Transcript Lecture on chemistry solutions

Basic Chemical Calculations,
Determining Chlorine Dose in
Waterworks Operation
Math for Water Technology
MTH 082
Lecture 1
Handout on water chemistry, Basic Science Chemistry Ch 3,4,7
Chapter 3 and 12- Math for Water Technology Operators
Lbs/day formula, Dose Demand Residual
Week Reading
2-3 assignment:
Objectives
Chapter 3 and 12- Math for Water Technology Operators
Chemistry Chapter 3,4 and Chapter 7 Chemistry Chemical Dosage Problems
(Basic Science Concepts and Applications)
1. Review Temperature
2. Learn to calculate basic chemical solutions
3. Understand new formulas for liquid and solid
chlorine application
Temperature Conversions
Celsius to Fahrenheit
1. Begin by multiplying the Celsius temperature by 9.
2. Divide the answer by 5.
3. Now add 32.
oF=
oC)
(9 *
5
+ 32
oF=
Convert 17oC to Fahrenheit
(9 *17)+32=62.6oF=
63oF
5
Fahrenheit to Celsius
1. Begin by subtracting 32 from the Fahrenheit #.
2. Divide the answer by 9.
4. Then multiply that answer by 5.
oC=
5 * (oF – 32)
9
Convert 451oF to degrees Celsius
oC=
5* (oF -32)=232.7oC= 233oC
9
Convert 88oF to oC?
o
88 F
• Given
• Formula: oC= 5
• Solve:
* (oF – 32)
9
oC= 5 * (88-32)/9
oC=
31
5 * (oF – 32)
9
oC= 5 * (88-32)/9
oC=
oC=
31
100%
31 OC
67 OC
17 OC
O
C
0%
17
O
C
67
O
C
0%
31
1.
2.
3.
Convert 16oF to oC?
o
16 F
• Given
• Formula: oC= 5
• Solve:
* (oF – 32)
9
oC= 5 * (16-32)/9
oC=
-9
5 * (oF – 32)
9
oC= 5 * (16-32)/9
oC=
oC=
-9
96%
-23 OC
-9 OC
26 OC
O
C
26
O
C
0%
-9
3
O
C
4%
-2
1.
2.
3.
Convert 35oC to oF?
• Given
• Formula:
• Solve:
35 oC
oF= (9 * oC) + 32
5
oF=
(9 * oC) + 32
5
oF= (9 * 35oC) + 32
5
oF=
95 oF
100%
57 OF
51 OF
95 OF
35 OF
O
F
35
O
F
0%
95
O
F
0%
51
O
F
0%
57
1.
2.
3.
4.
Solutions/Problems
• Mole (mol): chemical mass unit, defined to be
6.022 x 1023 molecules, atoms, or some other unit.
A mole of a substance is a number of grams of
that substance, where the number equals the
substances molecular weight.
• Molarity (mol/kg, molal, or M) denotes the number
of moles of a given substance per liter of solvent
Molarity (M)–
Moles of solute
Liters of solution
Molarity formula
• Grams=(formula weight, grams/mole)(liters)(M moles/liter)
• M (moles/liter) = _________grams____________
(formula weight, grams/mole)(final volume, liters)
Solutions/Problems
• Normality: a measure of concentration: it is equal to the
number of gram equivalents of a solute per liter of solution.
Depends on the valence or charge (old way)
Normality formula
• Grams=(equival. weight, grams/equival.)(liters)(N, equivalent./liter)
• N (equivl./liter) =
__________Grams ___________
(equival. weight, grams/equival.)(final volume, liters)
Formulas
• Percent Strength by Weight:
Weight of solute X 100
Weight of solution
Molarity formula
• Grams=(formula weight, grams/mole)(liters)(M moles/liter)
Normality Formula
• Grams=(equival. weight,
grams/equival.)(liters)(N, equivalent./liter)
_________ is defined as the
number of equivalents of solute
dissolved in one liter of solution.
Normality
Molarity
Alkalinity
Acidity
20%
ty
ci
di
A
lk
a
lin
ity
5%
A
ol
ar
ity
M
or
m
al
ity
10%
N
1.
2.
3.
4.
65%
The three most commonly used
coagulants in water treatment
are:
1. Aluminum hydroxide, lime and
sodium hydroxide
2. Aluminum sulfate, ferric chloride,
and ferrous sulfate
3. Lime, sodium hydroxide, and
chlorine
4. Soda, lime and chlorine
59%
36%
5%
So
..
an
d.
da
,l
im
e
so
di
um
..
m
e,
Li
lu
m
in
um
su
lfa
.
dr
o.
..
hy
A
um
lu
m
in
A
h.
..
0%
A chemical commonly used for
coagulation in water treatment
is:
Chlorine
Soda ash
Alum
Copper sulfate
86%
10%
su
lfa
te
lu
m
A
da
as
h
0%
C
op
pe
r
So
hl
or
in
e
5%
C
1.
2.
3.
4.
The chemical symbol for the
most common coagulant used in
water treatment, aluminum
sulfate (alum), is:
Al2(OH)6
Fe2(SO4)3
NH3(OH)7
Al2(SO4)3
100%
SO
l2
(
A
N
H
3(
O
H
3
2(
SO
4)
4)
3
0%
)7
0%
Fe
l2
(
O
H
)6
0%
A
1.
2.
3.
4.
Molecular Weights
• Step One: Determine how many atoms of
each different element are in the formula.
• Step Two: Look up the atomic weight of
each element in a periodic table.
• Step Three: Multiply step one times step two
for each element.
• Step Four: Add the results of step three
together and round off as necessary.
Solutions/Problems
Lime calcium oxide (CaO):
1 atom of calcium= 40 grams
1 atom of oxygen= 16 grams
Formula weight =
56 grams in 1 mole
So a .10 mole solution would contain how
many grams?
(0.10)(56 grams)= 5.6 grams
Determine the molar mass of ALUM
chemical formula Al2(SO4)3?
• Given
• Formula
• Solve:
2 Al, 3 S, 12 O FIND Al=26.98 g, S=32.06 g, O = 16 g
MM= Al 2(26.98 g) + S 3(32.06g)+ O12(16g)
MM= 2Al + 3S+ 12O
MM= 53.96 g + 96.18 g + 192 g
MM
Al2(SO4)3 =342.14 g
100%
14
g
2.
6
34
19
8.
2
g
0%
g
0%
g
0%
16
75 g
198.2 g
166 g
342.14 g
75
1.
2.
3.
4.
Determine the molar mass of sodium
hexametaphosphate chemical formula
(NaPO3)6 ?
• Given
• Formula
• Solve:
6 NA, 6 P, 18 O FIND Na=22.98 g, P=30.97 g, O = 16 g
MM= Na 6( 22.98g) + P 6(30.9g)+ O18(16g)
MM= 6Na + 6P+ 18O
MM= 137.88 g + 185.4 g + 288 g
MM
(NaPO3)6 =611.28 g
100%
g
0
.2
38
0.
70
61
1.
28
g
0%
g
0%
g
0%
50
70 g
611.28 g
700.38 g
50.20 g
70
1.
2.
3.
4.
M1V1=M2V2
M1V1 = M2V2
1 is starting (concentrated
conditions)
2 is ending (dilute conditions)
If we have 1 L of 3 M HCl, what is
M if we dilute acid to 6 L?
• Given
• Formula
• Solve:
M1 = 3 mol/L or 3 M, V1 = 1 L, V2 = 6 L
M1V1 = M2V2, M1V1/V2 = M2
M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M
V1 = 1 L
M1 = 3 M
M1V1 = 3 mol
V2 = 6 L
M2 = 0.5 M
M
20
5
M
0%
0.
2
M2V2 = 3 mol
0%
M
0%
M
19 M
2M
0.5 M
20 M
19
1.
2.
3.
4.
100%
What volume of 0.5 M HCl can be
prepared from 1 L of 12 M HCl?
• Given
• Formula
• Solve:
M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 L
M1V1 = M2V2, M1V1/M2 = V2
V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L
75%
25%
L
0%
1
L
12
24
L
0%
L
6L
24 L
12 L
1L
6
1.
2.
3.
4.
How many mL of a 14 M stock
solution must be used to make
250 mL of a 1.75 M solution?
• Given
M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL
• Formula V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M)
• Solve:
V1 = 0.03125 L = 31.25 mL
90%
10%
m
L
76
5
m
00
20
5
31
.2
5
m
L
L
0%
m
L
0%
7.
437.5 mL
31.25 mL
2000 mL
765 mL
43
1.
2.
3.
4.
Chlorine Concentrations
1.Sodium hypochlorite
5 to 15% available chlorine
2. Calcium hypochlorite
65-70% available chlorine
3. Chlorine gas
100% available chlorine
Determining Cl Concentrations from
Hypochlorite dosage
1. Disinfection requires 280 lb/day chlorine. If CaOCl
(65% available Cl) is used how many lbs day are required
(%concentration)(X lb/day)= total lbs/day required
Rearrange:
(X lb /day) = (280 lbs) = 430.77 lb/d CaOCL
(.65)
Hypochlorite Solution Feed Rate
1. Actual Dose=Solution Feeder dose.
(mg/L chlorine)(MGD)(8.34)= (mg/L chlorine)(MGD)(8.34)
% Dry Strength of Solution
1. Dry chlorine
% Chlorine Strength= Chlorine (lbs) X 100
Solution(lbs)
2. Dry chlorine
% Chlorine Strength= Chlorine (lbs)
X 100
Water, Lbs + Chlorine(lbs)
3. Dry chlorine
% Chlorine Strength= Hypo(lbs)(% available Cl)
100
X 100
Water, Lbs+ hypo(lbs) (% available Cl)
100
% Liquid Strength of Solution
1. Liquid chlorine
Lbs Cl in liq. hypo= Lbs of chlorine in hypochlorite solution
2. Liquid chlorine
(liq. Hypo lbs)(%Strength liq. Hyo) = (hypo Sol lbs) (% Strength liq. hyp)
100
100
3. Liquid chlorine
(liq. Hypo gal)(8.34)(%Strength liq. Hyo)=(hypo Sol gal)(8.34)(%Strength liq. Hyp)
100
100
A chlorinator setting of 20 lbs of chlorine per 24 hrs
results in a residual of 0.4 mg/L. The chlorinator
setting is 25 lb per 24 hrs. The chlorine residual
increased to 0.5 mg/L at this new dosage rate. The
average flow being treated is 1.6 mgd. On the basis of
this data is the water being chlorinated beyond
breakpoint?
• Given Residual 1= 0.4 mg/L, residual 2=0.5 mg/L, 1.6 mgd, new 5 lb/day
• Formula Lbs/d incr= Dose( flow)(8.34 lb/g)
Act increase in residual=New residual-Old residual
• Solve:
67%
Actual increase in residual was 0.5mg/L-0.4mg/L=0.1 mg/L
Not being met!
s
yesExpected was 0.37 but the actual was 0.1.
no
ye
1.
2.
33%
no
Dose= Lb/day increase/flow 8.34 lb/g
5 lb/day/1.6 mgd(8.34 lb/d)=
Residual = 0.37 mg/L
Specific Gravity, LBS, Gallons,
Solution Strength
Lbs  (% chlorine
sol . strength )( full strength chlorine )( 8 . 34 lbs / gal )( specific g ravity )( gal )
( lbs )
Gallons 
(% chlorine
% chlorine
sol . strength ) (full strenght
chlorine)
( 8 . 34 lb / gal )( Specific Gravity )
( lbs )
Sol . Strength 
( 8 . 34 lb / gal )( full strength chlorine )( Specific Gravity )( gallons )
( lbs )
Specific Gravity 
(% chlorine
sol . strength )( full strength
chlorine)
( 8 . 34 lb / gal )( gallons )
Team Scores
Today’s objective: Review basic chemistry and solution
making as it pertains to the waterworks industry
Calculate the chemical dosage using the standard
“pounds formula” has been met?
85%
Strongly Agree
Agree
Disagree
Strongly Disagree
15%
0%
ag
re
e
gr
ee
Di
s
is
a
ly
St
ro
ng
A
gr
ee
D
ly
Ag
re
e
0%
St
ro
ng
1.
2.
3.
4.