In the reaction of tin (IV) iodide in acetone and water, the solution turns a urine yellow.
Is this simply due to the hydrolysis of tin (IV) iodide:
$$\ce{SnI4 + 2H2O -> SnO2 + 4HI}$$
why is the acetone needed though?
and for the reaction with $\ce{KI}$:
$$\ce{SnI4 + KI -> ?}$$ I understand it forms iodine, but what happens to $\ce{Sn}$ and $\ce{K}$ - do they react with the acetone for instance?
If iodide is oxidized to iodine (the cause of the yellow color), then something needs to be reduced in turn. Potassium is already in its most stable oxidation state (+I), so tin will be reduced, from $\ce{Sn}$(IV) to $\ce{Sn}$(II).
$$\ce{SnI4 + 2H2O -> SnO2 + 4HI}$$
$$\ce{SnO2 + 2HI -> I2 + SnO + H2O}$$
With $\ce{KI}$, a yellow-brown solution of $\ce{KI3}$ will be formed, which contains the triiodide anion $\ce{I3-}$.
$$\ce{I2 + KI -> KI3}$$
When acetone is added, the yellow color of $\ce{I2}$ will slowly disappear as iodoacetone and $\ce{HI}$ are formed in an acid-catalyzed reaction (source):
$$\ce{(CH3)2CO + I2 -> CH3C(O)CH2I + HI}$$
