1. Henry’s Law
S1 S2 =
P1 P2

2. The question
If 0.80 g of sulfur dioxide at 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?

3. Pull out the important parts
0.80 g of sulfur dioxide, SO2
In 5.00 L
At 25◦C
At atm, P1
X g
In 1 L
At 9.00 atm, P2

4. Start assembling the equation
First, Calculate S1
Solubility of a gas is measured in g/L
So we have
S1 = 0.80g
5.00 L
= 0.16 g / L

5. Solve for S2 Another way of looking at the equation
S1 S2 =
P1 P2
Cross multiply to get an equation that is easy to use.

6. Solve for S2 Another way of looking at the equation
S1 S2 =
P1 P2
Cross multiply to get an equation that is easy to use.
S1 P2 = S2 P1

7. Solve for S2 Another way of looking at the equation
S1 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
(0.16g/L )(9.00 atm)
10.00 atm
= S2
= g/L

8. It asked for how much will dissolve in 1 L.
0.144 g L
= x g
1 L
Hopefully the answer is obvious that 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)
gL = x gL
Divide both sides by L to get g will dissolve in 0.75 L