S 2 P 1 P 2 Henrys Law If 080 g of sulfur dioxide at 1000 atm pressure P1 dissolves in 500 L of water at 250C how much of it will dissolve in 1 L of water at 900 atm ID: 577374
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Slide1
S1 S2=P1 P2
Henry’s LawSlide2
If 0.80 g of sulfur dioxide at 10.00 atm pressure (P1) dissolves in 5.00 L of water at 25.0°C, how much of it will dissolve in 1 L of water at 9.00 atm pressure (P2) and the same temperature?
The questionSlide3
0.80 g of sulfur dioxide, SO2In 5.00 L At 25◦CAt 10.00 atm, P1X gIn 1 L
At 25◦CAt 9.00 atm
, P2
Pull out the important partsSlide4
First, Calculate S1Solubility of a gas is measured in g/LSo we have S1 = 0.80g 5.00 L = 0.16 g / L
Start assembling the equationSlide5
Solve for S2Another way of looking at the equationS1 S2
=P1
P2
Cross multiply to get an equation that is easy to use. Slide6
Solve for S2Another way of looking at the equationS1 S2
=P1
P2
Cross multiply to get an equation that is easy to use.
S1 P2 = S2 P1Slide7
Solve for S2Another way of looking at the equationS1 P2 = S2 P1
We are solving for S2 so we’ll divide both sides by P1.Yes, we could have just multiplied both sides by P2. I am trying to think of ways to make this equation easy to remember- is it easier as S1P2 = S2P1 or as “S1 over P1 = S2 over P2”? Use what makes your brain happy.
S1P2
P1
= S2
(0.16g/L )(9.00
atm
)
10.00
atm
= 0.144 g/LSlide8
It asked for how much will dissolve in 1 L.0.144 gL=
x g1 L
Hopefully the answer is obvious that 0.144 g will dissolve in 1 L.
Just for fun, what if they asked how much would dissolve in 0.75L?
0.144 g
L
=
x g
0.75 L
Cross multiply to get (0.114 g) (0.75L) = (x g) (1 L)
0.0855
gL
= x
gL
Divide both sides by L to get 0.0855 g will dissolve in 0.75 L