WORKSHOP 1 - PHYSICOCHEMICAL PARAMETERS

In an acid-base titration, 3.9 ml of sodium hydroxide, NaOH at 0.1 N was used.

to neutralize 80 ml of a sulfuric acid solution, H2SO4Determine the
concentration of the acidic solution in mg/L of CaCO3.

1 eq CaCO31 mol CaCO3100g CaCO3

(
0.1 N NaOH0.0039 L )
1 eqNaOH 2 eq CaCO31 mole of CaCO3
Acidity, like g CaCO3 /L =
0.080 L
Acidez, = 0.24375 * 1000 = 243.75 mg CaCO3 /L

2. For the determination of iron in water, 15 ml of sample was taken diluted to 50.
ml and a transmittance of 45% was obtained. For the calibration curve, were prepared
standards with 0.0, 0.1, 0.2, and 0.3 mg of Fe, and absorbances of 0.0, 0.24 were observed
0.49 and 0.64, respectively. Determine the iron concentration of the sample.

Below is the graph of mg of iron vs absorbance with its respective equation.

of the line

Calibration curve
0.8

0.6

0.4

0.2 y = 2.17x + 0.017

R² = 0.989
0
0 0.1 0.2 0.3
mg of Iron

The calibration equation

= ( 2,17 ) ℎ + 0.017
Absorbance is related to transmittance as:

= −( log10 )

If the transmittance is 45%, then

Absorbance = - log10(0.45)

= 0.347
From the calibration curve, the mass of iron is determined:
 Absorbance - 0.017
ℎ =
2.17

hierro = 0.152

If a sample of 50 ml was analyzed, then the concentration will be:

0.152 mg Fe
[ ] = = 30.04 mg Fe/L
1L
50 ml
1000 ml

Multiplying by a dilution factor of 50/15, then the concentration of iron in the sample
mother of 15 ml will be:

50
[ ] = 3,04 mg Fe/L *
15

[ ̅
] = 10.13 /

3. In the determination of solids in water, 120 ml of sample and a crucible were used.
of filtration with an initial mass of 13.2845 g. After filtration and drying at 103 °C,
the mass was 13.3627 g. the calcination at 550°C resulted in a final mass of 13.3428
g. say what type of solids the test corresponds to and determine the percentage of
volatiles with respect to the total.

Total solids, TS = (mass at 103 °C - mass of the crucible) / Volume of the sample

( 13.3627 g - 130.2845 g )
= 651.6667 mg/L
0.120 L
Total fixed solids, SFT = (final mass at 550 °C - mass of the crucible) / Volume of the sample

( 13.3428 g - 13.2845 g )
= 485.8333 /
0.120 L
Therefore, the volatiles correspond to the difference between the total solids and the solids.
total fixed.

= −

SVT = 651.6667 - 485.8333= 165.8334 mg/L

And its percentage

165.8334
% = ∗ 100 = 25.4476%
651.6667
 A sample of 100 ml of water requires 7.2 ml of H2SO4-0.025 N to reach the
endpoint of phenolphthalein and an additional 3.9 ml to reach the endpoint of the
methyl orange. Determine the total amount and the value of each of the causes of
alkalinity.

Determination of Total Alkalinity

A * B * 50000
Alkalinity, mg CaCO33 /L =

Where;

A. Volume of acid used in the titration.

B. Normality of the acid.

Therefore, to calculate the total alkalinity (M), A is taken as the total volume.

( 7.2 +3.9 )*0.025 ∗ 50000

=
100 mL
= 138.75 mg CaCO33 /L

Now, the same is done, but until the phenolphthalein turning point (F).

7.2 mL * 0.025 N * 50000

F, mg CaCO3 /L =
100 mL

F = 90 mg CaCO3 /L

According to the relationships of alkalinity

F > M/2

Then, the contribution to alkalinity from the presence of OH groups.-

Alcalinidad, mg CaCO3 2F - M

( 138.75
Alcalinidad = 2 90- ) = 41.25 mg CaCO3 /L

2−
And the contribution to alkalinity from the presence of carbonates ( 3 )

Alcalinidad, mg CaCO3 /L = 2 (− )

(
Alcalinidad = 2 13-81.25= 97.5) mg CaCO3 /L
Source: [1] and [2].
 A water sample shows the following analysis: 86 mg/L of calcium, 27 mg/L of
magnesio
sulfates, and 173 mg/L of chlorides. Determine the total hardness, the carbonate hardness and the
carbonate hardness of water.

The hardness is mainly due to the calcium and magnesium in the sample and can be classified into:

a. Carbonate hardness. Composed of carbonates and bicarbonates.

b. Hardness in carbonated water. It is due to the cations associated with sulfate cations, chlorides, and
nitrates.

In addition, the alkalinity is due to the bicarbonates in the sample. Considering that these are also
two properties that are related, therefore, it is possible to determine the carbonate hardness and not
starting from the comparison between hardness and total alkalinity. Therefore, total hardness is obtained from the
concentration of calcium (PM=40.078) and magnesium (PM=24.305) in CaCO3, then

[ ] = [Ca ]+ Mg
mg 1 mmol Ca1 meq Ca 1 mmol CaCO3100 mg CaCO3
[Ca] =86
L 40,078mg 1 mmol Ca 1 mEq CaCO31 mmol CaCO3

[Ca] = 214,582 mg CaCO3

mg 1mmol Mg1 meq Mg 1 mmol CaCO3100 mg CaCO3
[Mg] =27
L 24,305 mg 1 mmol Mg1 mEq CaCO31 mmol CaCO3

[Mg] =111,088 mg CaCO3 /L

mg 1 mmol Mg1 mEq Mg 1 mEq CaCO31 mmol CaCO3100 mg CaCO3
[Mg27
]
L 24,305 mg1 mmol Mg 1 meq Mg 1 mol CaCO31 mmol CaCO3
So

= 325.66 mg CaCO3 /L
The total alkalinity will be equal to the amount of bicarbonate (MW=61) in the sample:

1 mg mmol HCO31 mEq HCO31 mEq CaCO31 mmole CaCO3100 mg CaCO3

[ ] 146
3=
61 mg 1 mmol HCO32 meq HCO31 mEq CaCO31 mmol CaCO3

[HCO3=
] 119.67 mg CaCO3 /L

So

119.67 mg CaCO3 /L

Since the total alkalinity is less than the hardness, then the carbonate hardness will be equal to the
alkalinity
 Carbonate hardness = 119.67 mg CaCO3 /L

Non-carbonated hardness = total alkalinity - carbonated hardness

Non-carbonated hardness = 205.99 mg CaCO3 /L

Source: [1] and [2].

6. Los resultados del análisis de un agua fueron. Alcalinidad = 98 mg/L soccer

62 mg/L 12°C
Determine the Langelier, Ryznar, and Aggressiveness saturation indices.

LANGELIER SATURATION INDEX (LSI)

Source: [3].

With TDS in mg/L. So, replacing

1
= = 0.0830
( -)
10(log10160

B = -13,12log1012 (+ 273,2 + 34.55

) = 2338

= (1062 mg/L CaCO33)-0.4 = 1392

= (1098 mg/L CaCO₃31,991

)

(
= 9,3+ 0.0830 +2338 − 132,083= 8,338 )

( ) -1.5338 = -1.838

According to the classification of the saturation index, for LSI < 0 it is water with a tendency to be
corrosive
 RYZNAR INDEX (RSI)

Source: [3].

According to the calculation method for the index, the pHsat will be equal to the pHsat of the LSI, then,
the index will be

RSI = 2 *8,338
( -6,5
) =10,176

According to the saturation index classification, for RSI >9 it is intolerably water.
corrosive

AGGRESSION SATURATION INDEX

For the calculation of the aggression index

=( + ( )
∗ )
Where;

AI. Aggressiveness index

A. Total alkalinity (mg/L CaCO)3)

H. calcium hardness (mg/L CaCO)3)

So, replacing we have that the aggression index is

( ∗ 62 )) = 10.28
= (6.5 + log98

Where values of the aggression index between 10-12 are approximately corrosive.

Source: [4].

7. The results of the water analysis are organic nitrogen = 12.0 mg/L N, nitrites =
2.4 mg/L N
7 days of incubation = 326 mg/L. Assuming k = 0.16/d, determine the standard BOD.
the ultimate carbonaceous BOD and the ultimate combined BOD.

Standard BOD calculation (BOD5)

From the mathematical formulation, the oxidation stage is defined as shown below:

=−

And after integrating, changing the base, and clearing, we have

= (1 − 10− )

Where;

y. BOD exerted at time t, mg/L

L. Remaining BOD for t = 0, last BOD, mg/L
k. Constant of the oxidation rate, decimal base, d-1
t. Time, d

So, based on the information provided by the DBO7(326 mg/L) we can calculate the BOD
remaining at time t = 0

= −
( 1 - 10 )
326 ⁄
= [ -1 ( 7 )] )
= 352.76 ⁄ 2
(1 − 10− 0.16

Therefore, the standard biochemical oxygen demand (BOD5)

 5 -1] )( = 296.85
)
= 352.76(1 − 10−[0.16 ⁄ 2

And the carbonaceous BOD (20 days)

20 ] )-1=( )
= 352.76(1 − 10.16 352.54 ⁄ 2

To calculate the combined ultimate BOD, the ultimate carbonaceous BOD and BOD are added.
nitrification (amount of oxygen to oxidize ammonia plus the amount of oxygen to oxidize
the nitrites). So, to calculate the ultimate BOD by nitrification we have the following reaction
global ammonia oxidation

3+ 2O2→ 3+ + 2

And the nitrites oxidize according to the following reaction

2NO2+ 2→ 2NO3

Then, the amount of oxygen to oxidize ammonia is calculated as follows:

2 31 mmol of NH32 mmol of O232mg of O2

= 8,3 = 31.19
17.031 mg NH31 mmol of NH31 mmol of O2

And for the calculation of the amount of oxygen to oxidize the nitrites

2 21 mmol of NO21 mmol of O2 32mg of O2

= 2,4 ∗ ∗ ∗ = 0.83
46 mg NO2 2 mmol of NO21 mmol of O2

Therefore, the final combined BOD will be

= 31.19 + 0.83+ 352.54 = 384.56 ⁄2

REFERENCES

[1] A. Londoño, G. Giraldo, and Á. Gutiérrez, Analytical methods for evaluation

Physical chemistry of water. Blanecolor Ltda Publishing 2010.
[2] R. Baird and L. Bridgewater, Standard methods for the examination of water and
wastewater(no. 27). Washington: American Public Health Association, 2017.
[3] M. P. E. Gonzales, M. Hernandez, "Calculation of the Langelier indices (simplified)"
Saturation (SM2330 revised), Ryznar, Pockorius, Larson-Sckol, Hardness and Alkalinity
Canary Water Center Foundation, 2010.
[4] M. F. Davilet al., "Survey of Corrosion and Scaling Potential Produced Water from Ilam"
Water Treatment Plant, "World Applied Sciences Journal, p. 2, 2009.