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What Is the volume of NaOH used reach thc equivalencc point? What is thc rolume of NaOH uscd to reach thc 50P0 ` (half-way cquivalcuce point)? Ahet the pKa ofthc un...

Question

What Is the volume of NaOH used reach thc equivalencc point? What is thc rolume of NaOH uscd to reach thc 50P0 ` (half-way cquivalcuce point)? Ahet the pKa ofthc unknown acid?What is the molccular weight of the unknonz acdiKhat is th: Unknoun 1cid?

What Is the volume of NaOH used reach thc equivalencc point? What is thc rolume of NaOH uscd to reach thc 50P0 ` (half-way cquivalcuce point)? Ahet the pKa ofthc unknown acid? What is the molccular weight of the unknonz acdi Khat is th: Unknoun 1cid?



Answers

What mass of $\mathrm{NaOH}$ is needed to neutralize $34.50 \mathrm{~mL}$ of $0.1036 M \mathrm{HCl}$ ? If the $\mathrm{NaOH}$ is available as a $0.1533 \mathrm{M}$ aqueous solution, what volume will be required?

So the equivalence point occurs when the number of multi asset added is equivalent to the number of moles of base. So in this case we have a volume base 0.225 mueller's. No. And since we have a mono predict based since there's only one hydroxide, the contribution is one. So we can write our number of moles of acid is equivalent to more clarity. That has some volume of acid with a certain contribution factor equals the polarity based times the volume of base required. So in the first case we have a monochromatic assets of the contribution of the the Royal Factors one and the more clarity, His 0.315, The Volume 20 ml. And we have the sufficient information to find the volume base required. And by substituting all our values in, we see that 28 mm is required. In our second part, we have polarity uh 0.250 A volume of 30. And once again we have a monochromatic assets that the contribution to the growth factor will be one. And by substituting all our values in the on base is 33 mm in their third case, instead of being given a concentration, were given a mass of six g of a certain substance, which is 99.7% pure, which means that this is about since 99.7% pure is relatively similar. It's going to be still about six g that of the actual compound we're interested in. And once again, we have a mon operatic acid. So the growth factor will be one and were given that this is acetic acid. So we need to convert this into moles. So acetic acid has a molar mass of 24 plus 4, 28 28 plus 16 plus another 16. So the molar mass would be 60 g per mole. So we can find in this case that the number of moles that acid is 0.100 moles. Yeah. So by finding this, we can see that from our equation above. We can directly solve in this case. Since the polarity comes, the volume is just the number of motive. A certain compound added. Then we divide this by .225. Yeah. And we can find that the okay, we can find that the volume of base required is equivalent to 0.4 for military leaders in this case. Or this is equivalent to 440. Mhm. Yeah. And this is our final answer. Sure.

When calculating polarity of sulfuric acid by knowing smalls and dividing it by its leaders. So polarity sulfuric acid is simply most sulfuric acid divided by leaders. We can calculate the most sulfuric acid by taking the volume of sodium hydroxide that was required to react with sulfuric acid, converting its leaders and then converting the volume, or the leaders of sodium hydroxide to moulds, sodium hydroxide using the polarity of sodium hydroxide. Then we recognize the story. Geometry of the reaction has given to one to mull sodium hydroxide required for everyone. Mold that sulfuric acid that then gives a small sulfuric acid in our numerator and the denominator. The volume that was tight traded was 20 milliliters. Sulfuric acid will convert that to leaders by dividing by 1000 that then gives us similarity of 0.234 molar sulfuric acid

Okay. So we're going to react a strong acid with a strong base and we're just going to go ahead and make some water this right to start with the we know that the P. H. Is 13.68. So our H plus concentrations the ph of our Noh solution. So the H plus is 2.1. I'm saying the -14 Moller. And if we change that to hydroxide kW over H plus Will give us .48 moller solution of R. O. H minus. Mhm. For our strong acid. Okay, are folkloric acid? We know we've got .128 more And .0350 leaders. So that's gonna give us .00448 moles of H plus. Now when we titrate that means it's going to require the same amount of O. H minus. So if we date that and divide it by arm polarity which is 0.48 moller which is moles per liter. We'll get our leaders of O H minus that. We needed to add 2.0094 l Or 9.4 mm. That's the amount of O. H minus we needed to add. And then we want to know what the ph of the equivalence point where all the ph is seven. Okay, Ph is seven at the equivalence point for all strong acid, strong based it rations. Okay, so now we're going to try to find the concentration of a couple ions. So for any O. H. Which is where you're an A plus is coming from. We said we had 48 moller. Okay, that was up here and We're going to dilute that right? We started with 94 ml And we'll divide that by the total volume which was we took 35 mm and added 94. So 44.4 mm is our total volume after the titillation. Okay, so this gets diluted, So this comes out to be .10 moller. So our concentration of our sodium ion mm There's no one there is going to be .11 More. And then for our concentration of our hcl Oh right. We know that we had zero 448 moles. Right? That's where the each class came from. Mm And that was going to be in .0444 leaders. Right. Our total volume. So that's going to give us .11 moller Hcl four. But really we just want the per chlorate. Okay, so cielo four minus

Questions 15 points 16 is a rather lengthy problem. Has three parts three different type Trish in calculations. In addition to also calculating the equivalence point value which is necessary in order to make sense out of the three additional penetration volumes to give you to solve for the equivalence point volume, you will take the initial volume of the analyze that you have multiplied by its concentration to figure out the moles of the h o. C. L that you have then recognized that the reaction of HCL with an a O. H is a 1 to 1 molar reaction. You will then have at this point the molds of strong base that you need to add using the polarity of the strong base. You can calculate the moles of strong base needed, man using the polarity, you could calculate the leaders of strong base needed. Then you can multiply by 1000 in order to get the volume milliliters of strong base needed for the equivalence point. Volume is 40.0 no leaders. So after the addition of 10 millilitres, we know that we are pre equivalents because 40 is the equivalence when we are pre equivalents with a weak acid being traded with a strong base. We have a solution. So we need to know the moles of weak acid left in solution and the moles of base that we perform. Moles of weak acid left in solution is going to be equal to the molds of weak acid. We started with polarity times volume multiplied by the moles of weak acid that reacted most of weak acid that reacted will be equal to the molds of strong base form. So polarity times following you. Then the molds of week based form will be equal to the moles of strong days. Added for the molds of strong base. Added again is 10 litres multiplied by the concentration that strong based 100.4. Now, knowing the moles of weak acid and it's confident base left in solution, we can use the Henderson Hassle Baulch equation. P H is equal to peek a negative lock into K, a value that was given to us, plus the log of the molds of the base over the most of the acid, which just calculated appear well. Then set Calculate that ph of 6.98 There we go to Part B, which asked us to calculate the pH of the equivalence point pH in the equivalence point for a weak acid strong based filtration which this is always equal to p k p K. Then simply the negative log of Kay You, which was given to us which in this case, a 7.46 Then at the equivalence point, we need to recognize that all of the weak acid has reacted with strong base and has now become weak base. So all we have in solution is the base. If all we have in solution at the equivalence point is the face, then all we have to do is carry out a base pH calculation. A week based pH calculation requires us to no concentration of the weak base. The concentration of the base is going to be equal to the molds of strong base added, or it's also equal to the molds of weak acid. We started with most weak acid we started with is going to be the volume 100 mil leaders multiplied by the concentration 1000.16 You then have to divide that by the new total volume. We started out with 100 mil. Leaders calculated earlier that the equivalence point volume was 40 No leaders. That then gives us a concentration of weak base at 1.14 times 10 to the negative too. The pH From that the hydroxide concentration for a solution containing a weak base is always equal to okay, be value for the weak base which we can get from the K. A value for the weak acid. K B is the kw value which is 1.0 times 10 the negative 14 divided by K. So this right here is our K B value multiplied by the concentration of the weak base and then we square. So that gives us five 5.71 times 10 to make it to five Moeller concentration for hydroxide get pH from hydroxide when you take the negative log of the hydro Liam concentration which we get from the hydroxide concentration by dividing it into kw. So this right here represents the hydro nian concentration that we will take the negative log in order to get our ph of 9.76


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