Post on 26-Dec-2015
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).