Lecture Notes Chapter 4, acids-bases part 2 PDF

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LECTURE NOTES Chapter 4 ACIDS AND BASES – part 2 1. Salts of weak acids and bases, hydrolysis When we think of bases, we would normally think of strong bases such as NaOH, Ca(OH) 2, or maybe a weak base such as ammonia. Perhaps surprisingly, salts of weak acids, for instance Na2CO3 and Na3PO4 also are bases. Sodium phosphate in fact is almost a strong base. It is the ingredient in hard surface cleaners called TSP: trisodium phosphate. Conversely, ammonium chloride, NH 4Cl, the salt of the weak base ammonia is a weak acid in solution. Na3PO4 is a base because the phosphate ion is the conjugate base of a weak acid, and ammonium chloride is an acid because the ammonium ion is the conjugate acid of a weak base. Anions of weak acids are bases, cations of weak bases are acids. In order to understand the reason for this behaviour, let’s return to the weak acid equilibrium. HA + H2O  A + H3O +

K a=

[ A− ][ H 3 O+ ] [ HA ]

HA/A is the conjugate acid base pair. The acid HA has donated its proton, and A , its conjugate base, now in turn has the ability to take up a proton again, accepting the proton from water as any other base: −

[ HA ][ OH ] K b= − [A ]

A + H2O  HA + OH When we dissolve a salt such as sodium acetate, which we can write as NaAc, in water, the salt, as all salts, will be fully dissociated: NaAc(s)  Na+(aq) + Ac(aq) We know Ac (short for CH3COO) as the conjugate base of acetic acid, HAc (short for CH3COOH), and therefore the acetate ion Ac is a proton acceptor. When sodium acetate, NaAc is dissolved in water the resulting actetate ion will accept a proton from water just like any other base:  HAc+OH Ac +H2O −

K b=

[ HAc ][ OH ] − [ Ac ]

When we compare this expression for K b, the base dissociation constant of the basic ion Ac , to Ka of its conjugate acid: −

K a=

+

[ Ac ][ H ] [ HAc ]

HAc  Ac + H+ we find the relation between K a of a weak acid and Kb of its conjugate base: KaxKb = Kw Kw K b= Ka

(1)

Kw K a= Kb

(2) Using the definition of pK = logK, eq. 1 is often written as: pKa + pKb = pKw (3) For instance, acetic acid, HAc has K a = 1.8x105, or pKa = 4.74. Then, from eq. 2 or 3 we can calculate the base dissociation constant of its conjugate base, the acetate ion Ac : Kb = 5.6x1010, pKb = 14  4.74 = 9.26. We note that Ac  is a weaker base than HAc is an acid, but it is nevertheless a base, and for instance a solution of sodium acetate (NaAc) in water will be a basic solution. Just as the acetate ion, as a base, makes sodium acetate solutions basic, so does the ammonium ion, the conjugate acid of the weak base ammonia make ammonium chloride solutions acidic: NH4Cl(s)  NH4+ + Cl (complete) NH4+ + H2O NH3+ + H3O+

K a=

+

[ NH 3 ][ H 3 O ] [ Ac− ]

For ammonia (K b = 1.8x105, pKb = 4.74, by accident of nature the same numerical values as for acetic acid, but of course ammonia is a base), and the same equations 1, 2 and 3 hold for this conjugate acid base pair: Ka(NH4+) = 5.6x1010, pKa = 9.26. Some other examples of weak acid/base conjugate pairs are given below in table 3. bases such as NaOH and Ca(OH) 2. Even the carbonate ion, CO32 is a stronger base than ammonia.

Table 1 A comparison of some conjugate acid-base pairs Kb pKb Ka pKa Conjugate base HSO4 9.1E13 12.04 1.1E-2 1.96 SO42 H3PO4 7.1E3 2.15 H2PO4 1.4E12 11.85 HNO2 7.2E4 3.14 NO2 1.4E11 10.86 HF 6.6E4 3.18 F 1.5E11 10.82 HCOOH 1.8E4 3.74 HCOO 5.5E11 10.26 CH3COOH 1.8E5 4.74 CH3COO 5.5E10 9.26 H2CO3 4.4E7 6.36 HCO3 2.3E8 7.64   H2PO4 6.3E8 1.6E7 7.20 HPO4 6.80 HClO 2.9E8 7.54 ClO 3.5E7 6.46 HCN 6.2E10 9.21 CN 1.6E5 4.79 + NH4 5.6E10 9.26 1.8E5 NH3 4.74 HCO3 4.7E11 10.33 CO32 2.1E4 3.67 CH3NH3+ 2.4E11 10.62 CH3NH2 4.2E4 3.38 CH3NH2CH3+ 1.8E11 10.74 CH3NHCH3 5.5E4 3.26 HPO42 4.2E13 12.38 PO43 2.4E2 1.62 Acid

Exercise: name all species in columns 1 and 4 of table 1, and write the equilibrium for each conjugate pair. Table 1 is worth some extra attention. In the table, the acids, in the first column, are listed in order of decreasing acid strength: K a decreases (and therefore pK a increases) from the almost strong acid HSO 4 to the almost imperceptibly weak acid HPO42. The weaker the acid, the stronger its conjugate base. Therefore in the table, the base strength increases from top to bottom: SO 42 is so weak a base that we don’t notice its basic character at all, and PO 43, the conjugate base of the very weak acid HPO4 is practically a strong base. Note also that NH4+, the conjugate acid of NH3, is listed in the acid column, and so are the other conjugate acids of amines, i.e., the methylammonium ion CH3NH3+ and the dimethylammonium ion, CH3NH2NH3+. It is interesting to note that the phosphate ion, PO 43 is the strongest base after the strong Equally interesting is the observation, derived directly from eq. 3 that a weak acid with pKa>7 will have a conjugate base with a stronger base strength, pK b...