Question 16.17: Calculate (a) Kb for the fluoride ion, (b) Ka for the ammoni...

Analyze We are asked to determine dissociation constants for F^{-}, the conjugate base of HF, and NH^{+}_{4}, the conjugate acid of NH_{3}.

Plan We can use the tabulated K values for HF and NH_{3} and the relationship between K_{a}   and   K_{b} to calculate the dissociation constants for their conjugates, F^{-} and NH^{+}_{4}.

Solve
(a) For the weak acid HF, Table 16.2 and Appendix D give K_{a} = 6.8 × 10^{-4}. We can use Equation 16.40 to calculate K_{b} for the conjugate base, F^{-}:

TABLE 16.2       Some Weak Acids in Water at 25 °C
Acid	Structural Formula*	Conjugate Base	K_{a}
Chlorous (HClO_{2})	H—O—Cl—O	ClO^{-}_{2}	1.0 × 10^{-2}
Hydrofluoric (HF)	H—F	F^{-}	6.8 × 10^{-4}
Nitrous (HNO_{2})	H—O—N=O	NO^{-}_{2}	4.5 × 10^{-4}
Benzoic (C_{6}H_{5}COOH)		C_{6}H_{5}COO^{-}	6.3 × 10^{-5}
Acetic (CH_{3}COOH)	H-O-\overset{\overset{O}{||} }{C} -\overset{\overset{H}{|} }{\underset{\underset{H}{|}}{C}} -H	CH_{3}COO^{-}	1.8 × 10^{-5}
Hypochlorous (HOCl)	H—O—Cl	OCl^{-}	3.0 × 10^{-8}
Hydrocyanic (HCN)	H—C≡N	CN^{-}	4.9 × 10^{-10}
Phenol (HOC_{6}H_{5})		C_{6}H_{5}O^{-}	1.3 × 10^{-10}
*The proton that ionizes is shown in red.

(b) For NH_{3}, Table 16.4 and Appendix D give K_{b} = 1.8 × 10^{-5}, and this value in Equation 16.40 gives us Ka for the conjugate acid, NH^{+}_{4}:

K_{a} × K_{b} = K_{w}          (for a conjugate acid-base pair)      [16.40]

TABLE 16.4          Some Weak Bases in Water at 25 °C
Base	Structural Formula*	Conjugate	K_{b}
Ammonia (NH_{3})	H-\overset{\cdot \cdot }{\underset{\underset{H}{|} }{N}} -H	NH^{+}_{4}	1.8 × 10^{-5}
Pyridine (C_{5}H_{5}N)		C_{5}H_{5}NH^{+}	1.7 × 10^{-9}
Hydroxylamine (HONH_{2})	H-\overset{\cdot \cdot }{\underset{\underset{H}{|} }{N}} -\overset{\cdot \cdot }{\underset{\cdot \cdot }{O}} H	HONH^{+}_{3}	1.1 × 10^{-8}
Methylamine (CH_{3}NH_{2})	H-\overset{\cdot \cdot }{\underset{\underset{H}{|} }{N}} -CH_{3}	CH_{3}NH^{+}_{3}	4.4 × 10^{-4}
Hydrosulfide ion (HS^{-})	H-\overset{\cdot \cdot }{\underset{\cdot \cdot }{S}} : ^{-}	H_{2}S	1.8 × 10^{-7}
Carbonate ion (CO^{2-}_{3})		HCO^{-}_{3}	1.8 × 10^{-4}
Hypochlorite ion (ClO^{-})	:\overset{\cdot \cdot }{\underset{\cdot \cdot }{Cl}} -\overset{\cdot \cdot }{\underset{\cdot \cdot }{O}} : ^{-}	HClO	3.3 × 10^{-7}
*The atom that accepts the proton is shown in blue.

Check The respective K values for F^{-} and NH^{+}_{4} are listed in Table 16.5, where we see that the values calculated here agree with those in Table 16.5.