By using Equation (38), [P04 ] may also be eliminated in favor of Ksp,FeP0 to produce

This will incorporate the iron into Equation (36) to indicate the use of the ferric salt. Finally, substituting the whole result into Equation (36), produces

7h2P04 K H2P04 K HP047FeIII[ Fe ] ^H3P04 K H2P04K HP047FeIII[ Fe ]

7ro4, YAl, Yh, 7hp^, and YH are, respectively, the activity coefficients of the phosphate, aluminum, hydrogen, hydrogen phosphate, and dihydrogen phosphate ions. KOT0 , KH P0 , and are, respectively, the equilibrium constants of the hydro gen phosphate and dihydrogen phosphate ions and phosphoric acid. Y^rn is the activity coefficient of the ferric ion.

As shown, Equation (43) still expresses [ spP0 FeIII] in terms of two variables,

[Fe ] and [H ]. We want it expressed only in terms of the K constants and the hydrogen ion concentration. This is where an expression needs to be found to eliminate [Fe3+]. If this expression indicates that some compound of iron is coprecipitating with FeP04, then [Fe3+] could be eliminated using the K constant of this compound.

When two compounds are coprecipitating, they are at equilibrium at this particular instance of coprecipitation. They are at equilibrium, so their equilibrium constants can be used in a calculation. This instance of coprecipitating with FeP04 is, indeed, possible through the use of the compound ferric hydroxide which, of course, must still be investigated whether, in fact, it is possible. This is investigated next. The dissociation reaction of ferric hydroxide is

If the results of this investigation show that the molar concentration of iron needed to precipitate Fe(0H)3 is about the same as the molar concentration of the iron needed to precipitate FeP04, then Fe(0H)3 and FeP04 must be coprecipitating.

Let x be the {Fe3+}. Because the coefficient of 0H— in the previous reaction is 3, {0H-} is therefore 3x. Thus, substituting into the Ksp equation,

KSp,Fe(0H)3 — {Fe3+}{ 0H-}3 — x (3 x)3 — 1.1( 10-36) x — 4.5 (10-10) gmol/L — {Fe3+} Now, from Equation (37), Ksp,FeP0 — {Fe3+}{P0f} — 10 9. Letting y equal

to {Fe }, {P04 } is also equal to y. Substituting into the Ksp equation,

These two concentrations, x and y, are practically equal; thus, this justifies the use of ferric hydroxide. We can conclude that Fe(OH)3(s) will definitely coprecipitate with FePO4. Therefore, r„ 3+n {Fe3+} Ksp,Fe(OH)3 Ksp,Fe(OH)3YH[H ]

eiii YFeiii{OH } 7FemK w

Substituting in Equation (43) to eliminate [Fe3+],

Ksp,FePO4Kw Ksp,FePO4Kw

7po 4"- sp ,Fe(OH)3 7hpo 4^- HPO^^- sp,Fe(OH)3

Ksp,FePO4Kw Ksp,FePO4Kw

7H2PO4K H2PO4K HPO4 Ksp,Fe(OH)3YH[H+] K H3PO4 Kh2P04 KHPO4Kp,Fe(OH)3

From this equation, the optimum pH for the removal of phosphorus can be determined.

Example 12 in the coagulation treatment of water using copperas, FeSO4 • 7H2O, the following complex reactions occur.

Fe (OH )2(s) + OH— ^ Fe( OH) — K Fe < oh)3c = 10-51 (c) A molar mass balance on the ferrous iron produces

where [spFeii] is the total molar concentration of all species containing the iron atom. Express [spFeii] in terms of the concentration of hydrogen ion and equilibrium constants.

Solution: in (d), we need to express [Fe2+], [FeOH+], and [Fe (OH)— ] in terms of [H+] and pertinent equilibrium constants. First, consider [Fe2+]. {sp} = Y[sp] ^ [Fe2+] = {Fe2+}/YFeii, where Yen is the activity coefficient of ferrous iron, Fe2+. From (a), rc 2 + , rritjn 2 2+, Ksp,Fe(OH)2


YFeII 7Fen{ OH-}2

From the iron product of water, {OH}{H} = Kw, {OH} = Kw/{H} = Kw/7h[H+], where Y is the activity coefficient of the hydrogen ion. Therefore,


The steps involved to express [FeOH+], and [Fe(OH)-] in terms of [H+] and pertinent equilibrium constants are similar to the ones above. Thus, we will not go through the detailed steps, again, but write the results at once:

YFeOHc YFeOHc {OH-} ÏfeohcKw ÏfeohcKw

where YFeOHc and YFe (OH)3c are the activity coefficients of the complexes FeOH+ and Fe(OH)-, respectively. Substituting in (d),

Ksp,Fe(OH)2YH [H+] KFeOHYH [ H+] KFe(OH)3cKw (d) [ spFeii ] =-2-+ —-p— +-+- Ans


Acids and Bases

Acids are substances that can donate a proton, H+, and bases are substances that can accept this proton. The most fundamental reaction of an acid, HA, is its reaction with H2O,

As seen from this reaction, the acid HA donates its proton to H2O, producing H3O+. H3O+ is called a hydronium ion. Because H2O has accepted the proton, it is therefore a base. H3O+ that results from the acceptance of this proton is an acid. The water molecule is neutral, but it has the proton, an acid, "clinging" to it. H3O+ is called the conjugate acid of the base H2O. Now, when HA donated its proton, it transforms into A—. 0f course, A— can accept the proton from H30+ to form back HA. For this reason, A— is a base; it is called the conjugate base of the acid HA. The equilibrium constant for an acid, Ka, is

Note that the activity {H20} is equal to unity, because H20 is unionized. Ka is also called ionization constant for the acid.

As an acid, H30+ reacts in water as follows:

0n the left-hand side of the equation, H30+ donates its proton to H20, with this H20 forming the H30+ on the right-hand side of the equation. By virtue of the donation of its proton, the H30+ on the left transforms to the H20 on the right-hand side of the equation. H20 on the left is a base and H30+ on the right is its conjugate acid. 0n the other hand, the H20 on the right is the conjugate base of the acid H30+ on the left. The net effect of these interactions, however, is that nothing happens, although intrinsically, the mechanism just described is actually happening.

Although the reaction of the hydronium ion H30+ as an acid may seem trivial, it serves an important function. As shown in Table 2, it can serve as an indicator of the lower limit of the strength of strong acids. The strength of an acid can be measured in terms of its production of the hydronium ions. Strong acids ionize completely (or nearly so) producing 100% equivalent of the hydronium ions. Weak acids only ionize partially, which, correspondingly produce a relatively small amount of hydronium ions. They normally ionize to about 10% or less of the original acid. The table shows the relative strengths of acid-conjugate base pairs. Under the column on acids, perchloric acid, sulfuric acid, hydrogen iodide, hydrogen bromide, hydrogen chloride, and nitric acid are the strongest acids. As indicated, they ionize to 100% H30+ and, since they ionize practically completely, the activity of the acid molecule is also practically zero. The denominator in Equation (47) would therefore be zero and Ka would be infinity.

H20 can also act as an acid. This is shown by the following reaction:

H30+ is the conjugate acid of the second H20 on the left-hand side of the equation and 0H— is the conjugate base of the first H20 on the left-hand side of the equation. The second H20 is a base that reacts with the first H20, which is an acid.

Referring to Table 2, the acids from H30+ down to H20 are weak acids. These are the acids that ionize to only 10% or less to the hydronium ions. The hydronium ion and water then form the boundary limits of the weak acids. Above hydronium, the acids are strong; below water, the compounds do not exhibit any observable acidic

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