Page 373 - Modern Analytical Chemistry

1400-CH09 9/9/99 2:14 PM Page 356

356 Modern Analytical Chemistry

+
Making appropriate substitutions for the moles of Ag , KCl, and NaBr gives the
following equation.
g KCl g NaBr
MV Ag = +
Ag
FW KCl FW NaBr
Since the sample contains just KCl and NaBr, we know that
g NaBr = 0.3172 g – g KCl
and
g KCl 0 3172 g – g KCl
.
MV Ag = +
Ag
FW KCl FW NaBr
Solving, we find
g KCl . 0 3172 g - g KCl
(. 0 1120 M)(0.03614 L) = +
74.551 g/mol 102.89 g/mol
3-
2 -
´
. 4 048 ´ 10 - 3 = . 1 341 ´ 10 (g KCl) + 3.083 ´ 3 - - . 9 719 10 (g KCl)
10
3
-
3.691 ´ 10 (g KCl) = . 9 650 ´ 10 4 -
g KCl = . 0 2614 g

that there is 0.2614 g of KCl and
g NaBr = 0.3172 g – 0.2614 g = 0.0558 g
0.0558 g of NaBr. The weight percents for the two analytes, therefore, are
0 2614 g KCl
.
´ 100 = 82 41.% w/w KCl
0.3172 g

.
0 0558 g NaBr
´ 100 = 17 59.% w/w NaBr
0.3172 g

–
The analysis for I using the Volhard method requires a back titration. A typical
calculation is shown in the following example.

9
EXAMPLE .18
–
The %w/w I in a 0.6712-g sample was determined by a Volhard titration. After
adding 50.00 mL of 0.05619 M AgNO 3 and allowing the precipitate to form, the
remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL
–
to reach the end point. Report the %w/w I in the sample.
SOLUTION
Conservation of charge for this back titration requires that
–
+
Moles Ag = moles I + moles SCN –
–
–
+
Making appropriate substitutions for moles of Ag , I , and SCN leaves us with
g I –
MV Ag = + M SCN SCN
V
Ag
AW I –