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When 30.OmL of 3.0M HCI is reacted with 30.OmL of 3.OM NaOH would the resulting solution be acidic, basic, or neutral?When 20.OmL of 3.0M HCI is reacted with 30.OmL...

Question

When 30.OmL of 3.0M HCI is reacted with 30.OmL of 3.OM NaOH would the resulting solution be acidic, basic, or neutral?When 20.OmL of 3.0M HCI is reacted with 30.OmL of 3.0M NaOH would the resulting solution be acidic; basic, or neutral?When 30.OmL of 3.OM HzSO4 is reacted with 30.OmL of 3.0M NaOH would the resulting solution be acidic, basic; neutral?23_ When 30.OmL of 3.0M H,SO4 is reacted with 60.OmL of 3.0M NaOH would the resulting solution be acidic, basic, or neutral?24. When 30.OmL of 3.0M

When 30.OmL of 3.0M HCI is reacted with 30.OmL of 3.OM NaOH would the resulting solution be acidic, basic, or neutral? When 20.OmL of 3.0M HCI is reacted with 30.OmL of 3.0M NaOH would the resulting solution be acidic; basic, or neutral? When 30.OmL of 3.OM HzSO4 is reacted with 30.OmL of 3.0M NaOH would the resulting solution be acidic, basic; neutral? 23_ When 30.OmL of 3.0M H,SO4 is reacted with 60.OmL of 3.0M NaOH would the resulting solution be acidic, basic, or neutral? 24. When 30.OmL of 3.0M H,SO4 is reacted with 30.OmL of 6.0M NaOH would the resulting solution be acidic, basic; Or neutral?



Answers

The three flasks shown below depict the titration of an aqueous $\mathrm{NaOH}$ solution with $\mathrm{HCl}$ at different points. One represents the titration prior to the equivalence point, another represents the titration at the equivalence point, and the other represents the titration past the equivalence point. (Sodium ions and solvent water molecules have been omitted for clarity.) a. Write the balanced chemical equation for the titration. b. Label each of the beakers shown to indicate which point in the titration they represent. c. For each solution, indicate whether you expect it to be acidic, basic, or neutral.

So the question here is going to talk about acid base reactions here. So in this case, we have an acid based hydration where we have sodium hydroxide reacting with hydrochloric acid again, both being strong here. So in this case, they're going to go into reaction. And once we do, our substitution reactions, but we're gonna add is assault. So he inquired and water as our reaction products here. So again, for the first part, it wants us to find the moles of hot of protons in HCL. So essentially that's gonna be the as we recall, moles is gonna be the volume times the concentration. So 24.58 over 1000 times 0.102 for be here. It wants us to find a polarity, the moles rather of hydroxide in the n a o H solution. So again, find moles. We essentially need Thio know that in this particular case, um, as we are taking 23.91 minute leaders of sodium hydrogen and that neutralized, in other words, is gonna be at equivalents point of the penetration. We know that the moles of hcl so the moles of hcl is gonna be equal to the moles of N a. O. H. Therefore, to find the polarity, we're essentially going to take our moles here and divided by 23.91 over 1000 here as we have to convert that into leaders. Um, lastly, for part C here, it's asking us to obtain the polarity of N a O H from O H minus. So again, as we know that this is going to be one toe, one reaction C is gonna be equal to the answer that you get to be here.

Question Number five asks you to determine whether or not the final solution is acidic, basic or neutral after mixing certain volumes at particular concentrations of HCL and N a. O. H. To answer this question, you need to figure out how many moles of HCL you have and how many moles of N a. O. H. You have to calculate the moles of HCL that you have. In each case, we're taking 100 mL at 1000.2. Moeller 100 mL is 1000.100 leaders in the concentration of point to Moller. If we take the mole arat e times the volume that will give us malls HCL So we have 0.2 00 moles HCL. We then need to compare the moles of HCL to the moles of O H that we have added. If we have more Mel's Ohh! The solution is basic. If we have an equivalent number of molds of any ohh as HCL, the solution is neutral. And if we have fewer moles of any ohh than HCL than the solution is a city because we have extra acid that was not neutralized by the Ohh that we added So for part A, we add 20 mL, which is 0.20 leaders at a concentration of 0.50 Moeller. So that gives us zero 0.10 malls. Ohh! We compare that to what we have up here. We added fewer moles of N a O. H, which means we have excess hcl, so the solution is acidic. For Part B, we add 40 mil leaders, which is 0.40 leaders at 0.5 Moeller zero 0.50 Moeller That then gives us zero 0.20 malls N a o h. So we have equal moles sodium hydroxide as n a. O. H. The solution has reacted completely with all of the acid. There is nothing in excess, so it is neutral for the last one. We add 60 mL, or 600.60 leaders at 0.50 Mueller. And as you might guess now that we have added mawr ohh than neutral, we should end up with Mawr n a O. H. After the reaction has occurred and we have mhm five times 50.603 moles n a o. H. So because we have added 0.3 moles of any O. H. And we only had 00.2 moles of HCL than subtracting those two. We end up with 0.1 moles, Noh and excess. And if we have extra any ohh, then the solution is going to be basic.

So our thai tradition is between a strong acid and a strong base. So our net ionic equation it's just gonna be H plus plus O. H minus gives us water. And at the equivalent point other than water we'll just have our two spectator ions also present. Okay and then we're gonna see if we can figure out how much. Anyway H. We need to get to our equivalence point. So let's start with our information about our strong acid. 5.1988 Moller multiply that by its leaders And we'll see that we had 7.36 Times 7 -3 Malls of our strong acid. Therefore we must have also reacted that same amount of our strong base to R. O. H minus. And then if we take that and divide it by the polarity will get our volume. Okay because polarities moles per liter. So that will give us point 03457 leaders or 34 .57 ml. Okay so before the titillation occurs, all we have is our strong acid R. H. Plus concentration Was given as .1988 Moller. So if we take minus the log of that, we'll get our ph Which is 7016. Yeah, mm. Yes, halfway to the equivalence point one, half of the moles of each plus have reacted. So we're left with half of our moles left and we've added half the volume of R. O. H minus. Okay, so the moles of Rh plus that are going to be present Is going to be 73, 6 Times 10: -3 divided by two. So we'll have 3.68 Times 10 to the -3 moles of O of H. Plus that remain. They haven't reacted yet. And we used 34 .57 ml. Well, at this point we have only added half of that. So 17.28 ml. So if we add that to the 37 ml that we had of Rh I who will get our total volume at that point? Which is 54.28 ml or .054- eight leaders. Okay, so our concentration of H plus halfway to the equivalence point Is our moles 368. I'm Santa -3. And we'll divide that by our total volume and we'll get our more clarity as 0678 Moller. That's our age plus. So we take minus the log of that. We'll get our ph as 1.17. Okay. And then finally they ask us, what's the ph at the equivalence point, while the ph is seven, Okay, anytime you have a strong acid and a strong base, the ph of the equivalence point is going to be seven

Number 1 18 is simply a die protic acid titrate Asian calculation. The two pH is that air given really are not important to the calculation. We simply need to recognize that the sodium hydroxide has neutralized the dye protic acid with the addition of 31.50 mill leaders of the 0.984 Moeller sodium hydroxide solution. So because we recognized, then that the final failing indicator is going to change color close to Ph. Nine, the only purpose of the two pH values that they gave to us is recognizing that the second equivalence point is going to occur at Ph. 8.8 and therefore when final failing changes color were at the second equivalents. Point meaning sodium hydroxide has reacted with both the first and the second hydrogen ions on Milan IC acid. So we now know this, like geometry is 2 to 1. We need two moles of sodium hydroxide for every one mole of Milan ic acid to reach the second equivalents point so we can calculate the moles of sodium hydroxide, added convert two moles of melodic acid with the 2 to 1 relationship, then divide the most of Milan ic acid by the newt. The total volume which is going to be the original 25 mL of Milan ic acid before we added the sodium hydroxide. So this is not the some volume. This is just the volume of Milan ic acid before titrate Asian, and we get 250.62 Moeller melodic acid.


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