Help with buoyant forces and liquids (high school physics) I am stuck on this problem. Here is the scenario  

Suppose you send a balloon
  (filled with air) from your submersible craft to the surface of the
  ocean, $10.8 \text{ km}$ above, with the message “I’M OK!” written on the
  balloon. Knowing that the balloon will expand as it rises, you write
  very small and don’t fully inflate the balloon, leaving its initial
  radius at $5.0 \text{ cm}$. The density of sea  water is $1050 \text{ kg/m}^3$ and the
  density of air $1.23 \text{ kg/m}^3$.

I am asked to find the acceleration and the time it will take to reach the surface (we are told we can assume the acceleration will stay constant). 
So far I found that the buoyant force is equal to  $6.3*10^{-3}$. So the $F_{\text{net}}$ upwards is $6.3*10^{-3}$ - the weight of the balloon which I got $5.38$. I was trying to find acceleration with $F=ma=(density*volume)a$ and solving for $a$ but I'm getting a really large number. 
I'm still confused, do I use the density of the water or density of the air? And why?
 A: Uniform acceleration is not realistic in this problem. 
Why even include the text about the expansion of the balloon, if it is not going to be taken into account?
The only way the balloon could continuously accelerate would be if it experiences no drag from the water. But even if we go with this greatly simplifying assumption, it is still far, far from realistic that the acceleration is constant.
There is an immense change in pressure from 10km under the seam, to the surface. Assuming the balloon doesn't burst, it will undergo a very large change in volume and density as it decompresses. 
The density of water does not change much with depth due to its large bulk modulus, so for the sake of simplicity, it can be assumed to be constant.
The larger the balloon gets as it rises to the surface, the more water it displaces. Yet the mass of the balloon stays constant, since it contains the same amount of rubber and air.
The more water the expanding ballon displaces, the more buoyant force it experiences (without getting any heavier). Therefore, the more acceleration it experiences. 
You can feel the change in buyoant force due to compression of air even over much smaller depths than 10 km! If you simply dive into a pool, you can feel the loss of buoyancy as your lungs compress, making it easier to reach the bottom.
The simplification in the problem statement means that you're probably being asked only to calculate the force that the balloon initially experiences at the original depth, which is simply the difference between the weight of the baloon, and the buoyant force, which is the weight of the volume of water which it displaces, and then use this force in a distance/acceleration formula to figure out time.
However, there is another serious, serious problem with this question is that the density of air is not 1.23 $kg/m^3$ at a depth of 10 km below the ocean's surface! (Unless you are in a craft that maintains regular atmospheric pressure, in which case any balloon that you inflate and send out through an air lock will instantly shrink).  1.23 $kg/m^3$ is the density of air at standard temperature and pressure (STP)!
A: $F_{\text{net}}=m_{\text{baloon}} a_{\text{baloon}} = \rho_{\text{water}} V_{\text{baloon}} g - m_{\text{baloon}} g$. Why use density of water? Because of Archimedes' principle. 
