examples of chemistry laws l. 1) concentration of a solute in a solution, the concentration of a...
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Sample Question 26.When 70 ml of 3.0-Molar Na 2 CO 3 is added to 30 ml of 1.0-Molar NaHCO 3 the resulting concentration of Na + is: a)2.0 M b)2.4 M c)4.0 M d)4.5 M e)7.0 MTRANSCRIPT
Examples of Chemistry Laws
L. 1) Concentration of a solute • In a solution, the concentration of a solute i is given by the
following formula: Concentrationi = Quantityi / Volumesolution
L. 2) Composition of strong electrolyte solutions • When an aqueous solution contains one or more strong
electrolytes, each one of them dissociates into their component ions. The quantity (in moles) of each ion in solution is given by the following formula: Quantityion = Quantityelectorlyte Coefficiention
Examples of Chemistry Laws
L.3) Conservation of mass• When mixing several solutions, the quantity of each solute
in the resulting solution is equal to the sum of the quantities of that solute in each of the original solutions.
L.4) Conservation of volume• When mixing several solutions, the volume of the resulting
solution is equal to the sum of the volumes of the original solutions (within scope).
Sample Question
26. When 70 ml of 3.0-Molar Na2CO3 is added to 30 ml of 1.0-Molar NaHCO3 the resulting concentration of Na+ is:a) 2.0 Mb) 2.4 Mc) 4.0 Md) 4.5 Me) 7.0 M
Answer to Sample Question• What is the concentration of Na+ in the final solution? Apply Law L.1. Need both the quantity
of Na+ and the final solution’s volume.
• What is the volume of the resulting solution? Use the conservation of volume law (Law L.4.) Volumefinal_solution = Sum of Volumesolution
i
So, we have Volumefinal_solution = 0.07 lit. + 0.03 lit. = 0.10 lit.
• What is the quantity of Na+ in the final solution? Use the law of conservation of mass (L.3). QuantityNa+ = Sum of QuantityNa+
i for each original solution i. Need QuantityNa+i
• What is QuantityNa+1? Use Law L.2, Concentration of ions in a strong electrolyte solution:
QuantityNa+1 = Quantityelectorlyte
1 CoefficientNa+electrolyte
CoefficientNa+electrolyte = 2, for electrolyte = Na2CO3. Need Quantityelectrolyte
1
• What is Quantityelectrolyte1? Use the concentration of solute formula (Law L.1.):
Quantityelectrolyte1 = Concentrationelectrolyte
1 Volumesolution1
So, we have Quantityelectrolyte1 = 3.0 M. 0.07 lit. = 0.21 moles
So, QuantityNa+1 = 0.21 x 2 = 0.42 moles.
Answer to Sample Question
• What is QuantityNa+2? Use Law L.2, Concentration of ions in a strong electrolyte
solution: QuantityNa+2 = Quantityelectorlyte
2 CoefficientNa+electrolyte
CoefficientNa+electrolyte = 1, for electrolyte = NaHCO3. Need Quantityelectrolyte
2
• What is Quantityelectrolyte2? Use the concentration of solute formula (Law L.1.):
Quantityelectrolyte2 = Concentrationelectrolyte
2 Volumesolution2
So, we have Quantityelectrolyte2 = 1.0 M. 0.03 lit. = 0.03 moles.
So, QuantityNa+2 = 0.03 x 1 = 0.03 moles.
So, we have QuantityNa+ in the final solution = 0.42 moles + 0.03 moles = 0.45 moles.
So, with QuantityNa+ and Volumefinal_solution, we have:ConcentrationNa+ = 0.45 moles / 0.10 lit. = 4.5 M.
So, the correct answer is (d)
Sample Question – Correct Answer
26. When 70 ml of 3.0-Molar Na2CO3 is added to 30 ml of 1.0-Molar NaHCO3 the resulting concentration of Na+ is:a) 2.0 Mb) 2.4 Mc) 4.0 Md) 4.5 Me) 7.0 M
A Simple Example
• When 70 ml of 3.0-Molar Na2CO3 is added to 30 ml of 1.0-Molar NaHCO3 the resulting concentration of Na+ is:
a) 2.0 Mb) 2.4 Mc) 4.0 Md) 4.5 Me) 7.0 M
Question Representation
volume
Mix
Aqueous Solution Aqueous Solution
Mixture
Na+
raw material
Na2CO3
3.0 M 0.07 lit
NaHCO3
0.03 lit
volume
1.0 M
conc.base base conc.
result
has-part
conc.
Question 26context
??
output
Background Knowledge
Chemistry laws:1. Concentration of a solute2. Composition of strong electrolyte solutions3. Conservation of mass4. Conservation of volumeetc.
Law 1: Concentration of a Solute
The concentration of a chemical in a mixture is the quantity of the chemical divided by the volume of the mixture.
Divide the quantity by the volume:<Quantity> / <Volume> = X *molar
Therefore, the concentration of <Chemical> in <Mixture> = X *molar
Explanation Template
Mixture
volumeconc.
Volume*liters
Concentration*molar
has-part
Chemical
Quantity*moles
quantity
Compute-Concentration Methodcontextinput output
Note: when this law is applied, the quantities
are automatically converted to the units-
of-measurement specified here
Law 2: Composition of Strong Electrolytes
Strong Electrolyte
Anion
has-part
Quantity*moles
quantity
Quantity*moles
quantity
Cation
Quantity*moles
quantity
Compute-Ions-in-Strong-Electrolytecontextinput output
Law 3: Conservation of MassConservation of Mass
contextinputoutput
Mix
Chemical1 Chemicaln
Chemical
raw-material
result
…
Quantity*moles
Quantity*moles
quantity quantity
Chemical
has-part
Quantity*moles
quantity
part-of
By the Law of Conservation of Mass, the quantity of a chemical in a mixture is the sum of the quantities of that chemical in the parts of the mix.
The quantity of <Chemical> in <Chemical1> is <Quantity1> *moles…The quantity of <Chemical> in <Chemicaln> is <Quantityn> *moles
Therefore, the quantity of <Chemical> = X *moles
Explanation Template
Law 4: Conservation of Volume
Mix
Chemical1 Chemicaln
Mixture
raw-material
result
…
Volume*liter
Volume*liter
volume volume
Volume*liter
volume
Conservation of Volumecontextinput
output
By the Law of Conservation of Volume, the volume of a mixture is the sum of the volumes of the parts mixed.
The sum of <Volume1>, …, and <Volumen> = <Volumeresult> *literTherefore, the volume of <Mixture> = <Volumeresult> *liter
Explanation Template
Step 1: Reclassify Terms
volume
Mix
Aqueous Solution Aqueous Solution
Mixture
Na+
raw material
Na2CO3
3.0 M 0.07 lit
NaHCO3
0.03 lit
volume
1.0 M
conc.base base conc.
result
has-part
Strong Electrolyte Solutionsuperclass
Strong Electrolytesuperclass
chemical
superclass
Step 2: Use Law 1 to Compute Concentration
Mixture
volumeconc.
Volume*liters
Concentration*molar
has-part
Chemical
Quantity*moles
quantityLaw 1
conc.
??*molar
volume
Mix
Aqueous Solution Aqueous Solution
Mixture
Na+
raw material
Na2CO3
3.0 M 0.07 lit
NaHCO3
0.03 lit
volume
1.0 M
conc.base base conc.
result
has-part??
*liters
volume
??*moles
quantity
The Search is non-deterministic
• Multiple laws might be used to compute a value for any property. For example, here’s another way to compute concentration:
pH = - log [H+], where [H+] is the concentration of H+
• Since this applies only to H+, this search path ends quickly
Step 3: Use Law 4 to Compute Volume
Mix
Chemical Chemical
Chemical
raw-material
result
…
Volume*liter
Volume*liter
volume volume
Volume*liter
volume
Law 4
.1
conc.
??*molar
volume
Mix
Aqueous Solution Aqueous Solution
Mixture
Na+
raw material
Na2CO3
3.0 M 0.07 lit
NaHCO3
0.03 lit
volume
1.0 M
conc.base base conc.
result
has-part??
*liters
volume
??*moles
quantity
Step 4: Use Law 3 to Compute Quantity
volume
Mix
Aqueous Solution Aqueous Solution
Mixture
Na+
raw material
Na2CO3
3.0 M
0.07 liters
0.03 liters
volume
1.0 M
conc. base
NaHCO3
base conc.
result
has-part
conc.
??*molar
.1*liters
volume
??*moles
quantity
Mix
Chemical Chemical
Chemical
raw-material
result
…
Quantity*moles
Quantity*moles
quantity quantity
Chemical
has-part
??*moles
quantity
part-ofLaw 3Na+Na+
??*moles
??*moles
has-part
quantity
Step 5: Use Law 2 to Compute Quantity of Ionic Parts
??*moles
quantity
Strong Electrolyte
Anion
has-part
Quantity*moles
quantity
Quantity*moles
quantity
Cation
Quantity*moles
quantity
Law 2
volume
Mix
Aqueous Solution Aqueous Solution
Mixture
Na+
raw material
Na2CO3
3.0 M
0.07 liters
0.03 liters
volume
1.0 M
conc. base
NaHCO3
base conc.
result
has-part
conc.
??*molar
.1*liters
volume
??*moles
quantity
Na+Na+
??*moles
??*moles
has-part
quantity
Step 6: Use Law 1’ to Compute Quantity
??*moles
quantityMixture
volumeconc.
Volume*liters
Concentration*molar
has-part
Chemical
Quantity*moles
quantity
Law 1’.21
volume
Mix
Aqueous Solution Aqueous Solution
Mixture
Na+
raw material
Na2CO3
3.0 M
0.07 liters
0.03 liters
volume
1.0 M
conc. base
NaHCO3
base conc.
result
has-part
conc.
??*molar
.1*liters
volume
??*moles
quantity
Na+Na+
??*moles
??*moles
has-part
quantity
Step 7: Wind out of Law 2 from step 5
Strong Electrolyte
Anion
has-part
Quantity*moles
quantity
Quantity*moles
quantity
Cation
Quantity*moles
quantity
Law 2
.42.21*moles
quantity
volume
Mix
Aqueous Solution Aqueous Solution
Mixture
Na+
raw material
Na2CO3
3.0 M
0.07 liters
0.03 liters
volume
1.0 M
conc. base
NaHCO3
base conc.
result
has-part
conc.
??*molar
.1*liters
volume
??*moles
quantity
Na+Na+
??*moles
??*moles
has-part
quantity
Step 8-10: Similar to steps 5-7
.03.21*moles
quantity
volume
Mix
Aqueous Solution Aqueous Solution
Mixture
Na+
raw material
Na2CO3
3.0 M
0.07 liters
0.03 liters
volume
1.0 M
conc. base
NaHCO3
base conc.
result
has-part
conc.
??*molar
.1*liters
volume
??*moles
quantity
Na+Na+
??*moles
.42*moles
has-part
quantity
Step 11: Wind out of Law 3 from Step 4
Mix
Chemical Chemical
Chemical
raw-material
result
…
Quantity*moles
Quantity*moles
quantity quantity
Chemical
has-part
??*moles
quantity
part-ofLaw 3
.45
.21*moles
quantity
volume
Mix
Aqueous Solution Aqueous Solution
Mixture
Na+
raw material
Na2CO3
3.0 M
0.07 liters
0.03 liters
volume
1.0 M
conc. base
NaHCO3
base conc.
result
has-part
conc.
??*molar
.1*liters
volume
??*moles
quantity
Na+Na+
.03*moles
.42*moles
has-part
quantity
Step 12: Wind out of Law 1 from Step 2
Mixture
volumeconc.
Volume*liters
Concentration*molar
has-part
Chemical
Quantity*moles
quantityLaw 1
.21*moles
quantity
volume
Mix
Aqueous Solution Aqueous Solution
Mixture
Na+
raw material
Na2CO3
3.0 M
0.07 liters
0.03 liters
volume
1.0 M
conc. base
NaHCO3
base conc.
result
has-part
conc.
??*molar
.1*liters
volume
.45*moles
quantity
Na+Na+
.03*moles
.42*moles
has-part
quantity
4.5
Answer and ExplanationWhen 70 ml of 3.0-Molar Na2CO3 is added to 30 ml of 1.0-Molar NaHCO3, what is the resulting concentration of Na+?.
The concentration of a chemical in a mixture is the quantity of the chemical divided by the volume of the mixture. By the Law of Conservation of Mass, the quantity of a chemical in a mixture is the sum of the quantities of
that chemical in the parts of the mix. In the Na2CO3 strong-electrolyte-solution and the NaHCO3 strong-electrolyte-solution :
In the Na2CO3 : Multiply the concentration and the volume: 3 molar * 70 milliliter = 0.21 mole.
The quantity of Na+ in the NA2CO3 solution is 0.42 mole. In the NaHCO3 :
Multiply the concentration and the volume:1 molar * 30 milliliter = 0.03 mole.
The quantity of Na+ in the Na2CO3 strong-electrolyte-solution and the NaHCO3 strong-electrolyte-solution is 0.45 mole.
Therefore, the quantity of Na+ = 0.45 mole. By the Law of Conservation of Volume, the volume of a mixture is the sum of the volumes of the parts
mixed. The sum of 70 milliliter and 30 milliliter = 0.10 liter.
Therefore, the volume of the strong-electrolyte-solution strong-electrolyte-solution mixture = 0.10 liter. Divide the quantity by the volume:.
0.45 mole / 0.10 liter = 4.50 molar. Therefore, the concentration of Na+ in the Na2CO3 and NaHCO3 mixture = 4.50 molar.
When 70 ml of 3.0-Molar Na2CO3 is added to 30 ml of 1.0-Molar NaHCO3, the resulting concentration of Na+ is 4.50 molar