Ca1OH22 has a Ksp of 6.5 * 10-6. (b) If 50 mL of the solution from part (a) is added to each of the beakers shown here, in which beakers, if any, will a precipitate form? In those cases where a precipitate forms, what is its identity? [Section 17.6]

Video transcript

Welcome back everyone in this example, we have a 75 millimeter solution made by dissolving 10 g of barium hydroxide with the given K. S. P. Or so liability product constant of five times 10 to the negative third power in mL of water. This is added to each of the beakers below and we need to determine which beaker will have to precipitate that will form and what the identity of this precipitate will be. So Our first step is to figure out our concentration of our barium hydroxide. Now we're doing so because we recognize that anything with a K. S. P is a solid and in this case we're going to have our barium hydroxide solid in an equilibrium which dissociates into its ionic components where we formed the B. A. Two plus catalon and our hydroxide an ion. And for this to be balanced, we see we have on the react inside two moles of our hydroxide. So we're going to place a co a vision of two in front of our hydroxide on the product side. So we form these two ions and we need to not only find the concentration of our compound but we need to find the concentration of its ions. So first beginning with our ionic compound, we're going to beginning begin by getting its concentration with its given mass in the prompt as g of barium hydroxide. And we're going to go from g into moles by recalling that we have our molar mass. From the periodic table of this compound equal to a value of 171.35 g for one mole of our barium hydroxide. So now we're able to cancel out gramps we're left with moles but we recall that polarity is in terms of moles per leader. So in our denominator We're going to use another conversion where we're going to take that volume given in the prompt of 300 ml that our compound is dissolved in water for. And we're going to go from middle leaders to leaders by recalling that are prefix milli tells us that we have for one leader, 10 to the third power. Middle leaders were able to cancel out. Male leaders were left with moles per liter as our final units. Which is in other words our polarity. And this is going to give us a molar concentration equal to a value of 100. 95 molar. So this is our concentration of Ariana compound and now we're going to continue to find our concentration of its ions beginning with hydroxide. So we're going to begin with the concentration of the compound as a whole 1.95 moller of barium hydroxide. And we're going to use this relationship between the compound and the ion hydroxide present where we see we have a subscript of two of our hydroxide for one mole of our compound barium hydroxide. So we would have a 1-2 molar relationship. So one mole of barium hydroxide. And so this means that we actually want to interpret this polarity as units of moles per leader. And then in our numerator we have two moles of our hydroxide and ion. And so we're now able to cancel out units of moles of barium hydroxide. And we're left with moles of hydroxide per leader, which is in other words are polarity or molar concentration, which is going to give us a result of 20.39 moller. Moving on to our concentration of our barium caddy on, We begin with the same process, .195 moles per liter of our barium hydroxide. And looking at the mole ratio based on our formula, we see that we have for one mole of our compound barium hydroxide, A total of one mole of Barbarian Kati on based on the formula. And again, we're able to cancel out moles of barium hydroxide. We're left with most of our cat ion per leader and this is going to result in a concentration of .195 molar. Now we're going to move to consider our beakers when we add our barium hydroxide. And so we're going to recognize that hydroponic acid for beaker, one is a strong acid. And so when we add our barium hydroxide where we're going to recognize that it's going to be acting as a base, specifically a strong base. This is going to become neutralized. So this is going to result in a neutralization, which means that our concentration of our barium hydroxide is going to be decreased. So we'll fit that in as our concentration of barium hydroxide is going to therefore be lower because we're going to recall that in neutralization reactions. Our products are going to be assault and water meaning we won't form a precipitate. And so we're going to say that this will not form a precipitate. So moving on to our second beaker here, we have 75 mL of 750.50 molar sodium bromide. So we want to recognize that this sodium bromide is an ionic compound, it's going to dissolve to form a sodium carry on and A B. R minus an ion. Recall that bromine is in group seven A. And sodium is in group one A. On the periodic table and that corresponds to the charges. And so if we are to combine our sodium bromide with our barium hydroxide or barium hydroxide from the prompt, that would mean we have two ionic compounds reacting. And if we know that these are both ionic compounds, Ak salts, they're both just going to form products that are free floating ions. And so we would say that here for beaker to we would not have a precipitate formed since technically there won't be a real visible reaction. That happens since we're just going to have ion products when we add our barium hydroxide. Again, that is because this is a salt and so is our barium hydroxide, it's a solid A. K. A salt because they are both ionic compounds. So now let's consider beaker three for beaker three. We have 30.50 Moeller of barium bromide. And looking at our barium hydroxide, we should recognize that both of these compounds both have the barium kati on B. A. So we're going to do some work for beaker three below. Again from the prompt, we have 75 mL of our 750.50 moller, not sodium but barium bromide. And we stated that our common ion is are catty on B. A. Two plus. So we want to calculate the total amount of this common ion between our barium bromide and our barium hydroxide, meaning we're going to need to get our total volume for both of these compounds first. So we have 75 mL from our barium bromide added to 75 mL from our prompt of our barium hydroxide. This results in a total of 150 mL. So we're going to take into account first our concentration of hydroxide to get that out of the way. So we're going to take the concentration we determined above as 1500. moller. This is for 75 mL of our barium hydroxide Which as a whole is going to be dissolved in 150 ml of solution. And so We would be able to get rid of our units of ml leaving us with more clarity resulting in a concentration of .195 molar. Moving forward. We want our concentration of our barium caddy on first from our barium hydroxide. And so taking that concentration we determined from our first calculation, we have .195 moller of our compound barium hydroxide. This is for 75 ml of our barium hydroxide. Now because we need total moles of our barium catalon for both of these barium containing compounds, we're going to interpret our molar concentration and units of moles per liter since we know that they are equivalent because we want moles to be our final unit. We're going to divide or sorry, we're going to not divide but we want to cancel out milliliters. So we're going to multiply by the conversion factor to go from milliliters to liters where we have 10 to the third power milliliters for one leader. This allows us to get rid of middle leaders as well as leaders leaving us with moles. And this is going to result in a value of 0.14625 moles of our barium cadmium from barium hydroxide. Following the same process to find our barium cat animals from our barium bromide. We're going to begin with our info from the image of bigger three of barbarian bromide which gives us a concentration of 30.50 Moeller and we're going to multiply this by its volume being 75 mL. We want to get rid of middle leaders. So again going from middle leaders to leaders. We recall that we have 10 to the third power leaders for one leader milliliters for one leader we can get rid of male leaders and we're interpreting our polarity in terms of moles per liter allowing us to now cancel out leaders leaving us with moles as our final unit, resulting in a value of 0.375 moles of our barium Catalan from barium bromide. And now we want our total moles of our barium two plus caddy on. But we're gonna look at this in terms of concentration and so we're going to add up our molds together. We determined we have 0.14625 moles of our barium. We're adding this to 0.0375 moles of are very um carry on. And this is going to be then divided by the total volume. Which we determined to be 150 ml which we want to convert to leaders to get to our final unit for polarity. So again we have 10 to the third power milliliters for one leader we can now cancel out middle leaders were left with moles per liter as our final units and this is going to result in 0.3475 moller as our concentration of total barium Catalan from both barium bromide and barium hydroxide. Now that we have the concentration of both of our ions. We want to recall our reaction quotient Q. And recall that that is going to be based on our concentration of our ions. So we have barium two plus which is one of our ion products. And then we have as we stated hydroxide as one of our ion products. Recall that hydroxide when we have barium hydroxide, disassociate will need to form two moles of hydroxide. So we have our coefficient of two Now as an exponent here in our reaction quotient expression. And so now that we have this, we're going to plug in our concentrations which we determined above. So our total A concentration of barium cat and we determined is .3475 molar, multiplied by our concentration of hydroxide which we determined above as .195 molar. This is going to be squared and actually we do not need to include our units of polarity here because recall that reaction quotient has no units. And so when we plug in our calculations into our calculators were going to yield a reaction quotient of 0.01321. And so to determine whether we have a precipitate that forms from beaker three, we're going to recall the K. S. P. Given four hour barium hydroxide from the prompt which was given as 5. times 10 to the negative third power. And we should recognize that are calculated value for reaction quotient of our ions of barium hydroxide is just this value here which is actually greater than this power of 10 to the negative three. And so we would say that Q. Is greater than R. K. S. P. And so therefore the third beaker with barium bromide. When we have barium hydroxide added our barium hydroxide will precipitate. And so so far we can confirm that we have a precipitate that forms in bigger three and that is going to be our barium hydroxide. So we'll label that above. So moving on to beaker four, we have 75 molar of 750.50 moller barium bromide. So this is just a smaller concentration of our barium bromide. And we're going to follow the same process here. Since we recognize that we have the common ion barium kati on. So now we're doing beaker four. So beaker four was given as 75 mL of our 0.50 moller barium bromide. Now smaller concentration here, we recognize that we have our common ion again between our barium hydroxide and barium bromide as our barium two plus catalon. And we recognize that our total volume of our mixture when these two are added to or when barium hydroxide is added to beaker four is going to be mL plus 75 mL giving us our 150 millimeter total just as before. And so now we want to do the work of finding our concentration of hydroxide yet again. And since our volumes are the same, we know that that concentration should still be 0.1 95 solar. So we're just going to go right into our concentration of our barium caddy on from our barium bromide. Since we know that the concentration change to this lower amount being 950.50 moller. And so to calculate this, we're going to begin with that value given from the prompt, 0.50 moller, which we're interpreting as moles per liter. And we're going to multiply by the volume from the prompt 75 ml. And to cancel out male leaders, we go from middle leaders to leaders, we have 10 to the third power milliliters for one leader were able to get rid of leaders as well as male leaders leaving us with moles of our barium Catalan from barium bromide given as calculated as 0.375 moles. And from our calculation before we know that our concentration of our barium hydroxide from our sorry of our barium Catalan from barium hydroxide is this unit here. So this value won't change. We won't have to do that work for that again. So we're just going to carry this down below our concentration of barium kati on from barium hydroxide we know is still equal to 0.14625 moles. So this does change our total moles of our barium cadmium concentration, since our concentration of barium bromide is lower, we now have a total of 0.14625 moles plus 0.375 moles. And this gives us a total equal to I'm sorry, we need to incorporate our volume so we end up with polarity. So we're dividing by our total volume as 150 mL. Again, we want leaders as our final unit. So we have 10 to the third power. Leaders for one mL for one leader we cancel out male leaders. We are left with leaders most per leader. And this is going to yield a value of .1225 moller. So now comparing our reaction quotient of our ion products, barium caddy on Two hydroxide and ion which is raised to the power of two or the coefficient of two as a power. Now we're going to get 0.1225 moller which again we don't want to include those moller units because reaction quotient does not have units multiplied by .195 squared. This results in a value of 4.658 times 10 to the negative third power. And so comparing this to R. K. S. P. Given in the prompt for barium hydroxide as 5.0 times 10 to the negative third power. We see that our Q value is smaller than this value. And so we would say therefore our barium hydroxide will not precipitate, meaning that our final answer to complete this prompt since we know that beaker four, barium hydroxide does not precipitate. So beakers 12 and four will not precipitate. Whereas our addition of barium hydroxide to beaker three will result in a precipitation. Since we calculated that its reaction quotient is greater than the K. S. P. Of barium hydroxide. So answer three or beaker three is our final answer and the identity of our precipitate is our barium hydroxide. So everything highlighted in yellow is our final answers. I hope everything I reviewed was clear. If you have any questions, leave them down below and I will see everyone in the next practice video.