### Henry`s Law - Embrace Challenge

```Henry’s Law
S1
P1
=
S2
P2
The question
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?
Pull out the important parts
0.80 g of sulfur dioxide, SO2
In 5.00 L
At 25◦C
At 10.00 atm, P1
X g
In 1 L
At 25◦C
At 9.00 atm, P2
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
Solve for S2
Another way of looking at the
equation
S1
P1
=
S2
P2
Cross multiply to get an equation that is easy to use.
Solve for S2
Another way of looking at the
equation
S1
P1
=
S2
P2
Cross multiply to get an equation that is easy to use.
S1 P2 = S2 P1
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
= S2
(0.16g/L )(9.00 atm)
10.00 atm
= 0.144 g/L
It asked for how much will
dissolve in 1 L.
0.144 g
L
= xg
1L
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
=
xg
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
```