Henry's Law Calculations - Video Tutorials & Practice Problems
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To calculate solubility of a dissolves gas, Henry's Law Constant and partial pressure are used.
Henry's Law Calculations
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Henry's Law Calculations Concept 1
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The concentration, also referred to as the solubility of a dissolved gas, can be determined from its Henry's law constant and partial pressure. Now when I say Henry's law constant, that's the variable k sub h, and it represents the solubility of a gas at a fixed temperature in a particular solvent in molarity, which, remember, use the variable capital m. Also remember, we've discussed this before where concentration and molarity are synonymous with one another. Usually, professors will interchange them. You can say concentration or molarity and they mean the same thing. With this idea of Henry's law constant comes Henry's law formula, where s gas, which equals the solubility of a gas in molarity equals henry's constant, which is k h, and here it's in units of molarity over pressure. So our normal type of pressure is atmospheres, but remember there's a possibility of tors or millimeters of mercury as well. So always be on the lookout for the units that you'll see for Henry's law constant. Usually, you'll see it in molarities over atmospheres, but there can be times when you might see it in molarities over tors or molarities over millimeters of mercury. If it makes you feel more comfortable, just convert those pressure units all to atmospheres and use this version of Henry's law constant. Now and that's times p gas. P gas is the partial pressure of the gas in atmospheres. And again, units are always important. The units here for partial pressure in atmosphere is because we're using these units for Henry's law constant, molarity over atmospheres.
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Henry's Law Calculations Example 1
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Calculate the solubility of carbon dioxide gas, which is c o 2, when its Henry loss constant is 8.20 times 10 to the 2 molarities per atmosphere at a pressure of 3.29 atmospheres. Alright. So we need to find solubility of our gas, so that's s gas, which is c o 2, equals Henry's constant times the pressure of that gas. So we're going to plug in 8.20 times 10 to the 2, molarities over atmospheres, times 3.29 atmospheres. Again, you see that units will cancel out. And what do we have at the end? Molarity. So this would come out to be 2.70 times 10 to the 3 molar. So this would be the concentration of carbon dioxide for this particular example question.
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Henry's Law Calculations Concept 2
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So now we take a look at Henry's law, at least in terms of the 2.4. We're gonna say the 2.4 method Henry's law formula illustrates how changes in pressure can affect gas solubility. Now we're gonna say this formula, this version, is used when when dealing with 2 pressures and with 2 solubilities for a given gas. Here we're going to say with this formula the units from solubility can be in molarity, or they can be in other units that are in mass per volume. So hennieslawformula, the two point form is solubility 1, which is your initial solubility, over your initial pressure, p 1, equals solubility 2, so your final solubility, divided by your final pressure P 2. So again, we use this version when we're dealing with 2 solubilities or 2 pressures for any given gas.
The 2 Point Form of Henry's Law Formula is used when dealing with 2 pressures and 2 solubilities of a given gas.
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Henry's Law Calculations Example 2
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At a pressure of 2.88 atmospheres, the solubility of dichloromethane, which is c h two c l two, is 0.384 milligrams per liter. If the solubility decreases to 0.225 milligrams per liter, what is the new pressure? Alright. So here they're giving us one pressure, but they're asking for a new pressure. So since this is the first pressure that's given, this is p 1. And since this is the second pressure that's talked about, this is p 2. Connected to p 1 is this solubility, which we're going to say is s 1, and connected to the new pressure we need to find is this solubility, which we call s 2. So now we're just gonna solve. So we're gonna say here that s 1 over p 1 equals s 2 over p 2. So then we're gonna plug in the numbers that we know, so this is 0.384 divided by 2.88 equals solubility 0.225 milligrams per liter divided by p 2, which we don't know. You're going to cross multiply these 2 and cross multiply these 2. So when we do that, what we're gonna see is we're gonna have 0.384 milligrams per liter times p 2 equals so when I multiply those 2 together using 0.648 milligrams over liters times atmospheres. Divide both sides now by 0.384 milligrams per liter. So these units cancel out and we'll have p 2 equal 2 atmospheres. So here p 2 equals 1.69 atmospheres as my final answer.
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Problem
Problem
Henry's Law Constant for nitrogen in water is 1.67 × 10-4 M • atm–1. If a closed canister contains 0.103 M nitrogen, what would be its pressure in atm?
A
617 atm
B
1.72 × 10–5 atm
C
1.62 × 10–3 atm
D
778 atm
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Problem
At 0°C and 1.00 atm, as much as 0.84 g of O2 can dissolve in 1.0 L of water. At 0°C and 4.00 atm, how many grams of O2 dissolve in 1.0 L of water?
A
0.105 g
B
3.36 g
C
6.72 g
D
4.68 g
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Problem
Problem
The atmospheric pressure in a lab is calculated as 1.3 atm. If oxygen gas contributes 62% of this atmospheric pressure, determine its mass (in g) dissolved at room temperature in 25 L of water. The Henry's Law Constant for oxygen in water at this temperature is 5.3 × 10–5 M/atm.