Chemistry Reference
In-Depth Information
0.05005 g CaCO
3
1 mL 1
M
HCl
1 mL 1
M
NaOH
In the experiment, approximately 1.5 g of sample was weighed and
added to 100 mL of water in a conical flask and 50.0 mL of 1
M
hydrochloric acid was added by pipette. The mixture was boiled gently
for 2 minutes and cooled and the unreacted HCl was titrated with 1
M
NaOH using methyl orange as indicator. The entire procedure was repeated
omitting the sample and the % w/w CaCO
3
in the sample was determined.
Results
Weight of chalk
1.5961 g
Volume of 1
M
(
f
0.996) HCl
50.00 mL
Volume of 1
M
(
f
1.012) NaOH
18.50 mL
Since neither volumetric solution is factor 1.000, the experimental volumes
must be modified by the factor to obtain the factor 1.000 volumes.
Volume of HCl available
(50.0
0.996)
Volume of NaOH in excess
(18.50
1.012)
Therefore, the volume reacting with chalk is given by
(50.0
0.996)
(18.50
1.012)
31.08 mL
From the equivalent,
1 mL 1
M
HCl or NaOH
0.05005 g CaCO
3
Therefore,
0.05005) g CaCO
3
1.5554 g CaCO
3
However, 1.5931 g of sample was weighed. Therefore, the percentage of
calcium carbonate is
31.08 mL 1
M
solution
(31.08
1.5554
----—
100
97.6% w/w
1.5931
Using these same data, the calculation can be repeated, but this time
taking account of the blank determination. If an assay requires a blank,
then the concentration of the reagent (hydrochloric acid in this case) must
change in the course of the assay; therefore, the volume and factor of the
