CHEM 30A Indepdendent Project
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Uploaded by ConstableRainToad32
Independent Project
Maili Nguyen Tingle
Introductory Chemistry
Professor Irina Slobodov
December 1, 2023
1
Title:
Dilution to customized concentration of Hydrochloric Acid and Iron (III) Chloride
Introduction/Objective:
The ability to dilute solutions to the needed and appropriate concentrations will be reflected in this lab. Today, we will be working with the following elements HCL (hydrochloric acid) and
FeCl
❑
3
(Iron (III) Chloride). The solution in 100 mL should be homogenous and include 0.095 M FeCl
❑
3
and 0.125 M HCL in diluted proportions.
Lab Materials:
1.
A 100 mL volumetric flask
2.
Tablespoon
3.
50 ml graduated cylinder
4.
Digital scale
5.
50 mL beaker
6.
Protective eyewear
7.
Full length pants and full length shirt
8.
Isotonic water (H2O)
Protocol:
1.
The 50 mL graduated cylinder can hold up to 4.2 mL of 3 M HCl 2.
Add the 4.2 mL of 3 M HCl into the 100 mL volumetric flask
3.
Measure 2.6 g of FeCl
❑
3
∗
6
H
2
O
using a tablespoon and digital scale
2
4.
Add the measure 2.6 g of FeCl
❑
3
∗
6
H
2
O
into the 100 mL volumetric flask with the HCl
5.
Add H2O (pure water) into the 100 mL volumetric flask until the mark (100 mL). 6.
Swirl the volumetric flask to form a homogeneous solution
7.
Record the final concentration
Calculations
:
Required Volume of 3 HCl for dilution:
M
1
∗
V
1
=
M
2
∗
V
2
3.0
M HCl
∗(
V
1
)=
0.125
M
∗
100
mL
V
1
=
0.125
M
∗
100
3
=
4.2
mL
4.2 mL of HCl is required for the dilution
Required number of moles of solid FeCl3 * 6H2O:
0.095
M FeCl
3
=
x
0.100
L
=
0.0095
moles FeCl3
0.0095 moles FeCl3 is required
Mole to Gram conversion of Solid FeCl3 * 6H2O:
0.0095
FeCl
3
∗
270.38
1
mol FeCl
3
∗
6
H
2
O
=
2.6
gFeCl
3
∗
6
H
2
O
2.6 g FeCL3 * 6H2O is required
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Solid lead (II) nitrate is slowly added to 125 mL of a 0.190 M potassium hydroxide solution until the concentration
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