We can now VERY safely assume thatzis small. This yields z = [PO43-

] = 1.3 x 10-18

M and the

pH still is 1.64 !!!!!

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(ii) For Na3PO40.100 M we have a very similar process starting with the phosphate and working

through the steps in the order Kb1, then Kb2, then Kb3we will find . . .. .

from the first equilibrium

PO43-

(aq) + H2O(l) OH-(aq) + HPO4

2-(aq)

solving the quadratic once again!!!!

[OH-] = [HPO4

2-] = x = 0.037 M

[PO43-

] = 0.100 x = 0.063 M

btw, this yields pOH = 1.43 giving pH = 12.57

from the second base ionization step we find

HPO42-

(aq) + H2O(l) OH-(aq) + H2PO4

1-(aq)

[H2PO4

-] = y = 1.6x10

-7M = Kb2

which is such a small change as to not alter the results found above and pH = 12.57 is still valid!

from the third base ionization step we find

H2PO41-

(aq) + H2O(l) OH-(aq) + H3PO4(aq)

[H3PO4] = z = 6.5 x 10-18

M

which is once again such a tiny change that our prior assumptions are all OK and pH = 12.57 still

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