320
Stoichiometry IV: Equivalent Weight and Normality
When dealing with normality and equivalents, it is not necessary to have
balanced chemical equations on which to base your calculations.
PROBLEM:
An HC1 solution is standardized by titration of a pure Na 2 CO 3 sample. Calculate
the normality of the HC1 solution if 41.30 ml are required to titrate 0.2153 g of
Na 2 CO 3 .
SOLUTION:
Because there are two equivalents/mole of Na 2 CO 3 , the equivalent weight is
-, = 53.00 —^7
2 equiv/
equiv
The number of equivalents of Na 2 CO 3 in the weighed sample is
°'
2153g
= 4.062 x 10~
3 equiv Na 2 CO 3
. -. e
53.00 -
8
equiv
At the endpoint,
equiv of Na 2 CO 3 = equiv of HC1
4.062 x 103 equiv = (0.04130 liter) N
liter) (
N = 0.09835 ^r^- = 0.09835 N
liter
PROBLEM:
A 0.5280 g sample of impure CaCO 3 is dissolved in 50.00 ml of 0.09835 N HC1.
After the reaction is complete and the CO 2 completely removed by warming,
the excess HC1 is titrated with 6.30 ml of 0.1052 N NaOH. Calculate the percentage of CaCO 3 in the original sample.
SOLUTION:
In this problem, the HC1 is used partly by CaCO 3 and partly by NaOH, but at the
endpoint
equiv of HC1 = equiv of CaCO 3 + equiv of NaOH
Substituting the given data into the equation, we obtain
( 0.09835 £HHiX\ (0.05000 liter) = (equiv CaCO 3 ) + (0.00630 liter) ( 0.1052 ^H!^
\
liter/
\
liter/
equiv CaCO 3 = 0.004918 - 0.000663 = 0.004255 equiv