Page 196 - Basic physical chemistry for the atmospheric sciences

Basic physical chemistry
1 8 2
Kh( = Kw!Ka) for the hydrolysis of C2H302 - ( aq) has the
same value as that calculated above for NH3, namely ,
5 . 6 x 1 0 - 1 0- Therefore, the amount of C2H 02 - ( aq) hy­
3
d r olyzed is also 2.8 x 1 0 - 5 M. Therefore,
Fraction of NH4C2H302 hydrolyzed

Amount of NH4+ (aq) hydrolyzed + a mount of C2H302 ( aq) hydrolyzed
-
Original amount of N H 4C2H302
.
2(2 8 x 1 0 - 5 )
0.0050
= I . I x 10 - 2 or l . 1 %

5 . 1 6 . pH = I . I . Hint: Every mole of Na ( aq) releases I mole of
+
l
negative ions [e. g . , c1 - (aq) or O H - (aq)] . Similar y , every
2
mole of Ca + (aq) releases two moles of negative ion s , and
every mole of Cl ( aq) releases one mole of positive ions
-
[e . g . , Na+ (aq) or H + ( aq)] , and every mole of Sol - ( aq)
releases two moles of positive ions . To solve the problem,
find the total number of moles of negative charge released
and the total number of moles of positive charge released ,
and hence the net number of moles of charge released.
Since the only other charge carriers are H + (aq) and
OH - (aq) , the net number of moles of positive or negative
charge must be in the form of H + (aq) or oH - ( aq) ions ; this
will permit determination of the pH of the solution.
5 . 1 7 . pH = 8 . 9 5 . Solution: N H 4 + (from N H 4Cl) and NH3 are an
acid-base conjugate pair. From Eq. (5 .26)

[NHn [H 0+ ( aq)]
--- 3
[NH ] Ka(NH.t )
3
Therefore,

Since KaCNH.t ) = 5 6 0 x 1 0 - 10 and ,
.
[NHn
= 2 .00
[NH3]