salt water chemistry

Report
living with the lab
salt water mixtures
the world’s oceans contain an average of about 3.5% salt by weight
(35 grams of salt for every kilogram of seawater)
Image credit: NASA/GSFC/JPL-Caltech
© 2011 David Hall
living with the lab
concentrations used for fishtank project
we will calibrate our salinity sensor using small NaCl concentrations
• 0.00% weight NaCl – we will actually use deionized water (DI water)
• 0.05% weight NaCl
• 0.10% weight NaCl
• 0.15% weight NaCl
2
living with the lab
why not just use tap water?
• standard tap water contains dissolved ions such as sodium, calcium, iron,
copper and bromide
• the electrical resistivity of tap water can vary widely (15kΩ-cm is typical)
• ions have been removed from DI water resulting in a much higher
resistivity (18MΩ-cm is typical)
Did you notice the strange units on resistivity (Ω-cm)? The electrical resistance R of a
body (measured in Ω) is related to the electrical resistivity ρ (measured in Ω-cm) as


=
∙

A = cross sectional area
L = length over which resistance is measured
ρ = resistivity which is a material property
3
living with the lab
%  =


∙ 100% =
∙ 100%
 + 2 
 + 2 
calculating percent weight
A mixture contains 19 grams of water and 1 gram of NaCl. What is the weight
percent of NaCl?
 %  =

ℎ  
∙ 100%
∙ 100% =
 + 2 
ℎ  
=
 ∙ 
∙ 100%
 ∙  + 2  ∙ 
=

∙ 100%
 + 2 
=
1 
1 
∙ 100%
∙ 100% =
20 
1  + 19 
 = 9.81

2
 %  = 5%
4
living with the lab
Class Problem If you add 9.5 grams of NaCl to 5 gallons of water, what weight
percent of salt will the mixture contain?
Useful conversion factors:
Density of water = 2  = 1000 kg/m3 = 1 g/cm3 = 1 kg/L at 4oC (maximum density)
1 cm3 = 1 cc = 1 ml
1 L = 0.001 m3
1 gallon = 3.7853 L
5
living with the lab
Example
How much salt would you need to add to 2L of water to have a
concentration of 3.5 wt% NaCl?
unknown
%  =
3.5%

∙ 100%
 + 2 
unknown
3.5% =
2  = 2 ∙
1
= 2

mass of 2L of water

∙ 100%
 + 2
0.035  + 2 = 
0.035 ∙  +0.07 = 
1 − 0.035 ∙  = 0.07
 =
0.07
= 0.0725 = 72.5
1 − 0.035
6
living with the lab
recall the reactions at the electrodes
5V
e-
e-
10 kΩ
anode – oxidation
cathode – reduction
(loss of electrons)
e-
(gain of electrons)
e-
Cl-
Cl-
Na+
Cl2
Cl-
Na+
Cl-
Cl
Cl-
ClCl-
Na+
Na+
Cl
Cl-
reduction occurs at the
negatively charged cathode:
Na+
Na+
Na+
Cl-
Na+
ion migration
OHee-
2 2 () + 2 − → 2  + 2− ()
H
H2O
H
H2O
OH-
ClNa+
Cl-
Cl-
Na+
Na+
Na+
Na+ is a spectator ion
oxidation occurs at the
positively charged anode:
2  −  → 2  + 2 −
7
living with the lab
useful information for reaction calculations

Atomic Weight of Na = 22.99 
Atomic Weight of Cl = 35.45



Atomic Weight of NaCl = 58.44 
Avogadro’s Number =
6.022 x 1023

1 Coulomb = 6.24 x 1018 electrons
8
living with the lab

Atomic Weight of NaCl = 58.44 
Avogadro’s Number =
6.022 10 23

Class Problem Assume you have 5 gallons of water to which you add salt to
create a mixture with 0.2 wt% NaCl. Determine:
(a) the mass of the water
(b) the mass of the salt
(c) the number of moles of NaCl
(d) the number of Cl- ions
9
living with the lab
1 Coulomb = 6.24 x 1018 electrons
1A=
1 

Example
If a constant current of 0.1mA passes through the probes of the
conductivity sensor, how many H2 gas molecules would be formed over a 1 minute period?
HINT: Use the definition of an amp and a Coulomb along with the chemical reaction at the cathode.
2 2 () + 2 − → 2  + 2− ()
two electrons pass through the circuit for each H2 gas molecule formed
First, find the number of electrons that pass through the sensor per second.
−

= 0.1 ∙
∙

1000 

6.24 10 18  −

= 6.24 10
∙


14
−

Now, find the number of H2 molecules over a 1 minute period.
2  = 6.24 10
14
1 2 
 − 60
∙
∙ 1  ∙
= 1.87 10
2 −
 
16
2 
10

similar documents