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Student pertorms titration by combining 65.0 mL solution of 0.890 M sulfuric acid solution with _ 400 coba (III) hydroxide solution: Write the complete and balanced...

Question

Student pertorms titration by combining 65.0 mL solution of 0.890 M sulfuric acid solution with _ 400 coba (III) hydroxide solution: Write the complete and balanced equation for the neutralization reaction of sulfuric acid and cobalt (III) hydroxide_ DO NOT INCLUDE STATES OF MATTER: B. Calculate the volume of cobalt (III) hydroxide; In liters, that was used In this reaction: SHQW ALL_WORK FOR FULL CREDILI T Paragraph Arlal 3 Mashups(12pt)Path: ,

student pertorms titration by combining 65.0 mL solution of 0.890 M sulfuric acid solution with _ 400 coba (III) hydroxide solution: Write the complete and balanced equation for the neutralization reaction of sulfuric acid and cobalt (III) hydroxide_ DO NOT INCLUDE STATES OF MATTER: B. Calculate the volume of cobalt (III) hydroxide; In liters, that was used In this reaction: SHQW ALL_WORK FOR FULL CREDIL I T Paragraph Arlal 3 Mashups (12pt) Path: ,



Answers

A 0.500 -L sample of $\mathrm{H}_{2} \mathrm{SO}_{4}$ solution was analyzed by taking a 100.0 -mL aliquot and adding 50.0 $\mathrm{mL}$ of 0.213 $\mathrm{M}$ NaOH. After the reaction occurred, an excess of $\mathrm{OH}^{-}$ ions remained in the solution. The excess base required 13.21 $\mathrm{mL}$ of 0.103 $\mathrm{M}$ $\mathrm{HCl}$ for neutralization. Calculate the molarity of the original sample of $\mathrm{H}_{2} \mathrm{SO}_{4}$ . Sulfuric acid has two acidic hydrogens.

The reaction we're looking at is the following equation We have N h N 03 at n H is in equilibrium with n a n 03 at h tour with a 1 to 1 molar ratio. Just a note. So the concentration of H and 03 is what will consider first. So the moles of any wages concentration times volume generates the mullah value off 1.17 that is in Miller moles. So then the moles of H 03 is the moles of sodium hydroxide, which is the same value. So then we can determine concentration by most divided by volume. So we get a concentration of not point, not one for six. Next door, looking at the number of moles of sodium hydroxide that were in excess after the first edition. So the moles of sodium hydroxide needed where at that were added Sorry where 1.4 million moles, however, the molds that were needed were not 0.7 to 7 million moles, so the moles in excess was not. 70.275 millimeters by simply having the moles off added sodium hydroxide and subtracting. The moles are needed sodium hydroxide that tells us the miles of access

Yeah. Hi there. In this problem, we have a neutralization reaction. We are trying to determine the concentration of H two s 04 by neutralizing it with N A O h. Right. So I'm going to write a balanced equation for this. We haven't acid plus a base, and we're going to need to of the O. H for every h two s 04 to balance this. That is going to give us in a two s 04 and water. And we have 100 mL of this actually going to divide that by 1000 and make it leaders. We have 0.1 oo leaders of this and an unknown polarity, which is what we're trying to find the n a o h. We added 50 mL of that. So I want to put that in terms of leaders and its concentration. His 0.213 Moeller. Unfortunately, we added the n a o h. We added too much of it and we had to go back and neutralize that with HCL. So we had a had to add some hcl to that. So the first thing I need to do his figure out how much of this Any a wage was used just to neutralize the H two s 04 In other words, I'm going to have to take the moles of the H and subtract how many moles of HCL were added to that? So I need to calculate the moles of each of those first of all. So first of all, for the O. H remember that polarity equals moles per liter. So to figure out how many moles of any await I have, I am going to multiply the mole arat e times the volume. So for the O. H, that is going to be point to 13 Moeller times of volume of zero point 05 oo leaders That gives me 0.1065 moles. That's for the H for the HCL that we had to add to neutralize the excess n a o h its concentration with 0.103 Moeller and the volume of that that we added was 13.21 mL. So writing that in terms of leaders, that's 013 to 1 leaders calculating thes Together I get zero point 00 13 596 moles. So that's for the HCL. So this many moles of HCL neutralized that same number of moles of any Ohh. In other words, what I want to do is subtract thes two numbers because that will give me the moles of any ohh that reacted with the H two s 04 All right, so here I go. 0.1 065 minus zero point 00135 96 This is how many moles of the N E O. H were neutralized by the HCL. And this ends up being 0.0 9 to 904 moles. Thank you. Okay, now we have a value that we know the number of moles that were used to neutralize are 100 mL Bassett, So we can figure out the concentration of this asset. Now I am going to start off with 0.0 things number of moles of any O. H. Because this is going to be the same number of moles that is neutralized by the H two s 04 And that's gonna be the number of moles in that 100 mL sample. But what we first need to take into account is the mole ratio from the equation for every one mole H two s 04 Sorry. I found my nuh there for everyone. Mol of h two s 04 We needed two moles of the N a O h. Right now we can solve this. This is going to give us moles of H two s 04 Poor leader and doing the math. Here I get zero white zero 46 452 moles per leader for the H two s 04 or simply taking it down to significant figures. I get four point 64 times 10 to the negative second polarity or Mueller. Right? And that would be our answer for the H two s 04 Let me add H two s 04 to that. All right, so that is our answer. First determined how much of the n. A. O. H. Was neutralized by the extra HCL that we had it. Then use the remaining n a o. H to calculate the molar ity of the H two s 04 Thanks so much for watching

Hi there. In this problem, they are neutralizing some sulfuric acid with sodium hydroxide. So the first someone to do is write a balanced equation for that reaction. So what I have is H two s 04 thank you. Reacts with two n a. O. H to produce and a two s 04 and to H 20 This is a neutralization reaction where an acid reacts with a base to make a salt and water and to try to determine the mole Arat e of this h two s 04 we use 100 mL for dividing that by 1000.1 leaders. The mole arat E is what we're trying to find. So they add some sodium hydroxide to that to neutralize it, but end up adding too much sodium hydroxide. So now the solution is basic. So to correct that situation, they add some HCL to the endpoint or until we have perfect neutralization. So I also want to write the reaction for that. HCL reacts with n a O. H. And that produces in a C l and H 20 We know that we use 13.21 mL of the HCL to neutralize that excess. N a o h. I'm going to convert that toe leaders by dividing by 1000. So that gives me 0.0 13 to 1. Leaders of this and it's molar ITI is 0.103 This is all information given to us in the problem. The sodium hydroxide that was used 50 mL were used of that which, divided by 1000 gives us 10000.5 leaders and its concentration this point to 13 Moeller. Right? So the first thing I want to do is constantly is calculate how many moles of H. C l were used and how many moles of any a wage were used. So let me go ahead and do that. Starting with the HCL, I have 0.0 1321 Leaders of that have that solution. And if I calculated by the molar ity polarities of course moles her leader that will give me zero point 001 3601 rolls of hcl. All right. And then for the O. H. We have 50 mL or 0.5 leaders of the O. H, multiplying by the way, which is more clarity. Yeah, okay, gives me zero point 010 65 moles of anyway, H, this is how much anyway h is available. We can see that it is definitely mawr than the moles of HCL. So if we want to figure out how much anyhow wage was used by the HCL, we look back to the balanced equation and we see that as a 1 to 1 ratio. So if we used 0.136 So one moles of HCL, we're also going to have to have used 0.1 3601 moles of any ohh to neutralize that. Okay, so that's how much any of which was used to neutralize the HCL. And we can see how much anyway, which was available. So I'm going to subtract those two values to see how much any a wage remains because that is Thea Mount. That would neutralize our H two s 04 So I'm just going to take these two numbers and subtract 0.1065 moles minus 0.0 13 601 moles of anyway, H, remember this was the amount used. Looking in this first amount was the amount of any a wage available. The difference gives me zero 0.9 to 899 I will around a significant figures here in a few minutes when I get to the end of the problem moles of any of which. So let's think about this is this is the amount of any wage that was used to neutralize the H two s 04 Let's go ahead and do a little stuck geometry with that. If I have 0.0 9 to 8 99 moles of any O. H. Looking back to my first equation, I see that the mole ratio his two moles of any wage are needed for every one mole of the H two s 04 Doing this math gives me zero point 00 464 moles of H two s 04 That's the amount of H two s 04 that was neutralized by this base. So that's how many moles of H two s 04 were in this solution. We know that morality is defined as moles divided by leaders. So we now have the molds of H two s 04 Right? And we know from the problem that the amount we were testing was 100 mL or 1000.100 leaders. So taking moles divided by leaders is going to give us our modularity. Our polarity then is going to be four 0.64 times 10 to the negative, too. Final answer. This is theory, Jinnah. Concentration of our H two s 04 solution. All right. Thank you so much for watching, Okay?

Okay, so we have that h two s 04 reacts to produce any too. Plus, it's Joel. So we have that don't point Field 743 times 0.438 is equal to 0.3 to 6 moves of any. So it's more supply that by 1/2 to get the moves of H two s before a 0.1230 most of H 204 and then now calculating its polarity, we have the most of it to us before divided by its volume and military. So we multiply that by 1000 to return it into leaders, which is the 0.3561 polarity.


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