Page 710 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

Precipitation 665

Solution Solution
2
¼ [Mg ][OH ] , is the basis for
2þ
Equation 21.2, K Mg(OH) 2 State reactions:
the calculation. First, however, [OH ] must be calculated

from the relation, [H ][OH ] ¼ 10 14 ,atpH ¼ 11, [H ] ¼
þ

þ
10 11 ! [10 11 ][OH ] ¼ 10 14 ; and [OH ] ¼ 10 3 mol=L. CaCl 2 ! Ca 2þ þ 2Cl

H 2 O ! H þ OH
þ
2
[Mg ][OH ] ¼ 5.6 10 12
2þ
3 2
[Mg ][10 ] ¼ 5.6 10 12 NaOH ! Na þ OH
þ
2þ
[Mg ] ¼ 5.6 10 6 mol=L
2þ
The net equation is, combining Equations 1 and 3 and
5:6 mol 10 6 Mg 2þ 24,300 mg Mg 2þ
þ

2þ excluding spectator ions, Na and Cl :
L mol Mg 2þ
C(Mg ) ¼
¼ 0:14 mg Mg =L Ca 2þ þ 2[OH ] ! Ca(OH ) 2
2þ

Discussion Species in solution:
As seen, the residual concentration of Mg , that is, that
2þ
which is in equilibrium with Mg(OH) 2 particles, that is,

þ

‘‘sludge,’’ is very low. Ca ,Cl ,Na ,OH ,H þ
2þ
Mathematical relations:
21.2.1.3 Listing of Solubility Products
Equilibrium equations
Table 21.1 lists compounds and solubility products, categor-
ized by anions. From such a list, the precipitant compound 2 5:3
¼ [Ca ] [OH ] ¼ 10
2þ
may be selected.
K Ca(OH) 2
For example, to remove lead, Table 21.1 shows that K w ¼ [H ] [OH ]

þ
¼ 1.4 10 20 , which is the lowest listed. Removal
K Pb(OH) 2
involves merely raising the pH within the reactor, floccula- Electroneutrality
tion, and settling. The whole sequence is often done in the
industry by means of a ‘‘reactor clarifier.’’
2[Ca ] þ [Na ] þ [H ] ¼ [Cl ] þ [OH ]
2þ

þ

þ
In other cases, anions, for example, CN ,PO 4 , are

3
targeted for removal. The same approach is applied, that is,
finding a cation that when combined with the particular anion Mass balance
of interest, has a low solubility product. As seen in Table 21.1,
C ¼ 2[Ca ] þ 2[Ca(OH) 2 ] ¼ [Cl ]
2þ

phosphate is not difficult to remove because K Ca 3 (PO 4 ) 2 ¼
2.1 10 33 . Similarly, K Cu(CN) ¼ 3.5 10 20 , which means
that Cu(I) can remove CN . Derivative relations:

Substitute (8) in (7) to eliminate [Cl ]:

21.2.1.4 Solubility pC–pH Diagrams
For precipitation reactions, as with acid–base reactions, a
[Na ] þ [H ] ¼ [OH ]
þ
þ

graphical solution facilitates understanding and adds a visu-
alization component. As with an acid–base reaction, its con- This means that as Na þ is added, OH is added in the
struction requires an equilibrium statement, and equations for same amount.
mass balance and electroneutrality. The graphical solution
Express (5) in log form:
also has practical application in that whether precipitation
occurs or not is evident for any pH–pC coordinate. Example
log [Ca ] þ 2 log [OH ] ¼ 5:3

2þ
21.2 illustrates the method of constructing a pH–pC diagram
for calcium hydroxide.
or
Example 21.2 Construction of pH-pC Diagram p[Ca ] þ 2p[OH ] ¼ 5:3

2þ
for Ca(OH) 2
System point determination:
Given
A solution has calcium chloride, [CaCl 2 ] ¼ 0.1 mol=L and Substitute [Ca ] ¼ 0:1 mol=L in (11), that is,
2þ
is titrated with sodium hydroxide in which, [NaOH] ¼ 1.0
mol=L. p[0:1] þ 2p[OH ] ¼ 5:3
Required 5:3 1:0
¼ 2:15
Determine the pH versus p[Ca ] curve as titration occurs. p[OH ] ¼ 2
2þ