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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 solid ability of a dissolved gas, can be determined from its pennies law. Constant and partial pressure. Now, when I say Henry's law constant, that's the variable k sub h. And it represents the saw the ability of a gas at a fixed temperature in a particular solvent in malaria T, which remember, use the variable capital. And also remember, we've discussed this before, where concentration of polarity are synonymous with one another. Usually professors will interchange them. We can say concentration or polarity. They mean the same thing With this idea of Henry's law, Constant comes Henry's law formula. Where s gas? Which equals the Saudi ability of a gas in polarity, equals Henry's constant, which is K H. And here it's in units of polarity overpressure. So our normal type of pressure is atmospheres. But remember, there's a possibility of tours 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 polarities over atmospheres, but there can be times more. You might see it in polarities over tours or polarities over millimeters of mercury If it makes you feel more comfortable, just convert those pressure units all two 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 and atmosphere is because we're using these units for Henry's law, constant polarity over atmospheres.

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Henry's Law Calculations Example 1

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Calculate the scalability of carbon dioxide gas, which is CO2 when it's Henry lost. Constant is 8.20 times 10 to the two polarities per atmosphere at a pressure of 3.29 atmospheres. All right, so we need to find solid ability of our gas. So that's s gas, which is co two equals Henry's constant times the pressure of that gas. So we're gonna plug in 8.20 times 10 to the two polarities over atmospheres, Times 3 - nine Atmospheres. Again. You see that? Units will cancel out. And what do we have at the end polarity? So this would come out to be 2.70 times to the three Mueller. 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 Matt Henry's law formula illustrates how changes in pressure yeah, can affect gas solid ability. Now we're gonna say this formula this version is used when dealing with two pressures and with two psy abilities for giving gas Here, we're gonna say with this formula, the units from Saudi ability can be in polarity or they can be in other units. There are in mass per volume. So Henry's law formula the two point form is sly ability, one which is your initial suitability over your initial pressure. P one equals psy ability to so your final side ability, Divided by a final pressure P two. So again, we use this version when we're dealing with two side abilities or two 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 psy ability of deplore methane, which is ch two cl two is 20.384 mg per liter. If the cell viability decreases 2.225 mg/l, 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 one. And since this is the second pressure that's talked about, this is P two Connected to P. one. Is this solid ability, Which we're gonna say is S one And connected to the new pressure we need to find is this side building, which we call S two. So now we're just gonna solve, So we're going to say here, that s one over p one equals s two over p two. So then we're gonna plug in the numbers that we know. So this is .384, Divided by 2. Equals saw stability .225 mg/l, Divided by P two, which we don't know you're going to cross multiply these two And cross multiply these two. Yeah. So when we do that, what we're gonna see is we're going to have .384 mg/l, Times P two equals. So when I multiply those two together, he's 1.648 milligrams over leaders times. Atmospheres Divide both sides now, by .384 mg/l. Yeah. So these units cancel out, and we'll have P two equal two atmospheres. So here, p two equals 1. 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

Problem

At 0°C and 1.00 atm, as much as 0.84 g of O_{2} can dissolve in 1.0 L of water. At 0°C and 4.00 atm, how many grams of O_{2} 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.

A

1.4 × 10^{–3} g

B

6.9 × 10^{–5} g

C

0.055 g

D

0.034 g

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