Continuation of Acid-Base Chemistry

CALULATE THE pH OF A STRONG ACID

Compute the pH and equilibrium concentrations of all species in a 2 x 10-4 M solution of HCl.

1) Species: H+, Cl-, HCl0, OH-

2) Mass action laws:

14

2

10][

]][[

OH

OHHKw

30

10][

]][[

HCl

ClHK A

3) Mass balance: [HCl0] + [Cl-] = 2 x 10-4 M

4) Charge balance: [H+] = [Cl-] + [OH-]

Assumptions: HCl is a very strong acid so

[H+] >> [OH-] and [Cl-] >> [HCl0]

Now the only source of H+ and Cl- are the dissociation of HCl, so

[H+] = [Cl-]

(this is also apparent from the charge balance)

Thus, pH = - log (2 x 10-4) = 3.70, and [Cl-] = 2 x 10-4 M.

[OH-] = Kw/[H+] = 10-14/2 x 10-4 = 5 x 10-11 M

30

10][

]][[

HCl

ClHK A Mx

xHCl 11

3

240 104

10

)102(][

MONOPROTIC ACIDS

What are the pH and concentrations of all species in a 0.1 mol L-1 HF solution?

1) Write out important species:

2) Write out all independent reactions and their equilibrium constants

3) Write out mass-balance expressions.

4) Write out the charge-balance expression.

5) Make reasonable assumptions.

MONOPROTIC ACIDS

What are the pH and concentrations of all species in a 0.1 mol L-1 HF solution?

1) Write out important species: H+, OH-, HF0, F-.

2) Write out all independent reactions and their equilibrium constants:

HF0 H+ + F-

H2O(l) H+ + OH-

][

]][[10

02.3

HF

FHK

]][[10 14 OHHK

3) Write out mass-balance expressions:

0.1 mol L-1 = F = [F-] + [HF0]

4) Write out the charge-balance expression:

[H+] = [F-] + [OH-]

5) Make reasonable assumptions:HF is an acid, so [H+] >> [OH-] the charge-balance

becomes

[H+] [F-] = X

and the mass-balance becomes:

[HF0] = 0.1 - X

X

XK

1.010

22.3 01010 22.32.4 XX

6) Solve quadratic equation:

a

acbbX

2

42

a = -1; b = -10-3.2 = -6.31x10-4; c = 10-4.2 = 6.31x10-5

12

1031.6141031.61031.6 5244

xxx

X

X1 = -0.00825; X2 = 0.00765

[H+] = [F-] = 7.65x10-3 mol L-1; pH = -log [H+] = 2.12

[HF0] = 0.1 - 0.00765 = 0.0924 mol L-1

123

14

10318.11065.7

10]OH[

x

x

7) Check assumption: 1.318x10-12 << 7.65x10-3, so [OH-] << [H+].

What if we assumed [HF0] >> [F-], i.e., [HF0] 0.1? This might be valid because HF is a weak acid.

10-3.2 = X2/0.1X2 = 10-4.2

X = 10-2.1 = 0.00794[H+] = [F-] = 7.94x10-3 mol L-1; pH = 2.10

[HF0] = 0.1 - 0.00794 = 0.092 mol L-1

The above answer is only 8% different from 0.1. It seems in any case where KA < 10-3.2, the above assumption should be good!

POLYPROTIC ACID

What is the pH and concentration of all species in a 0.1 mol L-1 solution of H3PO4?

1) Species: H+, OH-, H3PO40, H2PO4

-, HPO42-,

PO43-

2) Mass action expressions:

H3PO40 H2PO4

- + H+

H2PO4- HPO4

2- + H+

HPO42- PO4

3- + H+

][

]][[10

043

421.21 POH

HPOHK

][

]][[10

42

240.7

2

POH

HHPOK

][

]][[10

24

344.12

3

HPO

HPOK

H2O(l) H+ + OH-

3) Mass-balance

0.1 mol L-1 = [H3PO40] + [H2PO4

-] + [HPO42-] + [PO4

3-]

4) Charge-balance

[H+] = [H2PO4-] + 2[HPO4

2-] + 3[PO43-] + [OH-]

5) Assumptionsa) Because H3PO4

0 is an acid [H+] >> [OH-]

b) Because H2PO4- and HPO4

2- are very weak acids and H3PO4

0 is only moderately weak:

[H3PO40] > [H2PO4

-] >> [HPO42-] >> [PO4

3-]

so, 0.1 = [H3PO40] + [H2PO4

-]

and [H+] = [H2PO4-] = X.

]][[10 14 OHHK

10-3.1 - 10-2.1X - X2 = 0

X

XK

1.010

21.2

1

12

11041010 1.321.21.2

X

X1 = 0.0245; X2 = -0.0324

[H+] = [H2PO4-] = 0.0245 mol L-1; pH = 1.61

[H3PO40] = 0.1 - 0.0245 = 0.0755 mol L-1

][

]][[10

42

240.7

2

POH

HHPOK

So [HPO42-] = 10-7.0 mol L-1

[PO43-] = 10-17.79 = 1.62x10-18 mol L-1

[OH-] = 10-14/10-1.61 = 10-12.39 = 4.07x10-13 mol L-1

][

]][[10

24

344.12

3

HPO

HPOK

0.7

61.1344.12

3 10

10][10

POK

GRAPHICAL APPROACH TO EQUILIBRIUM CALCULATIONS

Consider the monoprotic acid HA:

][

]][[10

05.5

HA

AHK A

CT = 10-3 = [HA0] + [A-]; so [A-] = CT - [HA0]

KA[HA0] = [H+][A-]

KA[HA0] = [H+](CT - [HA0])

KA[HA0] = [H+]CT - [H+][HA0]

KA[HA0] + [H+][HA0] = [H+]CT

][

][][ 0

HK

HCHA

A

T

CTKA - KA[A-] = [H+][A-]

CTKA = [A-]([H+] + KA)

][

]][[

AC

AHK

TA

A

AT

KH

KCA

][][

1) At pH < pKA, [H+] >> KA so [H+] + KA [H+]

[HA0] = CT([H+]/[H+]) = CT

log [HA0] = log CT

[A-] = CTKA/[H+]

log [A-] = log CT - pKA + pH

CT - [A-] = [HA]

2) pH = pKA; [H+] = KA so [H+] + KA = 2[H+]

[HA0] = CT[H+]/(2[H+]) = CT/2

log [HA0] = log CT - log 2 = log CT - 0.301

[A-] = CT [H+]/(2[H+]) = CT/2

log [A-] = log CT - log 2 = log CT - 0.301

3) pH > pKA; [H+] << KA so KA+ [H+] KA

[HA0] = CT[H+]/KA

log [HA0] = log CT + pKA - pH

[A-] = CTKA/KA = CT

log [A-] = CT

pH0 2 4 6 8 10 12 14

log

[i]

-10

-8

-6

-4

-2

HA0A-

OH-

H+

10-3 M NaAc

10-3 M HA

Speciation diagram for HA with pKA = 5.5 and CT = 10-3

SPECIATION DIAGRAM FOR A DIPROTIC SYSTEM

Consider H2S with pK1 = 7.0, pK2 = 13.0

ST = 10-3 M = [H2S0] + [HS-] + [S2-]

][1

][][

2

1

HK

KH

SHS T

1][][

][

221

22

KH

KKH

SS T

2211

02

][][1

][

HKK

HKS

SH T

1) pH < pK1 < pK2; [H+] > K1 > K2

TT S

HKK

HKS

SH

2211

02

][][1

][ log [H2S0] = log ST

1

2

1

][][

1][

][

KH

S

HK

KH

SHS TT

log [HS-] = log (STK1) + pH

21

2

221

22

][1

][][][

KKH

S

KH

KKH

SS TT

log [S2-] = log (STK2K1) + 2pH

2) pH = pK1 < pK2; [H+] = K1 > K2

2][][

1][

2211

02

TT S

HKK

HKS

SH

log [H2S0] = log ST - 0.301

2][

1][

][2

1

TT S

HK

KH

SHS

log [HS-] = log ST - 0.301

2221

22

][21

][][][

KH

S

KH

KKH

SS TT

log [S2-] = log (STK2/2) + pH

3) pK1 < pH < pK2; K1 > [H+] > K2

][][][1

][1

2211

02

HKS

HKK

HKS

SH TT log [H2S0] = log (ST /K1) - pH

TT S

HK

KH

SHS

][1

][][

2

1

log [HS-] = log ST

2221

22

][1

][][][

KH

S

KH

KKH

SS TT

log [S2-] = log (STK2) + pH

4) pK1 < pK2 = pH; K1 > [H+] = K2

][2

][][1

][1

2211

02

HKS

HKK

HKS

SH TT log [H2S0] = log (ST/2K1) - pH

2][

1][

][2

1

TT S

HK

KH

SHS

log [HS-] = log ST - 0.301

21

][][][

221

22 TT S

KH

KKH

SS

log [S2-] = log ST - 0.301

5) pK1 < pK2 < pH; K1 > K2 > [H+]

221

2211

02

][][][1

][

HKKS

HKK

HKS

SH TT log [H2S0] = log (ST /K1K2) - 2pH

][][1

][][

22

1

HKS

HK

KH

SHS TT log [HS-] = log (ST /K2) - pH

TT S

KH

KKH

SS

1][][

][

221

22

log [S2-] = log ST

pH0 2 4 6 8 10 12 14

log a i

-12

-10

-8

-6

-4

-2H2S

0HS-

S2-

H+OH-

7.0 13.0

Bjerrum plot showing the activities of reduced sulfur species as a function of pH for a value of total reduced sulfur of 10-3 mol L-1.

Easier approach:10-7 = [HS-] [H+] / H2S[HS-] / [H2S] = 10-7 / [H+]

10-12.9 = [S2-] [H+] / HS-

[S2-] / [HS] = 10-12.9 / [H+]

The lower the pH, the higher [H+], the smaller 10-7 / [H+], the smaller [HS-] / [H2S] and the higher [H2S]

The higher the pH, the lower [H+], the higher 10-12.9 / [H+], the higher [S2-] / [HS-] and the higher [S2-]

This type of diagram is called a Bjerrum diagram,after its originator. Although the previous examples were calculated in detail, it is possible to sketch the relationships based only on the Ks of the dissociation reactions. These will show the pH ranges of the dominant species; i.e. the species that is controlling the chemistry, except for a small region in the vicinity of the K value. This is illustrated in the following slide.

THE CO2-H2O SYSTEM - I

Carbonic acid is a weak acid of great importance in natural waters. The first step in its formation is the dissolution of CO2(g) in water according to:

CO2(g) CO2(aq)At equilibrium we have:

Once in solution, CO2(aq) reacts with water to form carbonic acid:

CO2(aq) + H2O(l) H2CO30

2

2

2

CO

COCO p

aK

THE CO2-H2O SYSTEM - IIIn practice, CO2(aq) and H2CO3

0 are combined and this combination is denoted as H2CO3*. It’s formation is dictated by the reaction:

CO2(g) + H2O(l) H2CO3*For which the equilibrium constant at 25°C is:

Most of the dissolved CO2 is actually present as CO2(aq); only a small amount is actually present as true carbonic acid H2CO3

0.

46.1* 102

32

2

CO

COHCO p

aK

BJERRUM PLOTS• These are used for closed systems with a specified

total carbonate concentration. They plot the log of the concentrations of various species in the system as a function of pH.

• The species in the CO2-H2O system: H2CO3*, HCO3-,

CO32-, H+, and OH-.

• At each pK value, conjugate acid-base pairs have equal concentrations.

• At pH < pK1, H2CO3* is predominant, and accounts for nearly 100% of total carbonate.

• At pK1 < pH < pK2, HCO3- is predominant, and

accounts for nearly 100% of total carbonate.• At pH > pK2, CO3

2- is predominant.

pH0 2 4 6 8 10 12 14

log

ai

-8

-7

-6

-5

-4

-3

-2

6.35 10.33H2CO3* HCO3- CO3

2-

H+

OH-

Common pHrange in nature

Bjerrum plot showing the activities of inorganic carbon species as a function of pH for a value of total inorganic carbon of 10-3 mol L-1.

In most natural waters, bicarbonate is the dominant carbonate species!

A Bjerrum plot shows the relative importance of the various species in an acid-base system under closed conditions (i.e., the total concentration of all species is constant). For example, for the CO2-H2O system, a Bjerrum plot shows the concentrations of H2CO3*, HCO3

-, CO32-, H+, and OH-, under the condition that the

sum of the concentrations of H2CO3*, HCO3- and CO3

2- is constant. The Bjerrum plot is constructed based partially on the concepts discussed in slide 6. That is: 1) At each pK value, conjugate acid-base pairs have equal concentrations; 2) At pH < pK1, H2CO3* is predominant, and accounts for nearly 100% of total carbonate; 3) At pK1 < pH < pK2, HCO3

- is predominant, and accounts for nearly 100% of total carbonate; and 4) At pH > pK2, CO3

2- is predominant. The Bjerrum plot is also constructed assuming that activity coefficients can be neglected. When pH < pK1, and H2CO3* is predominant, the concentrations/activities of the other carbonate species can be derived by rearranging the mass-action expressions for the dissociation reactions, and the mass-balance constraint that the sum of the concentrations of H2CO3*, HCO3

- and CO32- is constant. For example, rearranging

the equation given in the notes to slide 6 yields:log aHCO3

- = pH - pK1 + log aH2CO3*

At pH < pK1, the concentration of H2CO3* is approximately equal to the total concentration of all carbonate species, and is hence, approximately constant. Thus, the equation shows that, at pH < pK1, the concentration of bicarbonate increases one log unit for each unit increase in pH. Similar equations can be derived for all the carbonate species in each of the pH ranges of the diagram. For more details, consult Faure (1998) Principles and Applications of Geochemistry, Prentice-Hall (Chapter 9, pp. 123-124).