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Ch.20 - Nuclear Chemistry

Chapter 20, Problem 120

Imagine that you have a 0.007 50 M aqueous MgCl2 solu-tion, prepared so that it contains a small amount of radio-active 28Mg. The half-life of 28Mg is 20.91 h, and the initial activity of the MgCl2 solution is 0.112 mCi>mL. Assume that 20.00 mL of this MgCl2 solution is added to 15.00 mL of 0.012 50 M aqueous Na2CO3 solution and that the resultant precipitate is then removed by filtration to yield a clear filtrate. After a long break to go for a run, you find that the activity of the filtrate measured 2.40 h after begin-ning the experiment is 0.029 mCi>mL. What are the molar concentrations of Mg2+ and CO32- in the filtrate, and what is the solubility product constant of MgCO3?

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Welcome back, everyone. We have a laboratory student who prepared 0.1 55 molar potassium bromide solution with a small amount of radioactive bromine 77 with a half life of 57 point oh four hours. She recorded the activity of the solution as 0.1 27 militaries per millimeter and used 25 mL of this solution to precipitate an unknown medal. Why as Y bromide, she makes the potassium bromide solution with 25 mL of 250.0 to Mueller Y nitrate solution and remove the precipitate by filtration. The activity of the obtained filtration after five hours was 50.1 militaries per mil leader. Calculate the molar concentration of Y Catalan and bromide and the salt ability product of Y B R. So recall that because we have the radioactive decay of Romaine 77. Whenever we have radioactive decay we're always going to follow first order kinetics. And so we're going to recall the following rate law Where we would take the natural log of N where N represents our amount of browning 77 That remains after five hours of decay Divided by N0, which represents our amount of grooming 77 before the five hour period. So initially, this is set equal to negative one times our decay constant K. So K is the decay constant. And recall that we can calculate the decay constant by taking the Ln of two over our half life of our radio isotope K is then multiplied by the amount of time that passes where according to the prompt T is equal to five hours. So we're going to begin by getting our first step of the problem which is to solve for the activity of our potassium bromide. And why nitrate solution after five hours. And we're going to use our rate formula above. So what we have is the Ln of N. So we're solving for our numerator N and we're dividing this by N zero, which we're told that the activity of the solution initially is 00. 27 million curies per millimeter according to our prompt, This is then set equal to negative one times our rate constant or decay constant K. So the Ln of two will result in a numerator being 0.693. And our half life for the bombing 77 is given in the prompt as 57.04 hours. This is then multiplied by our time that passes for the decay being five hours as given in the prompt. So we're going to begin by simplifying the right hand side. So taking the product and the quotient here. And what we'll have is the same left hand side of our formula so far. So we're just gonna copy and paste that below so we can save time. So the left hand side is going to be equal to the product of the right hand side, which will result in a value of negative 0.607468. And as far as our units of ours, we have inverse hours which cancels out with ours here. And so we don't have any units on the right hand side. Now, our next step is to cancel out the lantern recall that we can use dealer's number as making both sides of our equation exponents to dealers number to cancel out Ln. And so now what we'll have is we'll just have our quotient here which will copy and paste down below, Which is set equal to the right hand side which results in a value of 0.9410615. And now we just want to get rid of that denominator. So we're going to multiply both sides by .127 militaries per millimeter. So again, .127 militaries per millimeter on both sides so that it cancels out on the left hand side. And what we're left with is our simple N is equal to a value of 0.1195148. Our units, our military's per millimeter as the activity of our solution of potassium bromide and y nitrate after five hours. Now, with this activity found we're gonna move on to our next step where we need to consider our moles of the compounds making up our solution. So we need to find most of potassium bromide and moles of why nitrate recall from the prompt that were given a concentration of potassium bromide being 0.1 molar and were given a volume used as 25 mL. And then for wide nitrate recalled that were given a concentration of 0.0 to 34 molar and also used a volume of 25 mL. Now, what we're going to use is still Akiyama tree. So we're gonna begin with the concentration and recall that we can interpret polarity or molar concentration in terms of moles per leader. And so we're going to interpret this concentration as 0.155 moles per liter of potassium bromide. And sorry, just so that this is clear, let's move this over a bit. Okay. So next, we want to get rid of the term leaders because we want to be left with moles are final unit. So we're given that volume in milliliters and we're going to multiply to convert from male leaders in the denominator to leaders in the numerator by recalling that are prefix milli tells us that we have 10 to the negative third power Of our base unit leader. So canceling out male leaders were left with leaders and we're going to be able to multiply now by 0.025 leaders as the volume used, which allows us to cancel out leaders leaving us with moles of potassium bromide, our first free agent, which is what we want. This is going to result in a value Equal to 3.8, 7, 5 times 10 to the negative 5th power moles of potassium bromide. And now we're going to move on to our second re agent which is our Y nitrate. So we're interpreting its molar concentration as 0.0 to moles of Y nitrate per leader where will multiply by the same volume used, which we converted to leaders as 0.0 point 025 liters. So canceling out leaders were also left with moles of ry nitrate. And in this case, in our calculators will get a result of 5.85 times 10 to the negative fifth power moles of Y nitrate. And because we see that this value is much larger than our moles of potassium bromide. We can say that potassium bromide is our limiting reagent. Whereas why nitrate is therefore in excess? So if we understand that our why nitrate is in excess, then we're going to now in our next step, assume that all of our Y plus cat eye on associates into our Y bromide precipitate, which is a solid. And so we're going to write out the full equation because we know that Ry catalon in the net ionic equation is reacting with our bromide an ion to give us Y bromide solid. And so we're going to set up an I C F tart for the final condition where we have our full precipitate formed as a solid. So plugging in our molar concentrations or rather molds of each of our re agents. From above, we solved a total of 5.85 times 10 to the negative fifth power, most of our white cat ion or Y nitrate. And for bromide, we solved a total of 3.875 times 10 to the negative fifth power moles of potassium bromide. And then for our Y bromide solid, it's a product initially. So it hasn't formed yet. So the initial concentration is zero most. Now for the change, we know that our potassium bromide being our bromide an ion is our limiting reactant. And so we're going to subtract from our limiting or by our limiting reactant. So by minus 3.875 times 10 to the negative fifth power on both of our reactant. And for our product, it would be plus this amount. So plus 3.875 times 10 to the negative fifth power. And so for the final condition, we would have a total of 3.875 times 10 to the negative fifth power moles of ry bromide solid, which forms for the condition of bromide at its final point, we would have zero. And then for Ry carry on what we would have from that difference as a result of 1.975 times 10 to the negative fifth power and we have moles. So with this understanding of how much we can, can we have left in the final conditions of our fill trait, we can now calculate our white cat ion that we have in excess. And so by taking the final concentration as 1.975 times 10 to the negative fifth power molds, we're going to divide by the total volume which we know we use 100.25 liters of both of our solutions of Y bromide. And we're sorry of why nitrate and potassium bromide. And so in our calculators, this will result in our excess Y carry on equal to a value of 3.95 times 10 to the negative fourth power moller. Now, with the smaller concentration, we're going to consider ourselves ability equilibrium of our excess Y bromide. And so we're going to consider its dissolution. So we have Y bromide which is in equilibrium with its ions being the white cat ion and the bromide, an ion. And so in this case, since we're considering an equilibrium, we're going to use an ice charts. So our letters R I C E now. And so for the initial condition, we're not going to recall, consider solids, we only consider Aquarius re agents. But for Ry carry on, we just determined that we have in excess of 3.95 times 10 to the negative fourth power moller. And for bromide, we have none left, it was limiting. So we have zero initially, the change is going to be plus X for our products. And for equilibrium, what we'll have is for Ry caddy on 3.95 times 10 to the negative fourth power plus X. And for bromide, we just have X. And so we need to figure out what this X value is to get our first two final answers. And so moving forward in our solution, just want to scroll down for more room. We're going to recall that we can use the following setup where we can consider the concentration of our bromide an ion Over the activity of our solution with Romain 77. But specifically what resulted for the filtration. And so actually, we're considering the, the precipitate Y bromide for our filtration. This is set equal to our concentration of bromide an ion again. And the activity of our roaming 77 in the initial solution Where we want to make note of the fact that according to our prompt, the activity of Ry bromide for our precipitate was equal to a value of 0.0156 million curies per mil leader. So this came from the prompt. And so we're going to plug in what we know into this formula so that we have that our concentration of bromide an ion divided by our activity of our bromide an ion infiltrate. As given from what we noted above in the prompt is 0.0156 million curies per millimeter. This entire quotient is for a fill trait and a set equal to our right hand side where our initial concentration of bromide, an ion with regard to potassium bromide in our initial solution as given in the prompt as the value 0.1 55 molar, which we're dividing by the activity of our initial solution, which we calculated in our first step As 0.1195, 14, 8 militaries per mil leader. And so we want to multiply both sides by the denominator, 0.1 56 million curies per millimeter so that it cancels out on the left hand side. And on the right hand side, We multiply by 0.0156 million curies per mil leader. And this is a zero here. So this will simplify in our next line to our concentration of our bromide and ion in the filtration equal to a value of 2.231803 times 10 to the negative fourth power which we will round to about three sig figs, which is our minimum number of sig figs based on the information in the prompt as 2.2 times 10 to the negative fourth power moller. And this is what is equal to X in R I start above. So this would be our first answer as our concentration of our bromide, an ion in the fill trade product of Y bromide. Now we're going to consider our Y plus catty on for the fill trait concentration. And so from our ice chart, we stated that that would be 3.95 times 10 to the negative fourth power plus X. And so we would have 3.95 times 10 to the negative fourth power moller and sorry, this should say moller in both of our steps here. This is plus R X which we determined to be 2.231803 times 10 to the negative fourth power moller. This sum will result in a value of R Y cada infiltrate concentration equal to 5. times 10 to the negative fourth power moller, which we will round to about three sig figs as our minimum number of sig figs being 5.97 times 10 to the negative fourth power moller. This is our second answer as our concentration of why carry on in our fill trade product. Why bromide? Now we want to recall our K S P which calculates the concentration of are soluble and ions as a product of one another. So we would have the concentration of ry catalon raised to its Mueller coefficient in our, in our equation which will write out next, which is multiplied by our concentration of bromide an ion which we raise to its Mueller coefficient from the equation. And you can see that our equation, we have coefficients of one because we have potassium bromide reacting with why nitrate? They're both Aquarius. And this will result in our products being our precipitate or are fill trait being Y bromide solid as well as potassium nitrate. And in our net ionic equation, we have our bromide an ion plus ry caddy on which forms are Y bromide solid. And so you can see we have coefficients of one in the net ionic equation as well as in our overall equation here. So now that we have our values for K S P plugging in our concentrations, we can say that R K S P is equal to 5.9731803 times 10 to the negative fourth power raised to the first power as our concentration of like a day on which is multiplied by the concentration of bromide, which we solved as 2.231803 times 10 to the negative fourth power raised to the first power. And this gives us a K S P equal to 1.208448, 207 times 10 to the negative seventh power, which will round to exactly three sig figs as 1.21 times 10 to the negative seventh power. And so based on this very lo que sp we can say therefore, Ry bromide solid is not very soluble in water. And so for our third final answer, we've successfully found R K S P of our fill trait. Why bromide as 1.21 times 10 to the negative seventh power? And note that K S P does not contain any units. So our final answers highlighted in yellow correspond to choice be in the multiple choice as our final answers. I hope everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.