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General Chemistry Principles, Patterns, and Applications, 2011

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Weak bases react with water to produce the hydroxide ion, as shown in the following general equation,<br />

where B is the parent base <strong>and</strong> BH + is its conjugate acid:<br />

Equation 16.17<br />

( ) + H2O( l) BH + ( aq) +OH - ( aq)<br />

B aq<br />

The equilibrium constant for this reaction is the base ionization constant (K b), also called the base<br />

dissociation constant:<br />

Equation 16.18<br />

Kb = K[H2O =<br />

] [ BH+][OH-] [ B]<br />

Once again, the concentration of water is constant, so it does not appear in the equilibrium constant<br />

expression; instead, it is included in the Kb. The larger the Kb, the stronger the base <strong>and</strong> the higher the<br />

OH − concentration at equilibrium. The values ofKb for a number of common weak bases are given in Table<br />

16.3 "Values of ".<br />

Table 16.3 Values of Kb, pKb, Ka, <strong>and</strong> pKa for Selected Weak Bases (B) <strong>and</strong> Their Conjugate Acids (BH + )<br />

Base B K b pKb BH + K a pKa<br />

hydroxide ion OH − 1.0 0.00* H2O 1.0 × 10 −14 14.00<br />

phosphate ion PO4 3− 2.1 × 10 −2 1.68 HPO4 2− 4.8 × 10 −13 12.32<br />

dimethylamine (CH3)2NH 5.4 × 10 −4 3.27 (CH3)2NH2 + 1.9 × 10 −11 10.73<br />

methylamine CH3NH2 4.6 × 10 −4 3.34 CH3NH3 + 2.2 × 10 −11 10.66<br />

trimethylamine (CH3)3N 6.3 × 10 −5 4.20 (CH3)3NH + 1.6 × 10 −10 9.80<br />

ammonia NH3 1.8 × 10 −5 4.75 NH4 + 5.6 × 10 −10 9.25<br />

pyridine C5H5N 1.7 × 10 −9 8.77 C5H5NH + 5.9 × 10 −6 5.23<br />

aniline C6H5NH2 7.4 × 10 −10 9.13 C6H5NH3 + 1.3 × 10 −5 4.87<br />

water H2O 1.0 × 10 −14 14.00 H3O + 1.0* 0.00<br />

*As in Table 16.2 "Values of ".<br />

There is a simple relationship between the magnitude of Ka for an acid <strong>and</strong> Kb for its conjugate base.<br />

Consider, for example, the ionization of hydrocyanic acid (HCN) in water to produce an acidic solution,<br />

<strong>and</strong> the reaction of CN − with water to produce a basic solution:<br />

Equation 16.19<br />

HCN aq<br />

( ) H + ( aq) + CN -( aq)<br />

Equation 16.20<br />

Saylor URL: http://www.saylor.org/books<br />

Saylor.org<br />

1447

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