Coagulation-Softening - ALUM SUMMARY General reaction: (1)...

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Coagulation-Softening 1 ALUM SUMMARY General reaction: (1) Al 2 (SO 4 ) 3 . 14H 2 O + 6H 2 O 2 Al(OH) 3 + 3H 2 SO 4 + 14H 2 O so 1 mol Alum destroys 6 mol AlK via (2) 3H 2 SO 4 + 6HCO 3 - 6H 2 CO 3 + 6H 2 O Alum produces Al(OH) 3 whose solubility is governed by pH via reactions (3) and (4): (3) Al(OH) 3 Al +3 + 3OH - K sp =10 -33 (4) Al(OH) 3 + OH- Al(OH) 4 - K =10 +1.3 can plot both (3) and (4) on a pC - pH diagram. This plot will show the solid and ionic or soluble phases for aluminum species. When from (3): K sp =10 -33 = (Al +3 )(OH - ) 3 taking negative logs (pK) pK sp = 33 = pAl + 3pOH from K w for water (pKw = pH + pOH) and substituting for pOH then 33 = pAl + 3(14 - pH) 33 = pAl + 42 - 3 pH and simplification gives: (5) pH = (pAl + 9)(1/3) = (pAl/3) + 3 Plot pH = 6 pAl = 9 pH = 8 pAl = 15 From (4), K = Al(OH - )4/OH - pK = p(Al(OH - )4) - pOH = -1.3 (6) p(Al(OH - ) 4 ) - 14 + pH = -1.3 pH = -p(Al(OH - ) 4 ) + 12.7 Plot pH = 6 p(Al(OH - ) 4 ) = 7.7 pH = 8 p(Al(OH - ) 4 ) = 4.7 Graphing (6) and (5) shows area for Al(OH) 3 formation and predicts optimum coagulation at pH= 5.5 for which pC is minimum at 10 -7.5 for optimum aluminum insolubility which is maximum coagulation or floc production. The figure shown on the following page shows other aluminum species than Al +3 and Al(OH) 4 - but is representative of the versatility of a pC - pH diagram, the amphoteric nature of aluminum species and the zone of solid formation or coagulation.
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Coagulation-Softening 2 FIGURE 4.1 Equilibrium Concentrations of Hydroxo Aluminum (III) Complexes 0 2 4 6 8 10 12 14 024681 0 1 2 1 4 pH - log concentration, pC [Al 3+ ] [Al 13 (OH) 34 5+ ] [Al 2 (OH) 2 4+ ] [Al(OH) 2+ ] [Al(OH) 4 - ] [Al 7 (OH) 17 4+ ] Al(OH) 3(s) Can predict effect on pH by Alum or required acid or base addition to get desired pH for reaction. Example: Assume alkalinity of 50 mg/L CaCO 3 , alum dose is 200 mg/l as Alum , original pH is 2.0. What is final pH, how much acid or base meq. are required to adjust pH to 5.5. Rx Al 2 (SO 4 ) 3 .14H 2 O + 6H 2 O 2Al(OH) 3 + 6H + + 3(SO 4 - ) 2 6H + + 6HCO 3 - 6H 2 CO 3 H 2 CO 3 H + + HCO 3 - pK = 6.3 200 594 6 1 10 0002020 10 3 269 mg LAlum mM mg Hm o l Alummol M mM mL H /. / . +− −+ == pH is 7.0, all AlK is HCO 3 - 50 50 10 10 3 3 3 mg L meq mg mmol meq M mM m LHCO // −− = free H + = 10 -2.69 - 10 -3 = 10 -3.00 with no base addition pH = 3.0 For pH=5.5 must know how much of carbonate species, C T , are in H 2 CO 3 , HCO 3 - , CO 3 -2 forms using α values. Remember (is fraction C T that is H 2 CO 3 ) For pH = 5.5,
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Coagulation-Softening 3 ) 10 ( 10 ) 10 ( 10 ) 10 ( ) 10 ( ) ( ) ( 3 . 10 3 . 6 5 . 5 3 . 6 2 5 . 5 2 5 . 5 2 1 2 1 + + + + + = + + = K K H K H H o α 0 10 10 16 . 1 10 ) 10 ( 10 ) 10 ( 10 ) 10 ( 14 . 0 10 16 . 1 10 ) 10 ( 10 ) 10 ( 10 ) 10 ( 86 . 0 7 . 5 11 6 . 16 3 . 10 3 . 6 5 . 5 3 . 6 2 5 . 5 2 1 2 11 8 . 11 3 . 10 3 . 6 5 . 5 3 . 6 2 5 . 5 1 1 = = + + = = + + = + x K K x H K o Since HCO 3 - = α 1 C T + C T = H 2 CO 3 + HCO 3 - + CO 3 -2 know HCO 3 - = AlK at pH 7 OriginalHCO mg LCaCO meg mg mmol meg M mmol ML C HCO pH x T 32 3 3 3 1 3 3 50 50 10 10 7 10 083 12 10 == = // (). . α (NOTE: if HCO 3 (alk) at pH 7 is used, must use α 1 at pH 7 for C T calculation) So at pH 3 all C T is H 2 CO 3 , (calculate α o C T ) (At pH 3) C T = H 2 CO 3 = 1.2 x 10 -3 m/L At pH 5.5 need H 2 CO 3 = α o C T = 0.86x10 -3 x 1.2 m/L ~ 1 x 10 -3 m/L HCO 3 - = α 1 C T = 0.14 x 10 -3 x 1.2 m/L = 0.20 x 10 -3 H 2 CO 3 ) Neutralized = 1.2 x 10 -3 m/L pH 3 - 1.0 x 10 -3 m/L pH 5.5 _____________________________ 0.2 x 10 -3 m/L Rx H 2 CO 3 + OH
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Coagulation-Softening - ALUM SUMMARY General reaction: (1)...

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