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Injn acid basc neutralization 116.47 mL ot 0.277 Msodium hydroxide Ihc molarity of the sulfuric acid soluticnz 'solution rejcts with 1236mL Round of sulfurk ac...

Question

Injn acid basc neutralization 116.47 mL ot 0.277 Msodium hydroxide Ihc molarity of the sulfuric acid soluticnz 'solution rejcts with 1236mL Round of sulfurk acid solutica the answer to Whatk dtrimal plares (Canvas c Amtot understand signific Jrt figures |W&Gott

Injn acid basc neutralization 116.47 mL ot 0.277 Msodium hydroxide Ihc molarity of the sulfuric acid soluticnz 'solution rejcts with 1236mL Round of sulfurk acid solutica the answer to Whatk dtrimal plares (Canvas c Amtot understand signific Jrt figures | W& Gott



Answers

In an acid-base titration, 45.78 $\mathrm{mL}$ of a sulfuric acid
solution is titrated to the end point by 74.30 $\mathrm{mL}$ of
0.4388 $\mathrm{M}$ sodium hydroxide solution. What is the
molarity of the $\mathrm{H}_{2} \mathrm{SO}_{4}$ solution?

So here we are given a specific titillation. So we know that for equivalence we require the number of moles of asked to be added to be equivalent to the number of moles of base or this is essentially age plus and minus. So we also have to be careful since we're using sulfuric acid, sulfuric acid is contributing to protons. And we also know that we can write the number of moles in terms of polarity and volumes. And we're using a monitor project based so we can ignore a factor. So this would just be the similarity of the base Times the volume of the base. And the polarity of the base were given is 2.12 molars. The volume is 35.08 mil leaders while our volume Of acid is 10. So now we only have one unknown. And we can directly evaluate for the amount of the polarity of acid. And as a result, we can find that the polarity of our certain acid is equivalent to about 3.72 molars. And this gives our final answer for the polarity of our certain by products sulfuric acid. And this is our final answer

Chapter six Problem 1 26 tells us about a sample of sulphuric acid, and it's centralization with sodium hydroxide. So silvery kassid or H two s 04 is has to acidic hydrogen ins in it. Uh, you can see that from the H two. So that means that when we react it with an a O. H, we'll need twice a smudge n a O. H as h two s 04 And this will give us the products of H 20 and in a two so four. So what's really important here is that we have to sodium hydroxide is for everyone sulfuric acid. So we're also told that we have 10 mils of sulphuric acid and we needed 35.8 mills of 2.12 Moeller sodium hydroxide and were asked for the polarity of sulphuric acid. So remember that mill Arat e is moles per leader and we already have the number of leaders here. So we really need to do is convert any O. H two moles of H two s 04 and then justified that for the number of leaders. So let's go ahead and get started remember that if we have polarity, we can convert two moles by multiplying by leaders. So 2.12 moles times 0.3508 Leaders tells us that we have 0.743 moles of any which next we can use our chemical equation and see that we have one mole of accent for every two moles of face. Therefore, we have zero went 0372 mol of H two s 04 And our last step is to divide this by leaders. So 0.372 moles divided pie 0.1 leaders That's the 10 Mills gives us a final polarity of 3.7 to Mueller.

The question is high attrition of 10 ml. Experience required 28.15 ml. Of 0.1 Mueller. Any which calculate the majority of exclusive. For So we are going to discuss how can we calculate the majority of H two so four We will be using mole ratio to calculate the moderator of extras for mole ratio of what? Malaysia of acid and base. Which are using in neutralization reaction. Okay, so what is given in the ocean? We have rolling of extra and so forth. That is the equal do in Yemen. If we convert it into a later then we will divide by how they and it will come out durable and durable. Later we have bowling of any which That is 28 .15 ML. And the cramp 0.0. Who is it 15 later. And more clarity of any. Which is given That is 0.1 morning. The reaction of H two s so full with any which will be plus extra. Now this equation is completely balanced situation. So from activation we can conclude that one mole of H2 and so forth needs to more of any questions for neutralization. So we can say that From more ratio mold of H two and so forth will be equal to mold of any which they went back to. Okay. And we also know that what is the relation between mole and maturity? More equality, more clarity into volume involving this in later. So put the value of mold in both the formulas Okay, similarity into volume of extra and so forth equal to modernity into volume of any language divide by two. Now put the values of all Arabs modernity or vegetables of what is unknown And volume of H two as a world leader is 0.01. It will do modernity of any which Is 0.1 and volume of any, which is 0.0 two, divided by two. And we will solve it. That will comes out 0.414 Mueller. That's true. That is the correct answer. Okay, thank you.

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


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