[lutesium]

Is there a a difference between KOH dissolved in Et-OH yielding Et-O^- K⁺ and Potassium metal Reacted with Et-OH yielding Et-OK??? I know that there's a difference but I have just one question:
If we evaporate the former solution to dryness what would we obtain??? KOH or the latter product (Et-OK) Can someone explain the mechanism please???

[NahiEw]

Think about what happens to the KOH as it is dissolved in EtOH (is it the K+ or OH- that's going to react with the EtOH?) and then think about what happens when you put potassium METAL into EtOH (look up what happens when you place potassium to water. You'll probably get the same reaction when you place it in EtOH).

For the second part, I suggest you to use LeChatlier's Principle to decide what happens to the reaction as you remove EtOH(l).

[Smrt guy]

EtOH is a slightly weaker acid than water.  Therefore when you dissolve KOH in water, there will be an equilibrium mixture containing mostly KOH, some KOEt.  Evaporating would likely yield the KOH back (both for acid/base reasons as well as b.p. reasons).