Tltratlon Reactions
197
moles of H 2 SO 4 - ( '
m
"'
e ^P,
4
,) (5 395 x lO'
3 moles NaOH)
V 2 moles NaOH/
= 2 698 x 10! moles H 2 SO 4
3 The moles of H 2 SO 4 used were dissolved in 42 67 ml of solution Therefore,
the molanty is
2 698 x 10
3 moles H 2 SO 4
„,„„ moles H 2 SO 4
molanty =
^T^r,
- 0 06322
0 04267 liters
liter
You have standardized the H 2 SO 4 solution and found it to be 0 06322 M H 2 SO 4
Analysis
Sometimes it is not possible to perform a "direct" titration as illustrated in the
last two problems In an indirect" titration, a known (but excess) volume of
standard solution is used to insure a complete reaction with whatever is being
analyzed for, then, at the end of the analysis, the unused quantity of the standard solution is determined by what is called aback-titration For example, such
a procedure is useful for the determination of a gas that can be bubbled through
the standard reaction solution Another example is illustrated by the next problem
PROBLEM
A 0 3312 g sample of impure Na 2 CO 3 is dissolved in exactly 50 00 ml of the H 2 SO 4
solution standardized in the preceding problem There is CO 2 gas liberated in this
reaction
Na 2 CO 3 + H 2 SO 4 -» Na 2 SO 4 + CO 2 t + H 2 O
CO$ + 2H
+ ^ CO 2 t + H 2 O
The CO 2 gas is totally driven off by gentle heating, so that it will not interfere with
the endpomt determination The unused portion of the H 2 SO 4 is then titrated The
back titration reaction is
2NaOH + H 2 SO 4 ^ Na 2 SO 4 + 2H 2 O
OH- + H
+ ^ H 2 O
The back-titration requires 9 36 ml of 0 1079 M NaOH Calculate the percentage of
Na 2 CO 3 in the original sample
SOLUTION:
At the final endpomt the H 2 SO 4 has been exactly used up by two reactions one
with NaOH, and the other with Na 2 CO 3 We know the moles of H 2 SO 4 and NaOH
involved because we know the volumes and concentrations of their solutions If
we subtract from the total moles of H 2 SO 4 the number of moles used by the
197
moles of H 2 SO 4 - ( '
m
"'
e ^P,
4
,) (5 395 x lO'
3 moles NaOH)
V 2 moles NaOH/
= 2 698 x 10! moles H 2 SO 4
3 The moles of H 2 SO 4 used were dissolved in 42 67 ml of solution Therefore,
the molanty is
2 698 x 10
3 moles H 2 SO 4
„,„„ moles H 2 SO 4
molanty =
^T^r,
- 0 06322
0 04267 liters
liter
You have standardized the H 2 SO 4 solution and found it to be 0 06322 M H 2 SO 4
Analysis
Sometimes it is not possible to perform a "direct" titration as illustrated in the
last two problems In an indirect" titration, a known (but excess) volume of
standard solution is used to insure a complete reaction with whatever is being
analyzed for, then, at the end of the analysis, the unused quantity of the standard solution is determined by what is called aback-titration For example, such
a procedure is useful for the determination of a gas that can be bubbled through
the standard reaction solution Another example is illustrated by the next problem
PROBLEM
A 0 3312 g sample of impure Na 2 CO 3 is dissolved in exactly 50 00 ml of the H 2 SO 4
solution standardized in the preceding problem There is CO 2 gas liberated in this
reaction
Na 2 CO 3 + H 2 SO 4 -» Na 2 SO 4 + CO 2 t + H 2 O
CO$ + 2H
+ ^ CO 2 t + H 2 O
The CO 2 gas is totally driven off by gentle heating, so that it will not interfere with
the endpomt determination The unused portion of the H 2 SO 4 is then titrated The
back titration reaction is
2NaOH + H 2 SO 4 ^ Na 2 SO 4 + 2H 2 O
OH- + H
+ ^ H 2 O
The back-titration requires 9 36 ml of 0 1079 M NaOH Calculate the percentage of
Na 2 CO 3 in the original sample
SOLUTION:
At the final endpomt the H 2 SO 4 has been exactly used up by two reactions one
with NaOH, and the other with Na 2 CO 3 We know the moles of H 2 SO 4 and NaOH
involved because we know the volumes and concentrations of their solutions If
we subtract from the total moles of H 2 SO 4 the number of moles used by the