weak acid calc

8/3/2019 Weak Acid Calc

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1

pH calculations:

To do pH calculations one must consider

both the nature and condition (amount of

ionization) of the species in the solution andthen calculate the concentration of the

hydronium ion (H+ or H3O+).


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Strong acid: [H3

O+] = concentration of acid

so: pH = -log [H3O+] = -log[acid]

Weak Acid: one must calculate the [H3O+]

from an equilibrium ionization expression.

HA + H2O H3O+ + A-

Ka = _[H

3O+][A-]

[HA]

These are equilibrium

concentrations.


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4

Titration Curve of a Weak Acid with a Strong Base

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50

Volume of Strong Base Added

equivalence pointbuffer region

note volume ratio

between these points

1

23

4

[H+]=[A-]

[H+]{[A-]

pOH=-Log[XS OH-]

[OH-]=[HA]

]HA[

]A][H[K

a

!

]HA[

]A][H[K

a

!

]A[

]HA][OH[K

b

!

Approximations:


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5

Problem: Calculate the pH of 25.0 mL of 0.10M

acetic acid (HOAc).

The Ka of HOAc = 1.8 x 10-5

Which region of a titration curve would this be?


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Problem: Calculate the pH of 25.0 mL of 0.10M

acetic acid (HOAc).

The Ka of HOAc = 1.8 x 10-5

Ka = [H+][OAc-]

[HOAc]

HOAc H+ + OAC-

IC

E

0.10 0 0-x +x +x

0.10-x x xKa = [x][x]

0.10-xSmall,drop, WHY?

Ka = x2

0.10x = 0.00134 = [H+]

pH = -log 0.00134 =2.87


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0

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10

12

14

0 5 10 15 20 25 30 35 40 45 50

Volume o f Strong Base Adde d

equivalence pointbuffer region note volume ratiobetween these points25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5

0.10 M NaOH

1.

2.

34

Some base has been added

This is also the buffer region

pKa = pH at this point (halfway point)


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Region 2 Theory and Calculations:

Base has been added and some of the acid has been

neutralized.

What has changed and WHY?

1. The moles of acid is decreased.

2. The moles of acid anion (A-) is increased.

HA + OH-p H2O + A-

3. The total volume has increased.

HA + OH- p H2O + A- + XS HA

note: the 1:1 ratio


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Region 2 Theory and Calculations:

Base has been added and some of the acid has been

neutralized. intial RXN

Equilibrium XS HA + H2O H3O+ + A-

Ka= [H+][A-]

[HA]

Leads to:

and [H+]{[A-] Why?

always(calculate from equilibrium)

HA + OH- p H2O + A-

2 sources


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Problem: Calculate the pH of a solution that is

prepared by combining 25.00mL of 0.10M aceticacid (HOAc) with 14.00mL of 0.080M NaOH.

(Ka of HOAc = 1.8 x 10-5).

Solution: HOAc + OH- p H2O + OAc- 1:1 ratio

1. Find initial moles of Acid:

= mol HOAc0.02500L1L

0.10mol HOAc 0.0025

2. find moles of OH-:

_____________________________= mol OH-0.01400LNaOH

1L

0.080molNaOH

1molNaOH

1molOH- 0.00112

Region 2


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Problem: Calculate the pH of a solution that is prepared by

combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of

0.080M NaOH. (Ka of HOAc = 1.8 x 10-5).

Solution:

1. We found initial mol of Acid: HOAc + OH- p H2O + OAc- 1:1 ratio

_____ = mol HOAc0.02500L

1L

0.10mol HOAc 0.0025

2. We found mol of OH-: which =s initial moles of OAC- anion.

_____________________________________= mol OH-0.01400LNaOH

1L

0.080molNaOH

1molNaOH1molOH- 0.00112

3. Subtraction gives what?No, not the numerical answer!

moles ofXS

acid (initially present for OUR

Eq. calculation).


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0.080M NaOH. (Ka of HOAc = 1.8 x 10-5).

Solution:

1. Find initial moles of Acid: HOAc + OH- p H2O + OAc- 1:1 ratio

_____ = mol HOAc0.02500L

1L

0.10mol HOAc 0.0025

2. find moles of OH-: which =s initial moles of OAC- anion.

_____________________________________= mol OH-0.01400LNaOH

1L

0.080molNaOH

1molNaOH

1molOH- 0.00112

3. Subtraction gives moles ofXS acid (initial for I.C.E.).

0.0025 - 0.00112 = 0.00138 mole acid INITIAL

4. and..we have 0.00112 mole OAc- INITIAL

for I.C.E.


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0.080M NaOH. (Ka of HOAc = 1.8 x 10-5).

Solution: F

ROM

PREVIOU

SS

LIDE.1. initial M of Acid: HOAc + OH- p H2O + OAc- 1:1 ratio

2. M of OH-: which =s initial M of OAC-

anion.

0.00138 mole/0.039L = 0.0354 M acid after RXN =INITIAL

0.00112 mole/0.039L = 0.0287 M OAc- after RXN =INITIAL**

[HOAc] [H+] + [OAc-]

I**

C

E

0.0354 0 0.0287

-x +x +x

0.0354-x x 0.0287+x

]HA[

]A][H[Ka

!

]x0354.0[]x0287.0][x[10x8.1 5-

!

Can we drop these?

x = 0.0000222 =[H+] pH = - Log [0.0000222] = 4.65


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Region 2 summary:

1. write balanced equation for initial weak/strong RXN.

2. find moles of acid

3. find moles of base

4. subtract to determine XS (and limiting reactant which equals

initial moles of salt formed)

5. Change XS to M and limiting Reactant to M

6. Write balanced Chemical Equilibrium Equation:

HA

H+

+ A

-

7. Use values from #5 a initial values in I.C.E. Chart

8. Work as Equilibrium problem.


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0

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8

10

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0 5 10 15 20 25 30 35 40 45 50



0.10 M NaOH

1.

2.

3.

4

Equivalence point


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Region 3. Theory and Calculations.

Mole of acid = moles of OH-

added. Equivalence pt.

HA + OH- p H2O + A- no XS of either

The titration curve indicates the pH to be above 7.

What causes this?

Hydrolysis of the A- anion: A- + H2O HA + OH-

Since this equilibrium involves OH- being formed,

it is a Kb problem.

]A[

]HA][OH[Kb

!


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17

Region 3 problem: Calculate the pH at the equivalence

point of a solution formed by the titration of 25.00mL of

0.10M acetic acid with 31.25mL of 0.080M NaOHKa(HOAc)=1.8 x 10

-5.

1. Find moles of Acid (which =s moles of base at Eq pt.)

25.00mL = mol HOac1000mL0.10mol HOAc 0.0025

2. Find mol of OH-

31.25mL __ = mol OH-

0.080mol NaOH1000mL

1molOH-

1mol NaOH0.0025

3. Since the # of moles are the same, this is at the Eq. Pt. and

therefore the initial moles ofOAc- ion = 0.0025 mol


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Region 3 problem: Calculate the pH at the equivalence

point of a solution formed by the titration of 25.00mL of

0.10M acetic acid with 31.25mL of 0.080M NaOHKa(HOAc)=1.8 x 10

-5.

moles of Acid = moles of base = 0.0025 = Eq. Pt.

New Volume = 0.05625L..and [OAc-] = 0.0025/0.05625=0.0444M

Question: What equilibrium(ia) gives rise to the pH?1. hydrolysis of the acetate anion:

OAc- + H2O HOAc + OH-

[OAc-] [HOAc] [OH

-]

IC

E

0.0444 0 0-x +x +x

0.0444-x x x

]OAc[

]HOAc][OH[Kb

!

]x0444.0[

]x[K

2

b

!


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Region 3 problem: Calculate the pH at the equivalence point of a solution

formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M

NaOH Ka(HOAc)=1.8 x 10-5.

moles of Acid = moles of base = 0.0025 = Eq. Pt.New Volume = 0.05625L...... and [OAc-] =0.0444M

Question: What equilibrium(ia) gives rise to the pH?

1. hydrolysis of the acetate anion:

OAc-

+ H2O HOAc + OH-

OAc- HOAc OH-

I

C

E

0.0444 0 0

-x +x +x

0.0444-x x x

]OAc[

]HOAc][OH[Kb

!

]x0444.0[

]x[K

2

b

!

Can we drop this x?

Also: KaKb=Kw=1.0 x 10-14 @25oC so Kb= 5.56 x 10

-10

How do we find Kb?


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Region 3 problem: Calculate the pH at the equivalence point of a solution

formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M

NaOH Ka(HOAc)=1.8 x 10-5.

moles of Acid = moles of base = 0.0025 = Eq. Pt.New Volume = 0.05625L...... and [OAc-] =0.0444M

Question: What equilibrium(ia) gives rise to the pH?

1. hydrolysis of the acetate anion:

OAc-

+ H2O HOAc + OH-

OAc- HOAc OH-

I

C

E

0.0444 0 0

-x +x +x

0.0444-x x x

]OAc[

]HOAc][OH[Kb

!

]x0444.0[

]x[10x56.5

2-10

!

x = 4.97 x 10-6 = [OH-]

pOH = 5.30

pH =14 - 5.30 = 8.70


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0

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8

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0 5 10 15 20 25 30 35 40 45 50



0.10 M NaOH

1.

2.

3.

4 XS base


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Region 4 theory and Calculations:

All the acid has been neutralized and XS base has been added

HA + OH- p H2O + A- + XSOH-

What is the dominating factor that controls the pH?

The XS strong base (OH-)

Calculations:

Find moles of XS OH- and then use the new

volume to find [OH-] and then the pOH and pH.


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Region 4 problem: Calculate the pH of a solution formed

by the titration of 25.00mL of 0.10M acetic acid with

40.00mL of 0.080M NaOH. Ka(HOAc)=1.8 x 10-5.

25.00mL = mol HOAc

1. Find moles of acid:

1000mL

0.10mol HOAc

2. Find moles of OH-:

40.00mL = mol OH-1000mL0.080molOH-

0.0025

0.0032

3. Subtract to find XS:

0.0032 - 0.0025 = 0.0007 mole XS OH-

4. Find OH- concentration, pOH, and pH:pOH = -Log[0.0007/0.06500] = 1.97 and pH = 12.03


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0

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4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50

Volume of Strong Base Added

equivalence pointbuffer region note volume ratiobetween these points

1

23

4

[H+]=[A-]

[H+]{[A-]

pOH=-Log[XS OH-

]

[OH-]=[HA]

]HA[

]A][H[K

a

!

]HA[

]A][H[K

a

! ]A[

]HA][OH[K

b

!

Approximations:

STUDY FORQUIZ


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Titration Curves for Three Different Weak Acids

pH

mL ofStrong Base Added

Ka = 1.4 x 10-5

Ka = 1.4 x 10-6

Ka = 1.4 x 10-7


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CO32- + H+ p HCO3

-

HCO3- + H+ p HCO3


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0

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10

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0 5 10 15 20 25 30 35 40 45 50



0.10 M NaOH

-

-

!

stotalLiter

molOHmolHA

stotalLiter

moleOH]H[

Ka

Region 2:acid + basepXS acid + saltHA+NaOH pHA(XS)+NaA

0

2

0

2

a]HA[

x

x]HA[

x

]HA[

]A][H[K $

$!

Region 1: only acid (HA)

-

$

!

stotalLiter

eHAinitialmol

x

]HA[]A[

]HA][OH[K

2

b

Region 3: mol acid=mol base

hydrolysis of the salt anion

A- + H2OHA + OH-

Region 4: etotalvolummolHAmolesOH

]OH[

!

pOH=-Log[OH-]


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0

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4

6

8

10

12

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0 5 10 15 20 25 30 35 40 45 50



-

-

!

stotalLiter

molOHmolHA

stotalLiter

moleOH]H[

Ka

Region 2:

acid + basepXSacid + saltHA+NaOHpHA(XS)+NaA

0

2

0

2

a]HA[

x

x]HA[

x

]HA[

]A][H[K $

$!

Region 1: only acid (HA)

-

$

!

stotalLiter

eHAinitialmol

x

]HA[]A[

]HA][OH[K

2

b

Region 3: mol acid=mol base

hydrolysis of the salt anion

A- + H2OHA + OH-

Region 4: XS base

etotalvolum

molHAmolesOH]OH[

!

pOH=-Log[OH-]

weak acid calc

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