00:02
This is a ksp question that has radioactivity associated with it in order to determine ksp.
00:09
The ksp reaction is mercury 1 iodide, dissociating into the mercury 1 ion that has a 2 plus charge because it's diatomic, plus two iodides.
00:22
This equation is provided, and the ksp is going to be equal to the mercury 1 concentration, multiplied by the iodide concentration, squared.
00:33
We know that the solid before added to the water has a radioactivity of 5 .0 times 10 to the 11 counts per minute per mole of iodine.
00:46
Now it's not every single iodine that is present, that is iodine 131, that results in radioactivity, but just enough so that we can relate the activity of the ones that are radioactive to.
01:03
This is the total moles of iodine, including the ones that are not radioactive.
01:11
So if we were to add an excess amount of mercury 1 iodide, we would get a significant amount that falls to the bottom and is undissolved and a very small amount that dissolves producing two iodides for every one mercury 1.
01:29
If the volume is 150 milliliters, or if it's not and we extract a hundred and 150 milliliters, 150 mil liters saturated volume of mercury 1 iodide has a count of 33, or has a decay rate of 33 counts per minute.
01:49
What's going to be helpful as we solve this problem is to recognize that the mercury 1 concentration, based upon the stoichiometry shown up here, is going to be one -half the iodide concentration, and it is with this information the 33 counts per minute that we can determine the iodide concentration because it is only the iodide that is radioactive.
02:14
So if we take the 33 counts per minute and we multiply it by this as a conversion factor, which tells us the counts per minute per mole of radioactive iodide, and now we're just referring to the moles of radioactive iodide that have introduced themselves into the solution, we can determine the number of moles of iodide that actually introduced themselves into the solution...