Page 335 - Modern Analytical Chemistry

1400-CH09 9/9/99 2:13 PM Page 318

318 Modern Analytical Chemistry

concentration of NH 3 is 0.0100 M. Using these values, we calculate that the con-
ditional formation constant is
16
4– a 2+ ´K f = (0.35)(0.0881)(2.9 ´10 ) = 8.9 ´10 14
K f ˝ = a Y ´ Cd
Because K f ˝ is so large, we treat the titration reaction as though it proceeds to
completion.
The first task in calculating the titration curve is to determine the volume of
EDTA needed to reach the equivalence point. At the equivalence point we know
that
Moles EDTA = moles Cd 2+
or

M EDTAV EDTA = M CdV Cd
Solving for the volume of EDTA
5
MV Cd (.500 ´10 -3 M)( .00 mL)
Cd
V EDTA = = = 25 .0 mL
M EDTA . 0 0100 M
shows us that 25.0 mL of EDTA is needed to reach the equivalence point.
Before the equivalence point, Cd 2+ is in excess, and pCd is determined by the
concentration of free Cd 2+ remaining in solution. Not all the untitrated Cd 2+ is free
(some is complexed with NH 3 ), so we will have to account for the presence of NH 3 .
2+
For example, after adding 5.0 mL of EDTA, the total concentration of Cd is
V
moles excess Cd 2 + MV Cd - M EDTA EDTA
Cd
C Cd = =
total volume V Cd + V EDTA
(. ´10 -3 M 50 0)( . mL - 0 0100) (. M)(5.0 mL)
5 00
= = 364. ´ 10 -3 M
50 0 mL + 5.0 mL
.
2+
To calculate the concentration of free Cd we use equation 9.14.
–4
–3
2+
C
[Cd ]= a Cd ´ Cd = (0.0881)(3.64 ´10 M) = 3.21 ´10 M
2+
Thus, pCd is
–4
2+
pCd = –log[Cd ] = –log(3.21 ´10 ) = 3.49
2–
At the equivalence point, all the Cd 2+ initially present is now present as CdY .
2+
The concentration of Cd , therefore, is determined by the dissociation of the CdY 2–
complex. To find pCd we must first calculate the concentration of the complex.
initial moles Cd 2+ MV
Cd Cd
[CdY 2 - ] = =
total volume V Cd +V EDTA
)( . mL
(. 500 ´10 - 3 M 500 ) - 3
= = . 333 ´10 M
. mL
50 . mL +0 25 0
Letting the variable x represent the concentration of Cd 2+ due to the dissociation of
2–
the CdY complex, we have
2
-
[ CdY ] . 333 ´10 -3 -x 14
K f ¢¢ = = = . 894 ´ 10
xx()
Cd
CC EDTA ()
x =C Cd = . 193 ´ 10 - 9 M