I suggest you read through the brilliant post from
@doug293cz to see how each bubble of CO2 produced affects things
Maths
This is the equation that all this centers on:
Final O2 Conc = 210000 ppm * ((Headspace volume L - 0.0001 L) / Headspace volume L)^ (Liters of CO2 produced / .0001 L)
From the assumptions I made, 5g 1.060 wort with FG of 1.012 you'll produce 499.35 L of CO2. Your headspace is either 7g or 12g. Plugging this into the equation:
210000 * ((26.5 L - .0001L) / 26.5 L) ^ (499.35 L/.0001 L) = .00137 ppm
210000 * ((45.4 L - .0001L) / 45.4 L) ^ (499.35 L/.0001 L) = 3.51 ppm
How this O2 in the headspace dissolves into solution is beyond me. Maybe somebody smarter than me can shed some light here. My take is that these calculations are all worst case scenario so real world is likely better than this however given enough time, agitation, etc. at a given temperature the O2 will be absorbed into the beer.
You can use that same equation to figure out how many liters of CO2 is required to hit 1 ppb in 2g of headspace. Turns out it takes 145.6 L of CO2 which is roughly when your SG drops from 1.060 to 1.046. At this point we can assume the fermenter is exhausting pure CO2. To purge/displace the serving keg full of sanitizer takes the volume of serving keg worth of CO2 which is 18.9 L. The SG when this has completed should be about 1.044. From there the remainder of the CO2 produced from 1.044 to FG of 1.012 is used to dilute the O2 in the dry hop keg. In my calculations I've assumed the CO2 filled fermenter does not mix with the gas in the dry hop keg... This will result in the dry hop keg with 4.25 ppb O2.