# Henry's law

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#### Opejay

##### Full Member
10+ Year Member
The Henry's law constants for N2 and O2 in water are 8.6E-4 /kPa and 4.4E-4 /kPa respectively There are approximately 1E5 Pa in 1 atm. Air is approximately 79% nitrogen and 21% oxygen. What is the approximate ratio of nitrogen to oxygen molecules dissolved in a glass of water that has been left exposed to 1 atm of air pressure for a long time?

#### Jepstein30

##### Full Member
10+ Year Member
The Henry's law constants for N2 and O2 in water are 8.6E-4 /kPa and 4.4E-4 /kPa respectively There are approximately 1E5 Pa in 1 atm. Air is approximately 79% nitrogen and 21% oxygen. What is the approximate ratio of nitrogen to oxygen molecules dissolved in a glass of water that has been left exposed to 1 atm of air pressure for a long time?

Hmm.. well first off I'm pretty sure you have the units wrong on the constants.. they should be kPa/M that way you end up with kPa after multiplying by concentration.

So henry's law: partial pressure = constant*concentration and we want to solve for concentration so: concentration = pressure/constant

Now let's work in kPa because of the constant's units. 1 atm has a partial pressure of .79 N2 and .21 O2 or 7900 Pa (7.9 kPa) and 2100 Pa (2.1 kPa), respectively.

Concentration of N2 = 7.9/8.6*10^-4 = (approx) 9200 M
Concentration of O2 = 2.1/4.4*10^-4 = (approx) 4700 M

Numbers don't seem very right so I prob made a mistake but my answer would be about 2:1.

#### LoLCareerGoals

##### Full Member
10+ Year Member
Henry's law is Solubility = const * Partial pressure over the solution.

Find: Solubility(N)/Solubility(O) = [const(N)*PP(N)]/[const(O)*PP(O)] = (8.6/4.4) * (79/21) = 8:1

#### Jepstein30

##### Full Member
10+ Year Member
Henry's law is Solubility = const * Partial pressure over the solution.

Find: Solubility(N)/Solubility(O) = [const(N)*PP(N)]/[const(O)*PP(O)] = (8.6/4.4) * (79/21) = 8:1

Weird. I wasn't sure what Henry's law was exactly so looked it up and.. wikipedia has it as I wrote it..

Partial pressure = constant * concentration

http://en.wikipedia.org/wiki/Henry's_law

which is it??

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#### LoLCareerGoals

##### Full Member
10+ Year Member
You can express it that way too, but constant in your equation will be 1/Henry's law constant.

Chemistry Central Science ed 10 textbook, page 541: Sg = kPg. (Sg solubility of gas, k Henry's law const, Pg gas partial pressure over solution).

Oh and an obligatory lolwikipidiaAsScientificSource.

#### Jepstein30

##### Full Member
10+ Year Member
You can express it that way too, but constant in your equation will be 1/Henry's law constant.

Chemistry Central Science ed 10 textbook, page 541: Sg = kPg. (Sg solubility of gas, k Henry's law const, Pg gas partial pressure over solution).

Oh and an obligatory lolwikipidiaAsScientificSource.

Ah I see, so it really just depends on the constant.

Wikipedia is a great source! Especially cause I can edit it if I'm wrong to make it like I'm right!

So: Solubility = constant * partial pressure

N2: 8.6*10^-4*79 = 640*10^-4 = 6.4 * 10 ^-2
O2: 4.4*10^-4*21= 80 *10^-4 = 8*10^-3

N2/O2=6.4*10^-2/8.8*10^-3 = 7.5

Cool! Thanks!

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#### LoLCareerGoals

##### Full Member
10+ Year Member
Why are you doing all that math? Numbers are made to be near factors of each other to see if you notice. This is MCAT and we don't have time for this. Sure if 2 answers are close to 8, then fine, but I mean seriously 8.6*79???

#### Jepstein30

##### Full Member
10+ Year Member
Why are you doing all that math? Numbers are made to be near factors of each other to see if you notice. This is MCAT and we don't have time for this. Sure if 2 answers are close to 8, then fine, but I mean seriously 8.6*79???

Not much math I just did 8 * 8 = 64 if you didn't notice. and 2 * 4.4 I write everything out to avoid stupid mistakes, I'm pretty fast with the math though so i'm good on time and stuff.