Addition of 50.00 mL of 2.238 m H2SO4 1solution density
= 1.1243 ...

Question

Addition of 50.00 mL of 2.238 m H2SO4 1solution density
= 1.1243 g>mL2 to 50.00 mL of 2.238 M BaCl2 gives a white precipitate.
(b) If you filter the mixture and add more H2SO4 solution to the filtrate, would you obtain more precipitate? Explain.

Accepted Answer

Hi everyone for this problem. It reads a 25 millimeter of 1.186 Malala hydrochloric acid solution with a density of 1. g per milliliter was added to a 25 millimeter of 1.564 moller, silver nitrate solution, Hydrochloric acid and silver nitrate react to form a white crystalline precipitate. The precipitate formed was filtered out. Then more hydrochloric acid was added to. The fill trait will more precipitate be formed, justify your answer. So the question that we want to answer here is whether or not more precipitate will be formed and we're going to justify that answer in the problem. We're told what the reaction is. So let's go ahead and write this out. We know that we have hydrochloric acid and that is reacting with silver nitrate and this is going to produce silver chloride plus nitric acid. Okay, so to answer the question of whether or not more precipitate will be formed. What we're going to need to do is compare the concentration of hydrochloric acid, two silver nitrate. Okay, so based off of the problem, that's right. What we know, we're told what the concentration of silver nitrate is and the problem, we're told it's 1. moller. And that's the concentration. But for hydrochloric acid we're told we have 1.186 Malawi Okay of hydrochloric acid. So malaria T and morality are two different things. We're going to want to convert our morality of hydrochloric acid to malaria T. So that we can compare the concentrations. Okay, so that's the first thing that we're going to want to do to solve this problem. The first thing that we're going to want to do is convert morality to malaria T. And let's just remember that morality is mole per kilogram of solvent and Malaria T is mole per leader of solution. Okay, so let's go ahead and write out what we know. So since we know we have 1.186 malakal of hydrochloric acid, This means we have 1. moles of hydrochloric acid per one kg of water, which is our solvent. Okay, so essentially we have what we need for the moles. Okay, Because we're converting from morality to mill arat e. So we know what the moles is. What we're missing is the volume of solution in leaders. So we need to figure out our volume of solution and leaders so that we can convert. Okay, so what we're going to do now is for the mass of the solution. The mass of the solution is going to equal the mass of hydrochloric acid plus the mask of water. Okay, that's the volume of that's the mass of the solution. So let's go ahead and calculate our mass of hydrochloric acid. Okay, so our mass of hydrochloric acid. We're going to go from moles to g. So we know we have 1. moles of hydrochloric acid. So let's convert this to g. We're going to need the molar mass in one mole of hydrochloric acid. The molar mass is 36. g. Okay, so our units for moles cancel and we're left with grams. And when we do this calculation, we get 43. g of hydrochloric acid. So we know how many grams we have. Now. Let's go ahead and calculate the mass of the solution for the mass of the solution. We're going to add our mass of hydrochloric acid plus 1000 g of water because morality is one kg of solvent. And so if we convert that kilograms, two g, it's going to be 1000 g. So the mass of the solution is going to be the mass of hydrochloric acid, which we just calculated. 43.239188 g of hydrochloric acid plus 1000 g of water. So this gives us a massive the solution equal to 1043. 188 g of solution. Okay, so now we can calculate the volume of the solution. Okay, so the volume of the solution is going to be So we're going to start off with our grams of the solution. Okay, we're gonna go from What we just calculated. So we have 1, 0. g of solution. And we want to go from grams of solution to volume of solution and we can do that Because in the problem we were told the density of the solution of yes, we're told the density of the hydrochloric acid solution is 1.129 g per mil leader. So that's where we're going to use that. Okay, so we have 1.129g per one millimeter. And I'll just right underneath here that this is the density. Okay, so our grams of solution cancel And we're left with male leaders. We need the volume and leaders. So let's go ahead and do that conversion and one millimeter There is 10 to the -3 leaders. Middle leaders cancel. And we're left with leaders. So let's go ahead and do this calculation. And when we do we get 0. 04 leaders. Okay. So now we know the volume of the solution. And remember we want to go to malaria T. Which is most per liter. So now we know what the leaders is. We know what the volume is and we already know what the moles is. It's going to be the same the as what was in the the numerator of our morality. Okay, so let's go ahead and calculate the molar itty. Okay, so our malaria T. So malaria T then Is going to equal 1. moles of hydrochloric acid over the volume of solution, which we just calculated is 0.92404 liters. So when we do this, the malaria t we're going to get for hydrochloric acid is 1.283 molar. This should be an eight. Alright, so 1.283 Molar. And this is going to be our concentration of hydrochloric acid. All right, So now we know both the concentrations. So our concentration of hydrochloric acid is equal to 1. Moller. And we know what the concentration of silver nitrate is. Because that was given in the problem, it is 1. Molar. Alright, so let's look at our reaction to see what the multiple ratio is. And the multiple ratio is one. We have one mole of hydrochloric acid and one mole of silver nitrate. Okay, so when we look after concentrations we'll see that the concentration of silver nitrate is larger. The concentration of silver nitrate is larger than the concentration of hydrochloric acid. Okay, so what this means is there is an excess of silver nitrate and what that means is more precipitate will be formed. Okay, so that's going to be our final answer. So our final answer is more precipitate will be formed because there is an excess amount of silver nitrate in the solution. That is it for this problem. I hope this was helpful

Addition of 50.00 mL of 2.238 m H2SO4 1solution density = 1.1243 g>mL2 to 50.00 mL of 2.238 M BaCl2 gives a white precipitate. (b) If you filter the mixture and add more H2SO4 solution to the filtrate, would you obtain more precipitate? Explain.

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