A student followed the procedure of this experiment to determine the solubility product of zinc(II) iodate, Zn(IO3)2. Solutions
of ZN(NO3)2 of known initial concentrations were titrated with 0.200 M KIO3 solutions to the first appearance of a white precipitate. The following data were collected.
Complete the table below and determine the solubility product constant.
[Zn(NO3)2]0, M Initial 0.226, 0.101 0.0452, 0.0118
[KIO3]0, M Titrant 0.200, 0.200, 0.200, 0.200
V0, mL of Zn(NO3)2 100.0, 100.0, 100.0, 100.0
V, mL of KIO3 titrant 12.9, 12.4, 13.0, 18.3
Complete the table below and determine the solubility product constant.
[Zn(NO3)2]0, M Initial 0.226, 0.101 0.0452, 0.0118
[KIO3]0, M Titrant 0.200, 0.200, 0.200, 0.200
V0, mL of Zn(NO3)2 100.0, 100.0, 100.0, 100.0
V, mL of KIO3 titrant 12.9, 12.4, 13.0, 18.3
1 answer:
Answer:
Explanation:
Complete the table below and determine the solubility product constant.
[Zn(NO₃)₂] 0, M Initial 0.226, 0.101 0.0452, 0.0118
[KIO₃] 0, M Titrant 0.200, 0.200, 0.200, 0.200
V₀, mL of Zn(NO₃)₂100.0, 100.0, 100.0, 100.0
V, mL of KIO₃ titrant 12.9, 12.4, 13.0, 18.3
What needs to be determined are the following for each column/sample
V0 + V, mL
[Zn²⁺] + [IO³⁻] ⇄ [Zn²⁺][IO³⁻]2
log [Zn²⁺][IO³⁻]2
Then there's the determination of Ksp itself
log Ksp = Ksp of Zn(IO3)2 =
Considering this, which of the ion products is closest to Ksp, and why?
Finally, the titration volumes for the first three samples don't vary greatly. However the last sample is considerably larger. Why is this to be expected?
The attached figures shed more light on the solution to this problem
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Considering the Dalton's partial pressure, the mole fraction of each gas is:
= 0.14
= 0.76
= 0.10
The pressure exerted by a particular gas in a mixture is known as its partial pressure.
So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:
where n is the amount of gases in the gas mixture.
This relationship is due to the assumption that there are no attractive forces between the gases.
Dalton's partial pressure law can also be expressed in terms of the mole fraction of the gas in the mixture. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of a component to the number of moles of all the components present.
So in a mixture of two or more gases, the partial pressure of gas A can be expressed as:
In summary, the total pressure in a mixture of gases is equal to the sum of partial pressures of each gas.
Mole fraction of each gas
In this case, you know that:
= 1.25 atm
= 6.73 atm
= 0.895 atm
= 1.25 atm + 6.73 atm + 0.895 atm= 8.875 atm
Then:
Substituting the corresponding values:
- 1.25 atm=
- 6.73 atm=
- 0.895 atm=
Solving:
In summary, the mole fraction of each gas is:
Learn more about Dalton's partial pressure:
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Answer : The heat your body transfer must be, 25.1 kJ
Explanation :
Formula used :
or,
where,
Q = heat = ?
m = mass of water = 500.0 g
c = specific heat of water =
= initial temperature =
= final temperature =
Now put all the given value in the above formula, we get:
Therefore, the heat your body transfer must be, 25.1 kJ
Answer : The value of work done for the system is 1144.69 J
Explanation :
First we have to calculate the moles of
Molar mass of = 65.01 g/mole
Now we have to calculate the moles of nitrogen gas.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react to give 3 mole of
So, 0.311 moles of react to give
moles of
Now we have to calculate the volume of nitrogen gas.
Using ideal gas equation:
where,
P = Pressure of gas = 1.00 atm
V = Volume of gas = ?
n = number of moles = 0.466 mole
R = Gas constant =
T = Temperature of gas =
Putting values in above equation, we get:
As initially no nitrogen was present. So,
Volume expanded = Volume of nitrogen evolved
Thus,
Expansion work = Pressure × Volume
Expansion work = 1.00 atm × 11.3 L
Expansion work = 11.3 L.atm
Conversion used : (1 L.atm = 101.3 J)
Expansion work = 11.3 × 101.3 = 1144.69 J
Therefore, the value of work done for the system is 1144.69 J