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The Islamic University of Gaza Faculty of Engineering Civil Engineering Department M.Sc. Water Resources

Water Quality Management (ENGC 6304)

Lect. 2: Chemical Water Quality

١ Chemical water – quality parameters • Water is called a universal • Chemical parameters of water are:

1. , 2. Alkalinity, 3. Hardness, 4. Fluorides, 5. Metals, 6. Organics, 7. Nutrients.

٢ of • Atom the smallest unit of element • Molecules of elements or compounds are constructed of atoms.

H + H → H 2

H 2 + O → H 2O • 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 ٣ • 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 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. NaCl ⇔ Na + + Cl − ٤ • 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 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.

٥ Example How many grams of calcium will be required to combine with 90 g of carbonate to form calcium carbonate?

Solution: 2- equivalent mass 12 + 3(16) 1- One equivalent of Carbonate (CO3 ) = = = 30g / equiv equivalence 2 40 2- One equivalent of Calcium(Ca2+) = = 20 g / equiv 2 3- The no. of equivalents of Carbonate must equal to the no. of equivalents of calcium. 90g 4- No. of equivalents of calcium = = 3 equiv 30g / equiv 5- So, we need 3 equiv of carbonate = 3 equiv X 20 g/equiv = 60 g.

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

(g / L)A ×(g / equiv)B = (g / L)A exp ressed as B (g / equiv)A

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

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

1- One equivalent of Calcium Carbonate (CaCO3) equivalent mass 40 +12 + 3(16) = = = 50g / equiv equivalence 2 23+ 35.5 2- One equivalent of Sodium Chloride (NaCl) = = 58 . 5 g / equiv 1 117(mg / L)NaCl ×50(g / equiv)CaCo =100(mg / L)NaCl as CaCo 58.5(g / equiv)A 3 3

3- One mole of a substance divided by its valence is one equivalent  2×10−3 mol / L = 2×10−3 equiv / L 1mol / equiv −3 thus,2×10 equiv / L×50(g / equiv)CaCo3 =100(mg / L)NaCl as CaCo3 ٨ Ionization of solids substances with crystalline structure

• Water is a reactant

2+ − CaO +H 2O → Ca + 2OH • Water is not a reactant + − NaCl + H 2O ↔ Na + Cl + H 2O

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

• Precipitation: the ionic components may be recombining into the solid form. ٩ • Dynamic Equilibrium is reached when the rate of dissolution and the rate of precipitation is exactly equal.

• Mass action equation y+ x− Ax By ↔ xA + yB Solid compound Ionic components [A]x[B]y K is an equilibrium constant for a = K given substance in pure water at given temperature [Ax By ] The brackets [ ] indicate . ١٠ EQUILIBRIUM SYSTEMS

RATE OF DISSOLUTION = RATE OF PRECIPITATION

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

[Ax By ] = Ks = const

x y [A] [B] = KK s = Ksp

Where Ksp is the 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) ١٢ SOLUBILITY PRODUCTS OF SOME COMMON SALTS AT 25 0 C

١٣ For example: 2+ 3- • Ca3(PO4)2 ⇔ 3Ca + 2PO4

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

2+ 3 3- 2 -27 [Ca ] [PO4 ] = 1x10 = Ksp = constant

What Does Solubility Product (ksp) Mean??

1. When [A]x [B]y < Ksp, the compound continue to dissociate and the solubilization continue to take place until [A+]a [B-]b = Ksp 2. When [A]x [B]y > Ksp, the compound will start to precipitate

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

BaSO 2+ 2- • 4 ⇔ Ba + SO4 The no. of moles of Ba2+ resulting from dissociation = x, and there 2- are x moles of SO4 , Then there are x moles of solid (BaSO4)

2+ 2- 2 -10 • [Ba ] [SO4 ] = (x) . (x) = x = Ksp = 1 x 10 (table 2-3) • x = 1x10-5 -5 Therefore, the solubility of BaSO4 is 1x10 M. 2+ - • CaF2 ⇔ Ca + 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

-4 • The solubility of CaF2 is 1.96x10 M. ١٥ Example: Calculating Concentrations From Ksp

Find the concentration of dissolved Al(OH)3 in a saturated solution. +3 - Al(OH)3 ⇔ Al + 3 OH Given: Ksp = 1.0 x 10-32

• Solution: • Ksp = [Al+3] . [OH-]3 +3 • Each mole dissolved Al(OH)3 gives one mole Al 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

١٦ Example

-12 The solubility product for the dissociation of Mg(OH)2 is 9x10 . Determine the concentration of Mg2+ and OH- at equilibrium,

Solution: 2+ − Mg(OH )2 ⇔ Mg + 2OH [Mg 2+ ][OH − ]2 = 9×10−12

• The no of moles of Mg2+ resulting from dissociation = x, • Then there are 2x moles of OH- x.(2x)2 = 9×10−12 ⇒ x =1.3×10−4 mol / L (Mg) 2x = 2.6×10−4 mol / L (OH ) ١٧ 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.

١٨ 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.

١٩ • 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

٢٠ Ion Balance The dissolved solids content of natural water is classified to: • Major constituents … (1-1000mg/L)

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

- 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)

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

٢١ Testing for ion balance

• 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?

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

٢٢ Ion Concentration Equiv, Equiv conc, (mg/L) (mg/mequiv) (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 2- (SO4 ) 60 96/2 1.25 (CL-) 89 35.5/1 2.51 7.86

٢٣ • Calculate the percent of error = (8.49-7.86)*100/7.86= 8% < 10% …. Accept analysis

0 2.75 4.23 8.49 (Ca2+) (Mg2+) (Na+)

- 2- - (HCO3 ) (SO4 ) (CL )

0 4.1 5.35 7.89

Bar diagram

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