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The Islamic University of GazaFaculty of Engineering

Civil Engineering DepartmentM.Sc. Water Resources

Water Quality Management(ENGC 6304)

Lect. 2: Chemical Water Quality

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Chemical water – quality parameters

• Water is called a universal Solvent

• Chemical parameters of water are:

1. Total dissolved solids,

2. Alkalinity,

3. Hardness,

4. Fluorides,

5. Metals,

6. Organics,

7. Nutrients.

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Chemistry of solutions

• Atom the smallest unit of element• Molecules of elements or compounds are

constructed of atoms.

2HHH →+

OHOH 22 →+

• Molecular mass is the sum of the atomic mass of all atoms in molecule

For example :• Atomic mass of oxygen (O) = 16 , for Hydrogen (H) =1• Molecular mass for water (H2O) =18

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• A mole of an element or compound is its molecular mass in gm.

For example :

one mole of oxygen (O2) = 32, water (H2O) =18

• One mole of a substance dissolved in sufficient water to make one liter of solution is called a one molar solution.

• The charged species is called ions. Produced when compounds dissociate in water.

• +ve ions called cations, -ve ions called anions

• Neutrality means the number of cations = the number of anions.

−+ +⇔ ClNaNaCl

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• The valence is the number of charges on an ion.

The valence of (Na+)=1,

The valence of (Cl-)=1

• The equivalence of an element is the number of hydrogen atoms that element can hold in combination or can replace in a reaction (= valence in most cases)

• An equivalent of an element is its gram molecular mass(mole) divided by its equivalence.

• A milliequivalent of an element is its milligrams molecular mass divided by its equivalence.

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ExampleHow many grams of calcium will be required to combine with 90 g of carbonate to form calcium carbonate?

1- One equivalent of Carbonate (CO32-)

Solution:equivg

eequivalencmassequivalent /30

2)16(312=

+==

2- One equivalent of Calcium(Ca2+) equivg /202

40==

3- The no. of equivalents of Carbonate must equal to the no. of equivalents of calcium.

4- No. of equivalents of calcium equivequivgg 3

/3090

==

5- So, we need 3 equiv of carbonate = 3 equiv X 20 g/equiv = 60 g.

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• The concentration of substance A can be expressed as an equivalent concentration to substance B as the following:

BasressedALgBequivgAequivg

ALg exp)/()/()/(

)/(=×

• Generally, the constituents of dissolved solids are reported in terms of equivalent calcium carbonate concentrations

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Example

333

33

)/(100)/(50/102,

/102/1

/102

CaCoasNaClLmgCaCoequivgLequivthus

Lequivequivmol

Lmol

=××

×=×

−

−−

1- One equivalent of Calcium Carbonate (CaCO3)

equivgeequivalenc

massequivalent /502

)16(31240=

++==

2- One equivalent of Sodium Chloride (NaCl) equivg /5.581

5.3523=

+=

33 )/(100)/(50)/(5.58

)/(117 CaCoasNaClLmgCaCoequivgAequivg

NaClLmg=×

3- One mole of a substance divided by its valence is one equivalent

What is the equivalent calcium carbonate concentration of:• 117 mg/L of NaCl?• 2x10-3 mol/L of NaCl?

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Ionization of solids substances with crystalline structure

• Water is a reactant

OHClNaOHNaCl 22 ++↔+ −+

−+ +→+ OHCaOHCaO 222

• Water is not a reactant

• Dissolution : the solid form may be dissociated into its ionic component

• Precipitation: the ionic components may be recombining into the solid form.

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• Dynamic Equilibrium is reached when the rate of dissolution and the rate of precipitation is exactly equal.

• Mass action equation

KBABA

yx

yx

=][][][

−+ +↔ xyyx yBxABA

Solid compound Ionic components

The brackets [ ] indicate molar concentration.

K is an equilibrium constant for a given substance in pure water at given temperature

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EQUILIBRIUM SYSTEMS

RATE OF DISSOLUTION = RATE OF PRECIPITATION

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• At equilibrium, the solid phase does not change concentration (precipitation = dissolution)

constKBA syx ==][

spsyx KKKBA ==][][

Where Ksp is the solubility product.

If the concentration of either or both of the ions is increased , the product of the ionic concentration will exceed the Ksp and precipitation will occur to maintain the equilibrium conditions.

See table 2-3 (solubility product values)

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SOLUBILITYPRODUCTS

OFSOME

COMMONSALTS

AT25 0 C

For example:

• Ca3(PO4)2 ⇔ 3Ca2+ + 2PO43-

The products of ionic concentrations of both ions is always a constant for a given compound at a given temperature, i.e.

[Ca2+]3 [PO43-]2 = 1x10-27= Ksp = constant

What Does Solubility Product (ksp) Mean??

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1. When [A]x [B]y < Ksp, the compound continue to dissociate andthe solubilization continue to take place until [A+]a [B-]b = Ksp

2. When [A]x [B]y > Ksp, the compound will start to precipitate

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Based on Ks value, we can actually calculate the solubility of any particular compound

• BaSO4 ⇔ Ba2++ SO42-

The no. of moles of Ba2+ resulting from dissociation = x, and there are x moles of SO4

2-, Then there are x moles of solid (BaSO4)

• [Ba2+] [SO42-] = (x) . (x) = x2= Ksp = 1 x 10-10 (table 2-3)

• x = 1x10-5

Therefore, the solubility of BaSO4 is 1x10-5 M.

• CaF2 ⇔ Ca2++ 2F-

• The no. of moles of Ca2+ resulting from dissociation = x, and there are 2x moles of F-, Then there is x moles of solid (CaF2)

•[Ca2+] [F-]2 = (x) (2x)2 = Ksp = 3 x 10-11 x = 1.96x10-4

• The solubility of CaF2 is 1.96x10-4 M.

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Example: Calculating Concentrations From Ksp

Find the concentration of dissolved Al(OH)3 in a saturated solution.

Al(OH)3 ⇔ Al+3 + 3 OH-

Given: Ksp = 1.0 x 10-32

• Solution:

• Ksp = [Al+3] . [OH-]3

• Each mole dissolved Al(OH)3 gives one mole Al+3 and three moles OH- ions therefore, if [Al+3] = X, [OH-] = 3X and

• Ksp = X. (3X)3 = 27 X4 = 1.0 x 10-32

• X = (1.0 x 10-32 / 27)1/4 = 4.4 x 10-9 M

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Example

• The no of moles of Mg2+ resulting from dissociation = x,• Then there are 2x moles of OH-

The solubility product for the dissociation of Mg(OH)2 is 9x10-12.

Determine the concentration of Mg2+ and OH- at equilibrium,

1222

22

109]][[

2)(−−+

−+

×=

+⇔

OHMgOHMgOHMg

Solution:

)(/106.22)(/103.1

4

4

OHLmolxMgLmolx

−

−

×=

×=⇒

122 109)2.( −×=xx

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Total Dissolved solids

The material remaining in the water after filtration for the suspended solid analysis is considered to be dissolved solids.

• Source

• Results from the solvent action of water on solids, liquids, and gases.

• Inorganic dissolved solids; minerals, metals, gases.

• Organic ; decay products of vegetation, organic chemicals and gases.

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Impacts

• Produce tastes, colors, and odor,

• Some chemicals are toxic or carcinogenic,

• Some dissolved solids may combine to form a compound of more dangerous than the original materials,

• Not all dissolved solids are undesirable.

Measurement

• See sec. 2-2 Suspended Solids

• By measuring the electrical conductivity of the water, Conductivity measures the ability of water to conduct an electrical current.

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• Conductivity is a good way to determine the ionic strength of water because the ability of water to conduct a current is proportional to the number of ions in the water

• Freshwater generally has low conductivity measured in microSiemens (uS)

• Marine systems have much higher conductivity measured in milliSiemens (mS) which can easily be converted to salinity

• Humans and other terrestrial animals require fresh water for survival as do plants and animals normally found in freshwater

Use• TDS parameter is important in the analysis of water and

wastewater to know more about the composition of the solids in water

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Ion BalanceThe dissolved solids content of natural water is classified to:

• Major constituents … (1-1000mg/L)

- Called common ions,

- Measured individually and summed on an equivalent basis to represent the approximate TDS,

- The sum of anions must equal the sum of cations (as a check)

sodium (Na+), calcium (Ca2+), magnesium (Mg2+), bicarbonate (HCO3-), sulfate (SO42-), chloride (CL-)

• Secondary constituents …. ….(0.01-10mg/L).Iron, potassium, carbonate, nitrate, fluoride, boron, silica

Testing for ion balance

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sodium (Na+) = 98mg/L,calcium (Ca2+) =55mg/L,magnesium (Mg2+)= 18mg/L,

bicarbonate (HCO3-)= 250mg/L, sulfate (SO42-) =60mg/L, chloride (CL-) = 89 mg/L

• The results of common ions for a sample of water are shown below,

• If 10% error in the balance is acceptable, should the analysis be considered complete?

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Ion Concentration(mg/L)

Equiv,(mg/mequiv)

Equiv conc,(mequiv/L)

(Na+) 98 23/1 4.26(Ca2+) 55 40/2 2.75(Mg2+) 18 24.3/2 1.48

8.49(HCO3

-) 250 61/1 4.1(SO4

2-) 60 96/2 1.25(CL-) 89 35.5/1 2.51

7.86

• Calculate the percent of error

= (8.49-7.86)*100/7.86= 8% < 10% …. Accept analysis

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(Ca2+) (Mg2+) (Na+)

(SO42-) (CL-)(HCO3

-)

0

0 4.1

4.232.75 8.49

5.35 7.89

Bar diagram