Page 187 - Modern physical chemistry

8.3 Conditions Determining Equilibria 179

This value may be corrected by using it together with the closest initial x in a repetition
of the calculation.

Example8.S

Calculate m H+ in 0.010 m alanine using ionization constants from tables 8.2 and 8.l.
Consider the Qy's to equal 1, The equation for ionization of the anion,

CH3 CHNH3 0HCOO- ( ) CH 3 CHNH3 COO + OH-,

is abbreviated as

while the equation for ionization of the cation,
CH 3 CHNH 3 COOH+ ( ) CH3 CHNH 3 COO + H+,

is abbreviated as

Molecule Z is called a zwitterion since in it H+ has migrated from the carboxyl to the
amino group, leaving the first region anionic and the second region cationic.
In the bulk of the solution, electrical neutrality prevails; thus the concentration of
positive charge equals the concentration of negative charge:

m ZH + + m H + = m ZOH - + m OH - .
But as long as the solution of alanine is concentrated enough, mH+ is small with respect
to 1nzH+ and mOIr is small with respect to 1nzoIr. The preceding equation then reduces to

m ZH + = m ZOH - •
From table 8.2 with each Q r = 1, one obtains

mzmow = 7.47 x 10-5 m
m zow

and
mzmH + = 4.57 x 10-3 m.
m ZH +
Dividing the second equation by the first and canceling the molalities that are equal yields
3
mH+ = 4.57 x 10- = 6.12 x 10.
mow 7.47 x 10- 5
But in the aqueous solution at 25° C, table 8.1 tells us that
m + m OH - = 1.008 x 10- 14 m 2 •
H
So we have
2
2
m H + =( m H + ~mH+mOW )=(6.12x10Xl.OO8x10-14 m )=6.17 x 10- 13 m 2
mow}