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General Chemistry Principles, Patterns, and Applications, 2011

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

The equilibrium constant expression for the ionization of HCN is as follows:<br />

Equation 16.21<br />

[ ]<br />

Ka = [H+][CN-] HCN<br />

The corresponding expression for the reaction of cyanide with water is as follows:<br />

Equation 16.22<br />

Kb = [OH-][ HCN ][CN-]<br />

If we add Equation 16.19 <strong>and</strong> Equation 16.20, we obtain the following (recall fromChapter 15 "Chemical<br />

Equilibrium" that the equilibrium constant for the sum of two reactions is the product of the equilibrium<br />

constants for the individual reactions):<br />

HCN aq<br />

( ) H + ( aq) + CN - ( aq)CN - ( aq) + H 2O( l) OH - ( aq) + HCN aq<br />

( ) + OH - ( aq)Ka = [H +][CN-] /[ HCN ]Kb = [<br />

OH-] [ HCN ] /[CN-]K = Ka ´ Kb = [H+][OH-]<br />

H + aq<br />

( )H 2O l<br />

In this case, the sum of the reactions described by Ka <strong>and</strong> Kb is the equation for the autoionization of<br />

water, <strong>and</strong> the product of the two equilibrium constants is Kw:<br />

Equation 16.23<br />

K a K b = K w<br />

Thus if we know either Ka for an acid or Kb for its conjugate base, we can calculate the other equilibrium<br />

constant for any conjugate acid–base pair.<br />

Just as with pH, pOH, <strong>and</strong> pKw, we can use negative logarithms to avoid exponential notation in writing<br />

acid <strong>and</strong> base ionization constants, by defining pKa as follows:<br />

Equation 16.24<br />

pK a = −log 10 K a<br />

Equation 16.25<br />

Ka =10 - pKa<br />

<strong>and</strong> pKb as<br />

Equation 16.26<br />

pK b = −log 10 K b<br />

Equation 16.27<br />

Kb = 10 - pKb<br />

( )<br />

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

Saylor.org<br />

1448

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