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ემზადებით გამოცდისთვის? მოემზადეთ ამ 3 გაკვეთილის დახმარებით შემდეგ თემაზე: Buffers, titrations, and solubility equilibria
იხილეთ 3 გაკვეთილი
ვიდეოს აღწერა
- [Voiceover] The goal is to calculate the solubility of copper II hydroxide. We're given the solubility product constant KSP, which is equal to 2.2 times 10 to the negative 20 at 25 degrees Celsius. So let's conceptualize this problem first. Let's say we have some copper II hydroxide which is blue, so let's say we put some copper II hydroxide, some solid copper II hydroxide into a beaker containing water. This is a slightly soluble ionic compound. So not everything we put into the beaker is going to dissolve. Let's say only a small portion of this dissolves, I'm gonna take my eraser here and I'm gonna take off a small bit of our solid there at the top and let's say that small amount turns into ions. So what ions would we have in solution? Copper II tells us we have copper II plus ions in solutions, so CU two plus and then we also have hydroxide ions in solution OH minus. So we'd have some hydroxide ions in solution two. Eventually we reach equilibrium right? So we have a solubility equilibrium where the rate of dissolution is equal to the rate of precipitation. So let's go ahead and write that out. What's the chemical formula for copper II hydroxide? We could use this simple little trick here of crossing over your charges to figure out the chemical formula is CU with parenthesis OH and a two here. That's our solids and we also have our ions. So CU two plus or ions in solution, we also have hydroxide ions in solution, OH minus. We need to balance this so we need a two in front of the hydroxide and everything else here would get a one. Alright so let's set up an ice table. So we have our initial concentration, our change, and then finally our concentration at equilibrium. Well before that small amount of copper II hydroxide dissolves, right so that small amount that I erased earlier, we didn't have anything for the concentration of our ions in solution. So that's our initial concentration of our ions, it's zero. Now let's think about the small amount of copper II hydroxide, the solid that dissolved. Alright so let's say that x is equal to the concentration of copper II hydroxide that dissolves. So we're going to lose a concentration of copper II hydroxide which we'll say is x. Look at your mole ratios for every one mole of copper II hydroxide that dissolves we get one mole of copper II plus ions in solution. So for losing, if we're losing x for the concentration of copper II hydroxide we're going to gain x for the concentration of copper II plus ions in solution. And for hydroxide ions, for every one mole of copper II hydroxide that dissolves we get two moles of hydroxide ions. Alright so instead of x it'd be 2x. Alright so we're going to gain 2x for the concentration of hydroxide ions. So at equilibrium, right our equilibrium concentrations of our ions would be x for copper II plus and 2x for hydroxide. Alright let's write our equilibrium expression, right? So KSP is equal to, we look at our products, right so we have CU two plus, we put the concentration of CU two plus and we raise the concentration to the power of the coefficient, and here our coefficient is a one. So raise this to the first power. Next our other product here would be the hydroxide ions, so OH minus and we raise that concentration to the power of the coefficient which in this case is a two. So we need to put a two here and once again we leave this pure solid out of our equilibrium expression. Alright let's plug in for KSP, the solubility product constant was given to us it's 2.2 times 10 to the negative 20. So let's plug that in, so this is equal to 2.2 times 10 to the negative 20 and this is equal to the concentration of copper II plus ions at equilibrium which is x. So we put that in, this is x to the first power times the concentration of hydroxide ions at equilibrium raised to the second power. So this would be 2x and then we have to square it here. And this is where some students get a little bit confused because if they say well you're doubling the concentration here, right, and then you're squaring it aren't you kinda like doing the same thing twice? But remember, these are two different things. This 2x is because of the mole ratios, right and we raise it to the power of the coefficient because that's what you do in an equilibrium expression. Alright so those are two different things, we're not doing the same thing twice. Alright, when we do our algebra on the right side we would have x times 4x squared. So that's equal to 4x cubed and this is equal to 2.2 times 10 to the negative 20. So we need to divide that by four, so we need to divide 2.2 times 10 to the negative 20 by a four, so you could do that in your head or on the calculator, 2.2 times 10 to the negative 20. Alright we divide that by four and we get 5.5 times 10 to the negative 21st. So we have 5.5 times 10 to the negative 21st is equal to x cubed. Alright so to solve for x we need to take the cube root of 5.5 times 10 to the negative 21st and unfortunately on this calculator it's a little bit trickier than on most calculators. In most calculators it's pretty straight forward and it's pretty easy to do. So let me show you one way to take the cube root of something on this TI-85 here. So we would put in, we'll take the cube root, so we put a three in here and then one way to find this is to go to 2nd catalog and then just move upwards here until you see the symbol. Alright so I don't see it yet and -- There it is, so right there is the symbol we want. So we're trying to take the cube root of, we want 5.5 times 10 to the negative 21st and that should give us the cube root, which is equal to, let's go ahead and round that to 1.8 times 10 to the negative seven. So this is equal to, x is equal to 1.8 times 10 to the negative seven, and this would be the concentration, right, this would be molar, this would be the concentration of copper II plus at equilibrium, right, let's go back up here. So x is equal to the concentration of copper II plus at equilibrium and notice it's also equal to the molar solubility of copper II hydroxide, right? That's how much copper II hydroxide dissolved, x. So we found the molar solubility of copper II hydroxide. Our question asked us for solubility, so maybe they meant molar solubility in which case we're done or maybe they meant solubility in grams per liter. So let's go ahead and do that now. So this is equal to, this is equal to the molar solubility. This is the molar solubility, which is moles over liters. What if they wanted grams over liters? Alright, you would need to have the molar mass of copper II hydroxide. Alright so you could look that up on your periodic table. So copper II hydroxide has a molar mass of 97.57 grams per mole. So our answer here for molar solubility, this would be moles per liter. So if we wanna go to solubility in grams per liter, let's look at our units and see what we'd have to do. We have 1.8 times 10 to the negative seven moles over liters. Alright if you look at the molar mass, alright if you wanna get to grams over liters all we have to do is multiply the molar solubility by the molar mass because the units for the molar mass are grams over moles. And if we multiply, the moles cancel, right, and we'll end up with grams over liters. So let's go ahead and do that calculation here. So we have, we rounded this to 1.8 times 10 to the negative seven, with it's molar solubility to get to the solubility in grams per liter we multiply that by 97.57 which is the molar mass of copper II hydroxide, and we get, if we round that to 1.8 times 10 to the negative five. Alright so this is equal to 1.8 times 10 to the negative five, and this would be grams over liters. So this is the solubility, in one liter of solution you could only dissolve 1.8 times 10 to the negative five grams. So copper II hydroxide is not very soluble at all. Alright so that's how to figure out the solubility if you're given the solubility product constant KSP.