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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
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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
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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
𝑳
𝑨
𝑅=
𝜌∙𝐿
𝐴
A = cross sectional area
L = length over which resistance is measured
ρ = resistivity which is a material property
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𝑤𝑡% 𝑁𝑎𝐶𝑙 =
𝑊𝑁𝑎𝐶𝑙
𝑚𝑁𝑎𝐶𝑙
∙ 100% =
∙ 100%
𝑊𝑁𝑎𝐶𝑙 + 𝑊𝐻2 𝑂
𝑚𝑁𝑎𝐶𝑙 + 𝑚𝐻2 𝑂
calculating percent weight
A mixture contains 19 grams of water and 1 gram of NaCl. What is the weight
percent of NaCl?
𝑤𝑡 % 𝑁𝑎𝐶𝑙 =
𝑊𝑁𝑎𝐶𝑙
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑁𝑎𝐶𝑙
∙ 100%
∙ 100% =
𝑊𝑁𝑎𝐶𝑙 + 𝑊𝐻2 𝑂
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑚𝑖𝑥𝑡𝑢𝑟𝑒
=
𝑚𝑁𝑎𝐶𝑙 ∙ 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
∙ 100%
𝑚𝑁𝑎𝐶𝑙 ∙ 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 + 𝑚𝐻2 𝑂 ∙ 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
=
𝑚𝑁𝑎𝐶𝑙
∙ 100%
𝑚𝑁𝑎𝐶𝑙 + 𝑚𝐻2 𝑂
=
1 𝑔𝑟𝑎𝑚
1 𝑔𝑟𝑎𝑚
∙ 100%
∙ 100% =
20 𝑔𝑟𝑎𝑚𝑠
1 𝑔𝑟𝑎𝑚 + 19 𝑔𝑟𝑎𝑚𝑠
𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 9.81
𝑚
𝑠2
𝑤𝑡 % 𝑁𝑎𝐶𝑙 = 5%
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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 = 𝜌𝐻2 𝑂 = 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
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Example
How much salt would you need to add to 2L of water to have a
concentration of 3.5 wt% NaCl?
unknown
𝑤𝑡% 𝑁𝑎𝐶𝑙 =
3.5%
𝑚𝑁𝑎𝐶𝑙
∙ 100%
𝑚𝑁𝑎𝐶𝑙 + 𝑚𝐻2 𝑂
unknown
3.5% =
𝑚𝐻2 𝑂 = 2𝐿 ∙
1𝑘𝑔
= 2𝑘𝑔
𝐿
mass of 2L of water
𝑚𝑁𝑎𝐶𝑙
∙ 100%
𝑚𝑁𝑎𝐶𝑙 + 2𝑘𝑔
0.035 𝑚𝑁𝑎𝐶𝑙 + 2𝑘𝑔 = 𝑚𝑁𝑎𝐶𝑙
0.035 ∙ 𝑚𝑁𝑎𝐶𝑙 +0.07𝑘𝑔 = 𝑚𝑁𝑎𝐶𝑙
1 − 0.035 ∙ 𝑚𝑁𝑎𝐶𝑙 = 0.07𝑘𝑔
𝑚𝑁𝑎𝐶𝑙 =
0.07𝑘𝑔
= 0.0725𝑘𝑔 = 72.5𝑔
1 − 0.035
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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 𝐻2 𝑂(𝑙) + 2𝑒 − → 𝐻2 𝑔 + 2𝑂𝐻− (𝑎𝑞)
H
H2O
H
H2O
OH-
ClNa+
Cl-
Cl-
Na+
Na+
Na+
Na+ is a spectator ion
oxidation occurs at the
positively charged anode:
2 𝐶𝑙 − 𝑎𝑞 → 𝐶𝑙2 𝑔 + 2𝑒 −
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useful information for reaction calculations
𝑔
Atomic Weight of Na = 22.99 𝑚𝑜𝑙
Atomic Weight of Cl = 35.45
𝑔
𝑚𝑜𝑙
𝑔
Atomic Weight of NaCl = 58.44 𝑚𝑜𝑙
Avogadro’s Number =
6.022 x 1023
𝑚𝑜𝑙
1 Coulomb = 6.24 x 1018 electrons
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living with the lab
𝑔
Atomic Weight of NaCl = 58.44 𝑚𝑜𝑙
Avogadro’s Number =
6.022 10 23
𝑚𝑜𝑙
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
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living with the lab
1 Coulomb = 6.24 x 1018 electrons
1A=
1 𝐶𝑜𝑢𝑙𝑜𝑚𝑏
𝑠
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 𝐻2 𝑂(𝑙) + 2𝑒 − → 𝐻2 𝑔 + 2𝑂𝐻− (𝑎𝑞)
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.
𝑒−
𝐴
= 0.1𝑚𝐴 ∙
∙
𝑠
1000 𝑚𝐴
𝐶𝑜𝑢𝑙𝑜𝑚𝑏
6.24 10 18 𝑒 −
𝑠
= 6.24 10
∙
𝐴
𝐶𝑜𝑢𝑙𝑜𝑚𝑏
14
𝑒−
𝑠
Now, find the number of H2 molecules over a 1 minute period.
𝐻2 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 = 6.24 10
14
1 𝐻2 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒
𝑒 − 60𝑠
∙
∙ 1 𝑚𝑖𝑛 ∙
= 1.87 10
2 𝑒−
𝑠 𝑚𝑖𝑛
16
𝐻2 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
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