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Ch.17 - Aqueous Ionic Equilibrium

Chapter 17, Problem 95a

Calculate the molar solubility of barium fluoride in each liquid or solution. a. pure water

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Hey folks, welcome back. So here we're going to consider the following sparingly soluble salt equilibrium. So here we have this reaction here showing a solid breaking up into ions. We have this KsB and we want to know what is the scalability of this compound of iron to hydroxide and water adds this temperature. And we want three significant figures. So notice that the concentration here or the scalability here is slightly weird units that they want. So basically scalability of this compound is going to be given by X. So we're going to go ahead and do a nice table and equilibrium. X is going to be the molars eligibility but it's going to be in moles per liter. Right? That is the number that we're going to get, or the the units that we're going to get here, we want grams per 100 mL. So obviously we're going to have to um convert moles into grams and then leaders into milliliters and then of course multiplied by 100. So that we can have that 100 mL on the bottom there. Okay, so let's go ahead and do a nice chart. So we're just looking at this equation right here. I don't want to rewrite it. So just imagine that it's down here. So we're going to have this. So this is going to be a reactant. Our reactant here is solid. So we're just going to ignore it because um and in equilibrium we ignore solids and liquids. Right, This is going to be iron and this is going to be O H minus. So let's let me go ahead and just write this out. So it doesn't get confusing. There we go. So initially we're not going to have any of the products formed yet, so it's just going to be zero and zero. Okay with change, we're going to lose reactions, gain products, so we're going to gain just one X for iron but two X for hydroxide because we have two moles of that right, in the balanced equation and then at equilibrium we're going to have what we gained basically or what we started with plus what we gained, which is just X and two X. Okay, so like we said Mueller scalability is going to be X. So we first want to go ahead and solve for X and then just convert to the correct units. Okay, so KSB here is going to help us find um X. Okay, so KSB is going to equal to just products most multiplied together. Okay, and um can you speak here is going to equal to x times two X. Right, But this two X is going to be squared because we have two moles of it. Right? So X here is technically square or to the first power, we just don't write it but two X here is going to be squared. Alright, so let's go ahead and plug in what we know. So we have KsB KsB is 8. Time stands with the -16. So that's KSB and that's going to equal to So here It's gonna equal to two x times for X Squared or four x cubed. Okay, so now we're just solving for X at this point. So we're going to go ahead and divide both sides by four. So put this number in Prentice's divided by four and that will give us X cubed. It's going to equal to 2.00 Times 10 to the -16. Okay, so we just want X by itself. So we need to take a a cube root of both sides to find X. Now some of the calculators of course will have that where you can actually plug in what route that you want? Uh some of them don't so if you don't have that when you need to do is Whatever number that you're trying to take a a square or a cube root off. So you're gonna take two times 10 to and then give 16. Put in parentheses. Right, so let's say our number right here, you're gonna take it to the power of one divided by three. Okay, so put that into your calculator. So this one right here this number Put it in parentheses and then take it to the power of 1/3. Um and that will be the cube root. Now if it was fourth route then it would be 1/4 for example. Right so that's a little trick that we can do. So let's go ahead and find X. So after Q brooding here, we're going to get 5.85 Times 10 to the -6. And of course it's in Mueller or most per liter. Right. This is not our answer because they want it um the one the units and grams for million liter. So now let's go ahead and go down here and do some Conversions. So we have 5.85 Times 10 to the -6. And of course it is moles of iron to um oxide divided by one leader solution. Right, So let's go ahead and I'll convert moles into grams first. Okay, so in one more of iron iron, two oxide. How many grams are there? Were not given the molar mass of this compound. So let's go ahead and calculate that. I'm going to go ahead and go down here and calculate. So we're going to have one iron. We have Two oxygen's and two hydrogen. So if you take a look at the product table, iron weighs about 55.84 g oxygen weighs 16. And we have two of them here. So this will be 32 And then hydrogen weighs 1.01 each. We have two of them here. So that will be point to .02. So go ahead and add that together and that will be 89. g per mole. So that's the molar mass of that solid compound. So in one mole we have 89 0. g of that compound. So I'm not writing that compound but just realize that molds here corresponds to iron to oxide and grams also. And then the small. So we're going to go ahead and cancel out the moles. And now we're going to have grams. Now we want to go ahead and convert this leader into militia leader. Okay, so we're going to put leader on the, on the top here and militia leader on the bottom millions of metric prefix. So we'll put one in front of that and Millie is going to be 10 to then give three liters. So now we can cancel out the leaders and we have milliliters. So let's actually go ahead and calculate this number and see what we get. So The, the answer here that we should get is 5. Times 10 to the -7. And of course it's in g per one millimeter. Right? It's grams per one millimeter. Now that's not what they want. They wanted per 100 mL. So all we need to do here is just multiply this The whole thing by 100. So it's going to multiply the top and it's gonna multiply the bottom and we are going to get As our final final answer, 5.25 times 10 To the negative five g per 100 ml. So kind of an odd unit here that they wanted instead of per one millimeter They wanted to per 100 mL. But there we go. We got that. So it's 5.25 times 10 to the negative five. Okay, negative to the negative six will just give you an answer per one millimeter, but they wanted to per 100. So just multiply the top and the bottom by 100 and there you are. You have your final answer. Alright folks, we're all done here. Please let us know if you have any questions and we'll see you in the next problem.