1-Concentration Units S12

8/13/2019 1-Concentration Units S12

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1-Concentration Units_S12.doc

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CONCENTRATION UNITS (SJ, p.14-23)

1) mg/L - mg of substance (solute) dissolved in 1 liter of H2O (solvent), (mass/volume)

1 mg 1 mg 1 g 103L 1 g-------= -------- --------- --------- = -------

L L 103mg 1 m3 m3

1 g 1 g 1 lb 1 m3 10

6 gal 8.34 lb

------- = -------- ------------- --------------- ------------ = ----------m3 m3 453.59 g 264.173 gal 1 MG 1 MG

2) ppm - parts per million, mg of substance (solute) dissolved in 1 kg of H2O (solvent),(mass/mass)

Mass fraction, ppmm- the mass of constituentiper unit mass of solution, with units

such as mg/kg or g/g

- This is commonly used to express the concentration of constituents in a solid phase.

- For example, a concentration of 1 ppmmin a solid phase means that there is 1 mg of the

given species per kg of solid.

1 mg 10-3g 1 g 1 kg------- = --------- = -------- = ---------- = 1 ppmm

kg 103g 10

6g 10

6kg

1 g 10-6

g 1 g 1 kg------- = --------- = -------- = --------- = 1 ppmm

g g 106g 10

6kg

Notes: a. Exception, volume/volume in air pollution.

b. 1 liter of H2O = 1 kg in normal environment *Thus, 1 mg/L = 1 mg/kg = 1 ppm

1 ug/L = 1 ug/kg = 1 ppb

1 ng/L = 1 ng/kg = 1 ppt

* Strictly speaking ppm = mg/L only for pure water at 4oC. However, at room temperature, the

effect of temperature on fluid density is negligible up to about 7,000 ppm dissolved solids.

3) M, molarity, moles/L

- number of moles of substance (solute) dissolved in 1 liter of H2O

* mole - quantity of compound that has a wt in grams equal to molecular weight of the compound


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Example: If 49 g of H2SO4is dissolved in 1 liter of H2O, what is molarity of the H2SO4solution?

For H2SO4(atomic mass; H = 1, S = 32, O = 16):gram MW = (1 x 2) + 32 + (16 x 4) = 98 g/mol

49 g 1 mol 0.5 mol

M = -------- ---------- = ----------- = 0.5 M1 L 98 g L

4) Logarithmic units

- If chemical concentrations range over many orders of magnitude, it is convenient to express

their values in logarithmic units.

- By convention, a lowercase ppreceding a symbol is used to designate the negative, base-10

logarithm (log10) of the value associated with the symbol

Examples:

pC = log{M} e.g., pH = log {H+} where {activity}

pC = log[M] e.g., pH = log [H+] where [molarity]

Example (FE exam) The pH of a 0.001 M HCl solution is

(A) 1

(B) 3(C) 5

(D) 7

(Solution)

Since HCl is a strong acid, it dissociate completely in water: HCl H+ + Cl

-

pH = - log10[H+] = log10{1/[H

+]} = log10{1/0.001} = 3

Answer is (B).

Note: Molarity (M) is the number of gram-moles of solute per liter of solution. To calculate pH,

the ionic concentration of H+

ions in moles per liter is needed. This is equal to the molarity forHCl.


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5) Mole fraction

- the number of moles of a species ias a fraction of the total number of moles of all substances inthat phase.

Number of moles of a speciesi

Mole fraction = ---------------------------------------------------Total number of moles of all substances

in the phase

For a mixture of molecules 1, 2, 3, , the mole fraction of molecule 1 is

n1

x1= -------------------------

n1+ n2+ n3+ . . . .

where n i= number of moles for molecule i

From Henrys law, the partial pressure is proportional to the mole fraction.

n1

p1 = x1P = ------------------------ Pn1+ n2+ n3+ . . . .

where p1 = partial pressure of molecule 1n i= number of moles for molecule iP = total pressure ( P = p1+ p2+ p3+ . . . . )

Example 1.1 (Benjamin, p. 10): A solution contains 8 g /Llead (Pb, atomic mass 207). Expressthis concentration as:

i) moles per liter,

ii) ppmm, and

iii) the mole fraction of Pb.

Assume the solution has a density of 1 g/mL.

i) mol/L

8 1

10

1

2073865 10

6

8

g

L

g

g

mol

gx mol L

= . / (M)

ii) mass fraction, ppmm

3

6 3 9 6 6

8 1 1 1 8 8 10 1

10 10 1 10 10 1 / 10

mg g kg L kg x kg ppm

L g g kg kg kg kg kg

= =


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38 10 0.008m mx ppm ppm

= =

iii) mole fraction

a) The molar concentration of Pb in the solution = 3.865 x 10-8

mol/L from 1)

b) The molar concentration of water is

(assuming that the solution is pure water, MW = 18 g/mol)

1000

1855556

g

L

mol

gmol L

= . /

c)

Mole fraction of Pb moles of Pb in solution

total moles of all species in solution

x M

x Mx

=

=

+

=

3 865 10

3 865 10 55 5566957 10

8

8

10.

( . . ).

Example(Lindeburg, Review Manual, p 35-5): A wastewater treatment plant uses chlorine gas as a

reactant. A tank is filled with 800 m3of 20Cwater, and chlorine is added at a dosage of 125 g per

cubic meter of water. (Assume all of the chlorine dissolves and none initially reacts chemically.)

If the atmospheric pressure is 1.0 atm, what is the theoretical partial pressure of the chlorine gas at

the tank surface immediately after the gas is added?

( A) 3.1 x 10-5

atm ( B) 2.3 x 10-4

atm ( C) 0.039 atm ( D) 0.11 atm

(Solution) Approach:

1) determine the masses of Cl2and H2O.

2) determine the number of moles of Cl2and H2O.

3) determine mole fraction of Cl2

4) calculate the partial pressure of Cl2

1) The mass of chlorine in the tank is

(0.125 kg Cl2/m3H2O) (800 m

3H2O) = 100 kg Cl2= 100 000 g Cl2

The mass of water in the tank is

(800 m3H2O) (1000 kg/m

3) = 8 x 10

5kg H2O = 8 x 10

8g H2O

2) The MW of Cl2= (2)(35.5) = 71 g/mol

The MW of H2O = (2)(1.00) + 16.0 = 18.0 g/mol


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Determine number of moles

100 000 g Cl2

nCl2 = ----------------------- = 1408 moles of Cl2

71 g Cl2/ mol Cl2

8 x 108g H2O

nH2O = -------------------------- = 4.44 x 107 moles of H2O

18 g H2O / mol H2O

3) Mole fraction is

n1 nCl2 1408 mol

x1= ------------------------- = --------------- = ------------------------------------------

n1+ n2+ n3+ . . . nCl2 + nH2O 1408 moles + 4.44 x 107 moles

= 3.17 x 10-5

4) From Henrys law, the partial pressure is proportional to the mole fraction.

p1 = x1P = (3.17 x 10-5

) (1 atm) = 3.17 x 10-5

atm

Answer is A.

6)meq/L- mg/L per equivalent weight (eq.wt or EW)

* Equivalent weight (eq.wt), mg/meq, g/eq

- the weight of an element or compound which, in given chemical reaction, has the same

combining capacity of 8 g of oxygen or 1 g of hydrogen.

MW g/mol g mg

eq.wt = ----------- = ---------- = ------ = ---------

z eq/mol eq meq

where MW = molecular or atomic weight (g/mole)

z = ion valence, ion charge for ions= # of protons (H

+) or OH

-that can contribute to

an acid/base reaction.

= change in valence of the ion or radiation in a redox

reaction.

Unit of z: eq/mol or meq/m mol


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mg/L mg/L

meq/L = ------- = ----------- = meq/L

eq.wt mg/meq

meq m mol

meq/L = z (m moles/L) = ---------- --------- = meq/Lm mol L

Example: Determine eq.wt of CaCO3

Atomic weight Ca = 40, C = 12, O = 16

gram MW = 40 + 12 + (16 x 3) = 100

CaCO3 Ca2+

+ CO32-

(z = 2)

2 H+

+ CO32-

H2CO3

MW 100

eq wt = -------- = -------- = 50 mg/meq

z 2

7) N, normality, eq/L - number of equivalents per liter

- mass of substance per L / eq wt

- refers to a reaction or assumed reaction

1 N = 1 eq/L = 1000 meq/L

Note: N = z M

eq mol eq

N = z (M) = ------- ------- = ----

mol L L

Note: What is the unit of z?

N eq/L eq meqz = ------- = ---------- = ------- = ----------

M mole/L mole mmole

eq

z = -------

mole


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Examples

Convert 100 mg/L CaCO3to M, N, meq/L. (MW = 100 g/mole)

100 mg 1 g 1 mol

----------- ------------- --------- = 1 x 10-3

mol/L = 0.001 M

L 1000 mg 100 g

mg/L 100 mg/L

meq/L = ---------- = -------------- = 2 meq/L

eq. wt 50 mg/meq

= 2 x 10-3

eq/L = 0.002 N

or

2 eq 0.001 mol 2 eq

N = z M = --------- -------------- = ------- = 0.002 N

mol L L

eq/L 0.002 eq/L

M = --------- = --------------- = 0.001 moles/L

z 2 eq/ mol

Examples

1. What is normality of the solution when 0.98 g of H2SO4is dissolved in 1 L of H2O?

H2SO4 (MW = 98) H2SO4 2 H+

+ SO42-

H 2 x 1 = 2

S 1 x 32 = 32

O 4 x 16 = 64

-------------------------

H2SO4 98

MW 98

eq. wt = --------- = ------- = 49 g/eq

z 2

0.98 g 1 eq 0.02 eq

------- --------- = ----------- = 0.02 N

L 49 g L


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2. Convert 24 mg/L of Ca2+

and 100 mg/L of Cl-to meq/L

-------------------------------------------------------------------------------------

AW Charge eq.wt mg/L

Ions Conc. or ------------

MW z MW/z mg/meq

Species FW

(mg/L) (g/mole) (eq/mole) (mg/meq) (meq/L)-------------------------------------------------------------------------------------

Ca2+

24 40.1 2 40.1/2 = 20 24/20 = 1.2

Cl- 100 35.4 1 35.4/1= 35.4 100/35.4 = 2.8

--------------------------------------------------------------------------------------

3. Convert 100 mg/L of Ca2+

to molarity (M); Atomic Weight of Ca2+

= 40 g/mole

100 mg mol g 2.5 x 10-3

moles

------------- -------- ---------- = -------------------- = 2.5 x 10-3

M

L 40 g 103

mg L

8) epm, equivalents per million - ppm of substance per eq.wt

ppm

---------- = epm

eq.wt

- useful unit for comparison of anion and cation concentrations.

9) % strength (%S)

wt of solute

% S = --------------------- x 100

wt of solution

where wt of solution = wt of solute + wt of solvent

1 g of substance is mixed with 99 g of water

1 g

% S = --------------- x 100 = 1 %

1 g + 99 g

Note:

1 g 1,000 mg 1 kg

1% = --------- = -------------- -------- = 10,000 mg/L

100 g 0.1 kg 1 L

Assume the density of the solution D= 1


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Example (FE exam) Water and SO3combine to sulfuric acid, H2SO4, according to the following

reaction.

H2O + SO3 H2SO4

How many grams of water must be added to 100 g of 20% olem (20% SO 3and 80% H2SO4) by

weight) to produce a 95% solution (by weight) of sulfuric acid?

(Solution)

- Olem solution is 80% (by weight) H2SO4, which is to be increased to 95% (by weight) by adding

H2O.

- The reaction takes place based on molar concentrations.

MW of

H2O 2 (1.0 g/mol) + 16 g/mol = 18.0 g/mol

SO3 32.1 g/mol + 3(16 g/mol) = 80.1 g/mol

H2SO4 2 (1.0 g/mol) +32.1 g/mol + 4(16 g/mol) = 98.1 g/mol

H2O + SO3 H2SO4

1 mol 1 mol 1 mol

MW 18.0 g/mol 80.1 g/mol 98.1 g/mol

1 4.45 5.45 normalized per MW of H2O

x 4.45x 5.45x for unknown mass of water x

For a 95% solution,

2 4

2 4 3

0.95 H SO

H SO SO

m

m m=

+

Note all H2O is converted to H2SO4.

( ) ( )2 4

2 4 3

800.95

80 20H SO

H SO SO

g new m

g new m g reacted m

+=

+ +

Let x be the mass of the water added.The mass of SO3reacted is 4.45x.The mass of new H2SO4 produced is 5.45x.

( ) ( )

80 5.45 80 5.450.95

80 5.45 20 4.45 100

g x g x

g x g x x

+ += =

+ + +

x = 3.33 g Answer is: 3.33 g of water must be added.


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10) Partial pressure for gas, atm

e.g., Pco2= 10-3.5

atm

11) Color Unit for color

12) TU, JTU, NTU for turbidity

Turbidity is a measure of suspended material in water

Absorptometry measure transmitted light

Jackson candle turbidity meter (unit: JTU)Range: 40-1000

Nephelometry measure scattered light

Hach turbidity meter (unit: NTU)Range: 0-50

Absorptometry Nephelometry

Photo tube

particle Photo tube

Light

source


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13) Radionuclides

Radioactivity

Conventional Units:Curie, Ci 1 Ci = 3.7 x 10

10disintegrations per sec

= 103mCi = 106Ci = 109nCi = 1012pCi

Systeme International (SI) Units:

Becquerel, Bq 1 Bq = 1 disintegration per sec

1 Bq = 2.7027 x 10-11

Ci1 kBq = 103Bq

1 MBq = 106Bq

1 GBq = 109Bq

Dose

Conventional Units:Rad 1 Rad = 100 ergs/gram = 0.01 J/kg

1 Rad = 0.01 Gy = 1 centigray (cGy)


Gray, Gy 1 Gy = 1 J/kg = 100 Rad

1 Gy = 100 cGy

Dose Equivalent

Conventional Units:

Rem 1 Rem = 103

mrem = 0.01 Sv


Sievert, Sv 1 Sv = 1 J/kg = 100 Rem

1 mSv = 100 mrem


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mg/L as something

e.g., mg/L as N; mg/L as P; mg/L as CaCO3

Example (SJ p. 15) If we were to analyze a solution of ammonium nitrate (NH4NO3), and find that100 mL of solution contained 36 mg NH4

+and 124 mg NO3-.

1) What are concentrations of NH4+and NO3-in mg/L?2) What is the total N concentration?

(Solutions)

1)

Dissolution reaction:

NH4NO3 > NH4+ + NO3

-1 mol 1 mol 1 mol

NH4+

36 mg NH4+

103

mL 360 mg NH4+

--------------- ----------- = ------------------

100 mL 1 L L

NO3- 124 mg NO3

- 103mL 1240 mg NO3-

--------------- ----------- = --------------------

100 mL 1 L L

2)

Calculate MW:

N = 14 g/mole = 14,000 mg/mole

NH4+

= 18 g/mole = 18,000 mg/mole since N=14 and H=1; (MW = 14 + 4(1) =18 g/mol)NO3

-= 62 g/mole = 62,000 mg/mole since N=14, O=16; (MW=14 + 3(16) = 62 g/mol)

as N

For NH4+-N: 360 mg NH4

+ 14,000 mg N/mol 280 mg NH4

+-N

------------------ ---------------------------- = ----------------------L 18,000 mg NH4

+/mol L

For NO3--N

: 1240 mg NO3- 14,000 mg N/mol 280 mg NO3

--N

-------------------- --------------------------- = -----------------------L 62,000 mg NO3

-/mol L

Total N = 280 mg N/L + 280 mg N/L = 560 mg N/L


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or

For NH4+ 360 mg NH4

+ 14 mg N/meq 280 mg NH4

+-N

----------------- ------------------------- = ----------------------

L 18 mg NH4+/meq L

For NO3- 1240 mg NO3- 14 mg N/meq 280 mg NO3--N-------------------- ------------------------- = -----------------------

L 62 mg NO3-/meq L

Total N = 280 mg N/L + 280 mg N/L = 560 mg N/L

Example Water contains 35 mg/L of calcium ion and 15 mg/L of magnesium ion. Express the

hardness as mg/L of CaCO3.

mg/L MW z eq.wt (mg/meq) meq/L

MW/z

--------------------------------------------------------------------------------------------------------------------CaCO3 100 2 100/2 = 50

--------------------------------------------------------------------------------------------------------------------

Ca2+

35 40.08 2 40.08/2 = 20.04 35/20.04 = 1.746

Mg2+ 15 24.30 2 24.30/2 = 12.15 15/12.15 = 1.235

--------------------------------------------------------------------------------------------------------------------

Total Hardness = Ca2+

+ Mg2+

= 1.746 + 1.235 = 2.981 meq/L

2.981 meq 50 mgTotal Hardness as CaCO3= -------------- ---------- = 149 mg/L as CaCO3

L meq

Note:

meq mg/L mg/L

------- = ---------- = ------------L eq. wt mg/meq


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Example(Sato): NBOD

NH3+ 2 O2 WNO3- + H2O + H

+

NH4+ + 2 O2 WNO3

- + H2O + H

+

Using the reactions above, calculate the theoretical NBOD of a wastewater containing:

a) 15 mg/L of ammonia (NH3)

b) 15 mg/L of ammonia as nitrogen (NH3-N)

c) 15 mg/L of ammonium (NH4+)

d) 15 mg/L of ammonium as nitrogen (NH4+-N)

(Solution)

a)

g O2 (2 mol)(32 g/mol) 3.765 g O2 3.765 mg/L O2

----------- = -------------------------- = --------------- = --------------------

g (NH3) (1 mol)(17 g/mol) 1 g NH3 1 mg/L NH3

3.765 mg/L O2

Th NBOD = (15 mg/L NH3) (----------------------) = 56.5 mg/L O2

1 mg/L NH3

b)


---------------- = -------------------------- = ---------------- = --------------------

g (NH3-N) (1 mol)(14g/mol) 1 g NH3 1 mg/L NH3

4.571 mg/L O2


1 mg/L NH3

c)

g O2 (2 mol)(32 g/mol) 3.556 g O2 3.556mg/L O2

----------- = -------------------------- = --------------- = --------------------

g (NH4+) (1 mol)(18 g/mol) 1 g NH4

+ 1 mg/L NH4

+

3.556 mg/L O2


1 mg/L NH3

d)


---------------- = ------------------------- = -------------------- = -----------------------

g (NH4+-N) (1 mol)(14g/mol) 1 g NH4+-N 1 mg/L NH4+-N

4.571 mg/L O2

Th NBOD = (15 mg/L NH3) (---------------------) = 68.6 mg/L O2

1 mg/L NH3

Note that the ThNBOD values for NH3-N and NH4+-N are the same.


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Example: Convert 1 mg/L Na2HPO4as P to moles /L (M)

(Solution)

Note that:i) mg/L Na2HPO4 as P (Na2HPO4-P) is the concentration of P.

Therefore,

1mg P mol g P 3.23 x 10-5

mol P

----------- ------------- ---------------- = --------------------- = 3.23 x 10-5

ML 30.97 g P 1000 mg P L

Note that:i) the concentration is given as P (1

stterm).

ii) one mole of P is 30.97 g (2nd

term)

iii) one gram P is 1000 mg P (3rd

term)iv) 1 M of Na2HPO4-P is 1 M of P

Na2HPO4 2Na+ + HPO4

-

1 M 2 M 1 M

Na=23; H=1; P=31; O=16

Na2HPO4= 142 g/mol (23x2 +1+31+16x4 =142)

NaH2PO4= 120 g/mol (23+2+31+16x4=120)

H3PO4= 98 g/mol (3+31+16x4=98)

Bar Diagram (Bar Graph) and Stiff Diagram

- purpose is visualization of the chemical composition.

- data can be expressed in meq/L (milliequivalents per liter).

a. Top row of the bar graph consists of major cations arranged in the order Ca 2+, Mg2+, Na+, K+.

b. Bottom row of the bar graph is aligned in the sequence of CO32-, HCO3

-, SO4

2-, Cl-, NO3-.

c. The sum of the positive meq/L equalsthe sum of the negative meq/L for a water in equilibrium

(if all ions were correctly analyzed).

Ion Balance or I Cations - Anions l

Charge Balance = --------------------------------- x 100

Cations + Anions

Charge Balance < 5% OK


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Bar Diagram - Example


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Jackson and Patterson (1982)

Pizometer HA1

(a) (1) (2) (3) (4) (5) (7)(3)/(4) (2)/(5)

Cation AW/MW z eq.wt Cummu.mg/L eq/mol mg/meq meq/L

Ca2+ 12.0 40.1 2 20.1 0.60 0.6Mg2+ 2.0 24.3 2 12.2 0.16 0.76Na+ 91.0 23 1 23.0 3.96 4.72K+ 1.8 39.1 1 39.1 0.05 4.77Fe2+ -- 55.8 2 27.9 0.00 4.77Mn2+ 0.03 54.9 2 27.5 0.001 4.77Sr2+ 0.00 87.6 2 43.8 0.000 4.77

Anion

HCO3- 12.2 61.0 1 61.0 0.20 0.2SO42- 15.1 96.1 2 48.1 0.31 0.51Cl- 170.0 35.4 1 35.4 4.80 5.32NO3- 3.5 62.0 1 62.0 0.06 5.37Others -- -- 5.37

(b) Ion Balance -5.9629


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Unit of (7):

mg/L

------------ = meq/L

mg/meq

Ion Balance or I Cations - Anions l

Charge Balance = --------------------------------- x 100Cations + Anions

[ 4.77 5.37 ]

= --------------------- x 100 = 5.9% (> 5%)4.77 + 5.37

Ion does not balance.


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Bar Diagram - Examples

Documents

1-Concentration Units S12