In nonaqueous solvents, it is possible to react HF to create H2F+. Which of these statements follows from this observation? (a) HF can act like a strong acid in nonaqueous solvents, (b) HF can act like a base in nonaqueous solvents, (c) HF is thermodynamically unstable, (d) There is an acid in the nonaqueous medium that is a stronger acid than HF.

Video transcript

Hi everyone for this problem. It reads iodine pencil fluoride undergoes auto dissociation into iodine. Touch a fluoride. Cat ion and iodine hexafluoride and ion antimony pencil fluoride reacts to form hex, a floral anti mon an ion which I just referred to as S B f six minus when dissolved in iodine pencil fluoride, which statement accounts for this observation and were given three statements. Okay, so the thing we're going to want to do here is discuss what is auto dissociation? Auto dissociation occurs when two things react with itself. So an example is the auto dissociation of water. So for water we have two molecules of water are going to react and produce an equilibrium that produces hydro ni um ion and hydroxide ion. Okay, and in this auto dissociation, our hydro knee um represents our acid and our hydroxide represents our base. Okay, and so for acids, salutes, acids are salutes that increase the concentration of our hydro ni um ion and base is salutes. That increase our concentration of hydroxide ions. So here, in the problem we're dealing with the auto dissociation of iodine Penta fluoride and they tell us it undergoes to produce iodine tetra fluoride, cat ion and iodine hexafluoride and ion. So let's go ahead and write out this auto dissociation. Okay, so we have two molecules of iodine pencil fluoride. Okay, and it's going to produce what was said in the problem. Okay, so let's go ahead and write that out. Okay, so here we can identify the acid in the base. Okay, so we said, acids are solids that increase the concentration of our cat ion. So this is going to be our acid and our base increases the concentration of our an ion. So we'll say this is our base. So that is our auto dissociation and this should be I F five. So go ahead and put that there as the correction. Okay, so the problem tells us that antimony pencil fluoride reacts to form S B F six minus when dissolved in aydin, pencil fluoride. So let's go ahead and reflect that here and then we'll be able to see which statement is correct. Okay, so we have antimony pencil fluoride reacts with iodine, pencil fluoride and we're told it reacts to form S. B. F six minus. Okay, and we're also going to have I have four plus. Okay, so we have here This SBF 6 - is going to be our base. Okay, are an ion which means what it's coming from. This is our conjugate base. So the acid that it's coming from, is this so we can identify our are antimony pencil fluoride to be the acid. Okay, so let's take a look at our statements statement, a antimony pencil fluoride can act like a base and iodine pencil fluoride. So here this statement is not correct because we see here that it's the acid statement. B antimony pencil fluoride can act like an acid and iodine, pencil, Flora this statement is correct because we labeled and defined what these are. So this is correct. And our last statement antimony pencil fluoride can audit associate to form SPF six My. So this is not true because in the problem we're told that it reacts to form the following, so this statement is incorrect. So that leaves us with statement B. Being the correct statement for this observation. And that is the end of this problem. I hope this was helpful.