Four solutions are prepared and mixed in the following order: (a) Start with 100.0 mL of 0.100 M BaCl2 (b) Add 50.0 mL of 0.100 M AgNO3 (c) Add 50.0 mL of 0.100 M H2SO4 (d) Add 250.0 mL of 0.100 M NH3.
Write an equation for any reaction that occurs after each step, and calculate the concentrations of Ba2+, Cl-, NO3-, NH3, and NH4+ in the final solution, assuming that all reactions go to completion.

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Accepted Answer

Hello everyone today. We have the following problem. Consider the following solutions that are combined in the following steps provide the reaction equation that takes place in each step. Then determine the concentrations of calcium hydroxide, silver chlorate, the ammonium ion. And then we have an ammonium in the final solution. Considering each reaction goes to completion. So We're gonna start with step one. So for step one we have an initial solution containing 200 ml of .150 moller of calcium hydroxide. Well first we need to address the volume and so the volume and the question is 200 mL. However, we need to convert that to leaders and we can do that using the conversion factor that one mL ego to 10 to the negative third Leaders. When our unit's canceled out, we are going to be left with 0.2 leaders. Moving on to step two, We're going to address the volume as well. So we have 100 ml. We're going to convert that to leaders. Using the same method We're going to get 0.1 L. Now we can focus on our reaction here. So for this reaction we're going to have calcium hydroxide reacting with silver chlorate. And that reaction is going to give us silver hydroxide and calcium chlorate to balance this out, we're just gonna put a two in front of the silver and then a two in front of the silver on the right. So from this reaction we need to find the number of moles for our calcium hydroxide, calcium hydroxide and the silver chlorate, silver and chlorate. So we're gonna move down here to step to to do that. So we need to find the moles of our calcium hydroxide. And so how can we do that? Well we can take our volume. 0.2 liters. We can multiply by the polarity polarity for this was 0. 50 moller, which is going to be 0.1 50 moles over one liter. This is going to give us 0.3 moles of our calcium hydroxide. Next we need to find the moles for calcium. That's gonna be the most calcium is equal to the volume that we have And on our volume. But the number of moles that we have that we just solved for. So we just solved for this 0.03 moles of calcium hydroxide. And so we're essentially gonna do a multiple ratio by saying that in this equation that we sold for, we have one mole of our calcium hydroxide. And that is going to equal one mole of calcium itself. When our units canceled, we were left with 0.03 moles of calcium. So it's about equal. Next we need to find the moles over hydroxide. That's going to be using what we just sold for for the calcium hydroxide. We're gonna use the multiple ratio that in one mole of our calcium hydroxide, we are left with two moles of hydroxide. And so when our units cancel out, we're left with zero 06 moles of hydroxide. We need to find the moles of our silver, correct? We can do that by taking the volume that we had which was 0.1 liters. And multiplying by that maturity, which was one mole over one leader giving us Our parity which was .150 giving us actually One leader times 0.150 moles over one leader giving us . moles of our silver quarry. From that we can find the molds of our silver and our molds of chlorate. So our molds of silver. We're going to take that 0.1 50 moles of our silver chlorate. We're going to multiply that by the multiple ratio that we had which we said there's one mole of silver chlorate Is going to give us one mole of silver. We were left with 0.0150 of silver. And to save on time this is going to be very similar to the number of moles for our chlorate. So the moles of chlorate, It's going to equal to 0.015 moles as well Because of that multiple ratio being 1-1 is going to be the same. Now we have step three. So what we're gonna do, we're gonna go here and so we have step three. And so we need to worry about the volume for that. So the volume in a step three Was said to be middle leaders. So we're gonna do is we're gonna take our volume. We're gonna turn that 100 middle leaders. And we saw this earlier, we say that's .1 L. And then we write out the reaction for that. So the reaction for that, it's going to be the following. We are going to have our we're going to take our calcium and react it with this bicarbonate or carbonate here. So we're gonna take our seat a two plus which is calcium two plus and react that with our carbonate to give us calcium carbonate and hydrogen protons as well. And so to balance this equation, we're gonna need to put a two in front of the protons there and we are done. And so we're gonna need to do next is we're going to need to do a similar process as we did in our step two. We're gonna need to find the moles of each of our compounds there. So to find the moles of our carbonate, We're gonna take the volume point with leaders and multiply that by the polarity which is . moles per one liter. And this is going to give us 0.15 moles of our carbonate. We're gonna do the same thing for our hydrogen. So the most for our protons is going to be. We're going to use that moles that we found, which was 0.15 moles and use the multiple ratio That in one mole of our carbonate, We have two moles of our protons. And this is going to give us 0.03 moles of our protons. Lastly, we're gonna need to find the moles of our carbonate. It's gonna be a similar process. 0.015 moles multiply the multiple ratio which is one mole of carbonate Equalling One mole the bicarbonate equally one mole of carbonate. That's gonna give us 0.015 moles of our great. And so we see here that essentially we have a one mole to one mole ratio. So we have one mole of calcium per one mole of carbonate. And so what we need to do is we need to find the moles of our calcium that remains. So the number of moles of calcium that remains that we can do that by second d 0.3 moles of calcium And subtracting that by the moles of carbon 8 0.01 of moles of carbonate. And this is going to give us 0.0150 moles of calcium remaining. Let's take a deep breath and move on to our 4th and final step. So Step six here, we're going to move on to step four. So in step four we're gonna assess the volume, the volume was 400 ml. However, we know that that's going to translate 2.4 L by using the conversion factor. And the reaction for step four. It's gonna we're gonna be taking those protons and reacting it with our methyl amine to give us our methyl ammonium ion. And so it's already balanced. So we do not need to add any more coefficients. And so we can now solve for the moles of our methyl amine. We're gonna take our volume which was .4 L and multiply that by the malaria, which was .150 moles of that methyl amine And divide that by R one leader to give us 0. moles of our methanol. I mean we see once again we have a one mole ratio are 1 to 1 ratio with moles of our protons and our molds of methyl amine. And so we need to find the number of moles that are remaining. So the moles of our mental mean that remains Can be found by our 0.06 moles subtracted by 0.03 moles To give us 0.03 moles of our method. I mean that are left over. So and we're almost done here. We need to find the total volume. So the total volume. That's going to be our 0.2 liters plus r 0.1 liters plus our 0.1 liters once again. And then with our 0.4 liters and that's going to give us 0.8 liters. And so we're gonna use that to find the concentrations of our ions that were asked of us so far calcium. That's going to be 0.0150 moles over .8 L, giving us 0.0188 molar. Didn't we have hydroxide, Which will be 0.0 450 moles over 0.8. That's going to be 0.0563 molar. Then we have our chlorate Which will be 0. moles over .8 L. And that's going to give us 0.0188 molar. And then we're going to have our methyl amine Which was 0.03 moles over one leader. And in actuality this is going to be the same for our method, ammonium ion 0.3 moles per one leader giving us 0.375 moller. And so with that we have solved our problem overall. I do hope that this helped. And until next time

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Key Concepts

Stoichiometry

Solubility Rules

Acid-Base Reactions