Not leakage, you don't want that. Anyway, base current will dominate at 0.5V (it's not really necessary to have a B-E resistor at all, if it's just for leakage).
Base resistors should be dimensioned so that, when forward biased, Ib is about 1/10 to 1/20 of maximum Ic, and B-E resistor current is about 1/2 to 1/10 of Ib.
The B-E junction acts like a very small battery, holding ~pAh of charge, with a nominal 'charged' voltage of 0.7V, holding up for a moment (in the 0.7 to 0.5V range, say), then dropping to 0 (or reverse) when the charge is gone. This charge is called stored charge, and is responsible for the storage time during saturated switching. The junction also has fairly rapid self-discharge, so that it dissipates on its own on the order of 10-20us.
So the B-E resistor actually draws current out of the base, and the magnitude of that current should be maybe 1/10th to 1x the forward base current.
While base voltage remains up (due to storage), collector current remains up. Which gives all the more proof that BJTs are voltage dependent: drive them with a pure current, and they take forever to turn off; drive with a voltage source (of modest series resistance), and they turn off nicely.
The B-E resistor is simply to reduce the Thevenin resistance at turn-off, speeding things up.
Another way to think of the resistor values: make a voltage divider, such that the open-circuit voltage is 1.4 to 5V or so, and so that the current into a load of 0.7V is whatever Ib is required for the switch.
Tim