Slide 1
1
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ALKALINITY
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Bicarbonate-carbonate
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Slide 2
2
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Alkalinity is…
…the measure of the ability of a water to neutralize an acid.
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Slide 3
3
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Acidity
Most natural waters are buffered as a result of a carbon
dioxide(air)-bicarbonate (limestone – CaCO3) buffer
system.
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What is a buffer?
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Slide 4
4
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Buffer
Mixture of an acid (or base) and its conjugate base (or
acid)
Think of chemical equilibrium as a see-saw:
CO2 + H2O ↔ H2CO3
H2CO3 ↔ HCO3- + H+
HCO3- ↔ CO32- + H+
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CO2 + H2O ↔ H2CO3 ↔ HCO3- + H+ ↔ CO32- + 2 H+
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You need to put 2 fat kids on the see-saw!
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Slide 5
5
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Buffer
CO2 + H2O ↔ H2CO3 ↔ HCO3- + H+ ↔ CO32- + 2 H+
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CO2
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CO32-
If you have a big heavy weight at both ends of the
equilibrium, a small addition of acid or base from an
outside source can’t change the pH very much.
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Slide 6
6
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Reporting Alkalinity
Alkalinity can be reported in several ways – ways which are
not completely chemically accurate.
Alkalinity as
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= ml titrant * Normality of acid * 50,000
mL sample
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Slide 7
7
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What’s Normality?
Normality is like Molarity with the stoichiometry added right
in.
Normality (N) = equivalent moles of solute
L
What’s “equivalent” mean? It means you consider the
reaction.
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Slide 8
8
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1.5 M HCl
1.5 M HCl
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What’s HCl?
It’s an acid. What’s the relevant part of the acid?
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H+
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HCl + OH-  H2O + Cl-
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Slide 9
9
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1.5 M HCl
1.5 M HCl
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Since 1 HCl reacts with 1 OH-, there is one
chemical equivalent per molecule.
1.5 mole HCl * 1 H+ equivalent = 1.5 N HCl
L
1 HCl
HCl + OH-  H2O + Cl-
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Slide 10
10
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1.5 M H2SO4
1.5 M H2SO4
What’s H2SO4?
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It’s an acid. What’s the relevant part of the acid?
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H+
H2SO4 + 2 OH-  2 H2O + SO42-
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Slide 11
11
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1.5 M H2SO4
1.5 M H2SO4
Since 1 H2SO4 reacts with 2 OH-, there are TWO
chemical equivalents per molecule.
1.5 mole H2SO4 * 2 H+ equivalent = 3.0 N H2SO4
L
1 H2SO4
H2SO4 + 2 OH-  2 H2O + SO42-
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Slide 12
12
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Everyman eats 2 cupcakes
I had 500 cupcakes, I only have 300 left, how many men
came through?
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Suppose it was really 50 really hungry women who ate 4
cupcakes each?
If I only care about cupcakes eaten, it doesn’t matter:
50 hungry women=100 men
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Slide 13
13
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Metal + polyatomic
CaCO3 (aq) → Ca2+ + CO32-
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The carbonate is the “basic part”: it’s a negative ion with
lots of oxygen, it’s going to like H+
CO32- + H+ → HCO3-
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What do you think about HCO3-?
ALSO A BASE!
HCO3- + H+ → H2CO3
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Slide 14
14
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Moles! Moles! Moles!
I titrate 50.00 mL of calcium carbonate solution using a 1.5
M H2SO4 solution. Equivalence (2nd endpoint) is reached
after addition of 32.65 mL of acid. What is the
concentration of calcium carbonate in the original sample
in mg/L?
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Slide 15
15
1st thing we need?
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Balanced Equation
CO32- + 2 H+  H2CO3
OR
CO32- + H+  HCO3HCO3- + H+  H2CO3
You can do it all in one step, or you can do it in two steps.
But you aren’t done until all the base is gone.
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Slide 16
16
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Moles! Moles! Moles!
1.5 moles H2SO4 * 0.03265 L = 0.048975 mol H2SO4
1L
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0.048975 mol H2SO4 * 2 mol H+ = 0.09795 mol H+
1 mol H2SO4
0.09795 mol H+ * 1 mol CO32- = 0.048975 mol CO322 mol H+
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0.048975 mol CO32- 1 mol CaCO3 = 0.048975 mol CaCO3
1 mol CO32-
0.048975 mol CaCO3* 100.09 g * 1000 mg = 98039 mg/L
0.050 L
1 mol CaCO3 1 g
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98039 mg CaCO3 EQUIVALENTS/L
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Slide 17
17
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Alkalinity as mg/L CaCO3
= ml titrant * Normality of acid * 50,000
mL sample
= 32.65 mL * 3.0 N H2SO4 * 50,000
50 mL
=97950 mg/L
The expression in the book (or lab) is just the Moles!
Moles! Moles! solved for you.
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But this is just the TOTAL ALKALINITY.
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There are actually different types.
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Slide 18
18
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A base is a base is a base
If you titrate a solution with multiple bases, can you
tell what reacts with what?
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Essentially, you have 3 different types of bases in
the system:
OH- (strong monoprotic base)
CO32- (weak diprotic base)
and HCO3- (weak monoprotic base)
All 3 can be neutralized by addition of a strong
acid.
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Slide 19
19
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Strong vs. Weak
Strong – completely dissociates in water (or other solvent)
Weak – partially dissociates in water
HCl
100 molecules
= H+ + Cl100 molecules + 100
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HC2H3O2 (acetic acid) = H+ + C2H3O2100 molecules
2
+ 2 molecules
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Slide 20
20
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Different pH at endpoints
H+ + OH-  H2O (neutral)
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H+ + CO32-  HCO3(slightly basic)
HCO3- + H+  H2CO3
(acidic)
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HCO3- + H+  H2CO3
(acidic)
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Slide 21
21
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Different species – Different endpoints
H+ + OH-  H2O (neutral – EP1)
H+ + CO32-  HCO3(slightly basic – EP1)
HCO3- + H+  H2CO3
(acidic – EP2)
HCO3- + H+  H2CO3
(acidic – EP2)
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Slide 22
22
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Different endpoints at different pH values
EP1 – neutral to slightly basic (pH approximately 8)
EP2 – acidic (pH approximately 5)
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The key is that everything that happens at EP1 (endpoint
#1) happens BEFORE anything happens at EP2.
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Slide 23
23
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Different species – Different endpoints
H+ + OH-  H2O (neutral – EP1)
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H+ + CO32-  HCO3(slightly basic – EP1)
HCO3- + H+  H2CO3
(acidic – EP2)
HCO3- + H+  H2CO3
(acidic – EP2)
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Slide 24
24
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IT’S NOT WHAT IT IS…
…IT IS WHAT IT DOES!
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Bases accept protons from acids (neutralize acids). That’s
all they do.
So I refer to the amount of base based on the amount of
acid it neutralizes.
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Slide 25
25
3 “bases” – how much
H+
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do they eat?
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OH-
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Base
CO32-
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Slide 26
26
3 “bases” – how much
H+
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H+
OH-
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do they eat?
H+
H+
H+
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Base
H+
H+
CO32-
H+
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H+
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Slide 27
27
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1 CO32- = 2 OH- = 2/5 “Base”
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H+
OH-
H+
H+
H+
Base
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H+
H+
CO32-
H+
H+
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Slide 28
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28
So if I have 1.25 mol OH-:
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If I have 1.25 mol “Base”
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It’s not about what you really have. It’s about how much acid it absorbs relative
to carbonate!
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Slide 29
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29
My 3 types of base and their EP
EP 1
OH-
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EP 2
H+
H+
HCO3
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-
H+
CO32-
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H+
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Slide 30
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30
My titration
EP 1
OH-
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EP 2
H+
H+
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HCO3-
H+
EP2
CO32-
H+
EP1
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Slide 31
31
Suppose you have CO32- and OH- What
does EP1 look like?
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OH-
OH-
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CO32-
CO32CO32-
OH-
OH-
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CO32CO32-
CO32-
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Slide 32
32
Suppose you have CO32- and OH- What
does EP1 look like?
6 H+ to CO324 H+ to OH-
H+
H+
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H+
CO32-
CO32OH- H+
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CO32-
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OH- H+
H+
H+
OH-
H+
H+
OH- H+
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CO32CO32-
CO32-
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Slide 33
33
Suppose you have CO32- and OH- What
does EP2 look like?
6 MORE H+ to CO32EP1 10 H+ = 6 + 4
EP2 6 H+
H+
H+
CO32- H+
OH- H+
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H+
CO32- H+
CO32-
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H+
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OH- H+
H+
H+
OH-
H+
H+
OH- H+
CO32CO32-
H+
CO32-
H+
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H+
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Slide 34
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34
Suppose you have CO32- and HCO3EP1?
6 H+ to CO320 H+ to HCO3-
H+
H+
CO32-
CO32-
HCO3-
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H+
CO32-
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HCO3-
H+
H+
HCO3-
H+
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CO32-
HCO3-
CO32-
CO32-
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Slide 35
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35
Suppose you have CO32- and HCO3What does EP2 look like?
4 H+ to HCO36 MORE H+ to CO32EP1 6 H+
EP2 10 H+ = 6 + 4
H+
H+
CO32- H+
H+
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H+
CO32- H+
CO32-
H+
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H+
HCO3-
HCO3-
H+
H+
H+
HCO3-
H+
H+
HCO3-
CO32CO32-
H+
CO32-
H+
H+
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Slide 36
36
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I titrate a 25.00 mL water sample with 0.1250 M HCl. I
achieve the first endpoint at 22.5 mL of HCl and the
second after addition of another 27.6 mL of HCl.
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What can I conclude?
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Example
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Slide 37
37
Possible
EP1 vol
EP2 vol
Compare
CO32-
X
X
EP1 = EP2
HCO3-
0
Y
EP1 = 0
OH-
z
0
EP2 =0
CO32HCO3-
x
(x+y)
EP1<EP2
EP1 not 0
CO32-
(x+z)
X
OHHCO3OH-
???
????
EP1>EP2
EP2 not 0
????
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Slide 38
38
OH- is a strong base.
HCO3-
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is a weak acid
If I have more OH- than HCO3-, it completely
neutralizes it and I just have OHIf I have more HCO3- than OH-, then it partially
neutralizes it and I detect only HCO3-
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Slide 39
39
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Example
I titrate a 25.00 mL water sample with 0.1250 M HCl. I
achieve the first endpoint at 22.5 mL of HCl and the
second after addition of another 27.6 mL of HCl.
What can I conclude?
Carbonate and bicarbonate are both present.
How much?
0.1250 M HCl * 0.0225 L HCl = 2.813x10-3 mol HCl
2.813x10-3 moles CO320.1250 M HCl * (0.0276 – 0.0225 L HCl) = 6.375x10-4 H+
6.375x10-4 moles HCO3-
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Slide 40
40
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Units! Units! Units!
Carbonate and bicarbonate are usually measured as “mg equivalent CaCO 3/L”
So…
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Slide 41
41
What about the
CO32- + 2 H+ = H2CO3
HCO3- + H+ = H2CO3
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HCO3-?
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Slide 42
42
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Total Alkalinity.
I titrate a 25.00 mL water sample with 0.1250 M HCl. I achieve the first
endpoint at 22.5 mL of HCl and the second after addition of another 27.6
mL of HCl. What is the total alkalinity?
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Assume the second endpoint is reached and it was all CaCO3 in the
sample.
22.5 mL + 27.6 mL = 50.1 mL total HCl
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Slide 43
43
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Notice…
Total alkalinity = 12,535 mg CaCO3/L
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Carbonate alkalinity = 11, 260 mg CaCO3/L
Bicarbonate alkalinity = 1276 mg CaCO3/L
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11,260 + 1276 = 12536 mg CaCO3/L!!!!
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Slide 44
44
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Example
I titrate a water sample with 0.1250 M HCl. I
achieve the first endpoint at 22.5 mL of HCl and
the second after addition of another 27.6 mL of
HCl.
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What can I conclude?
Carbonate and bicarbonate are both present.
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Is this really true?
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Slide 45
45
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Example
I titrate a water sample with 0.1250 M HCl. I
achieve the first endpoint at 22.5 mL of HCl and
the second after addition of another 27.6 mL of
HCl.
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Carbonate and bicarbonate are both present.
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Is this really true?
No – any chemical species that behaves like
carbonate or like bicarbonate will look
identical!!!!!!!
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Slide 46
46
To be totally accurate, I should quote the levels as:
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“Bicarbonate and chemical equivalents”
“Carbonate and chemical equivalents”
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Slide 47
47
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Example
I titrate a 50.00 mL water sample with 0.1250 M
HCl. I achieve the first endpoint at 22.5 mL of
HCl and the second after addition of another 19.6
mL of HCl. What is the total alkalinity in mg
CaCO3/L?
What can I conclude about the species present?
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Slide 48
48
Possible
EP1 vol
EP2 vol
Compare
CO32-
X
X
EP1 = EP2
HCO3-
0
Y
EP1 = 0
OH-
z
0
EP2 =0
CO32HCO3-
x
(x+y)
EP1<EP2
EP1 not 0
CO32OHHCO3OH-
(x+z)
X
???
????
EP1>EP2
EP2 not 0
????
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Slide 49
49
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Example
I titrate a 50.00 mL water sample with 0.1250 M
HCl. I achieve the first endpoint at 22.5 mL of
HCl and the second after addition of another 19.6
mL of HCl. What is the total alkalinity in mg
CaCO3/L?
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What can I conclude?
Carbonate and OH- are both present.
BUT if I only care about total alkalinity I just
ASSUME it is all CaCO3!!!!
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Slide 50
50
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Total alkalinity:
22.5 mL + 19.6 mL = 42.1 mL
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Slide 51
51
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Hydroxide alkalinity
22.5 mL EP1 – 19.6 mL EP2 = 2.9 mL excess
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0.1250 M HCl (0.0029 L) = moles H+ 1 mol CaCO3 = mol CaCO3
2 mol H+
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