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The titration of 81.0 mL of 0.400 M HCOOH with 0.150 M LiOH; how many mL of LiOH are required to reach the halfway point?...

Question

The titration of 81.0 mL of 0.400 M HCOOH with 0.150 M LiOH; how many mL of LiOH are required to reach the halfway point?

the titration of 81.0 mL of 0.400 M HCOOH with 0.150 M LiOH; how many mL of LiOH are required to reach the halfway point?



Answers

A 20.0-mL sample of $0.28 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ requires $43.5 \mathrm{~mL}$ of an unknown LiOH solution for complete neutralization. What is the concentration of the $\mathrm{LiOH}$ solution?

Okay. Good. Answer this question. First thing we need to do a solve for the equivalence point volume. We're starting with 40 mill leaders of 400.25 molar. Jeff So 40 no leaders leaders, then the concentration of H F is 0.25 So 0.250 moles, everyone leader, our leaders will cancel. Then we know the reaction is 1 to 1. All these tight Trish in reactions are 1 to 1. The concentration of the sodium hydroxide being added is 0.20 zero's Moeller, 0.200 moles for one leader. And then course one leader is 1000. No. So for what times too divided, we end up with a quid looks point volume of 50. No leaders then wants us to calculate the ph first after the addition of 10 milliliters after the addition of 10 milliliters, we know we are pre equivalents, so we have a buffer solution. We can use the Henderson House the vault equation. If we can calculate the moles of acid that we have left, the moles have acid. That we have left will be equal to the molds of acid. We started with It's All you times. Its polarity gives us malls mine estimate. Lt's a base that we added 10 milliliters, well supplied by concentration of 0.0, which gives us 4 to 5 times two two 0.8 moles of each and then remove the face will be equal to simply the molds of Strong basically at it. 10 milliliters 0.10 leaders by its concentration 0.2 base, which in this case is F minus. Henderson Hassle Bulge equation allows us to get pH pre equivalent for these tight rations. PH equals negative log of the chest, which is 3.5. It's not giving you to look it up 2.5 times. 10 to negative 100 floor said Negative four. That's R K a value plus the log of the moles of base 0.2 Moles of acid little equals 5 to 4. Log that ratio this two point eight five then the next part is that half a quid once what half equivalents we know P h equals p k so pH simply equal to P. K. A value here. Five. Maybe a lot of that is 3.46 for part C at the equivalent at the equivalence point. All of the weak acid has been converted into weak base so we could get our hydroxide concentration simply knowing square root. Okay, B times the concentration of the weak base F minus. It's okay, baby will be equal to point times 10 to the next 14. That's Kate W over 3.5 times 10 to the negative four. So we have 2.86 times 10 to the negative 11 and then a fluoride concentration will be equal to the point. However many moles of a strong base, we added. That's how many moles of weak base we have. We added 50 milliliters 0.50 leaders at the concentration of 0.200 Moeller. So that's now how many moles of fluoride we have. We divide that by the new volume, new volume will be the mill leaders. We started with 40 plus a 50. We get two divided zero point 11 repeating So this go here and this here and when we do that, we get a hydroxide concentration 1.78 times 10 to the negative six so if we take the negative log of that, we get a p O. H. It is equal to 5.75 We subtract that from 14 to get our Ph. Ph is equal to 8.25 Then, for the last part, it wants to know the pH after the addition of 80 mil Levi's well, after the addition of 80 milliliters, we know that we are 30 mil leaders post equivalents. So for 30 milliliters post equivalents, we have an excess amount of 30 milliliters hydroxide. So the excess volume 1.30 leaders at the concentration of the bay's 0.200 divided by the new volume. The 50 mil leaders we've added. Plus the 40 milliliters started with will give us our hydroxide concentration. Hold on a second. Sorry, 80 mil leaders, we added, before you, we started with give us a concentration of 0.5 We take the negative log of that to get our p O H. Oh H is equal to one point 30 Subtract that from 14 on. Our PH is equal to 12.70

Here we have a reaction of agent or two with K O H, which forms water and K and R two. It worked high trading 23.4 milliliters 0.39 Miller H N 02 with 0.588 Miller, Kily. In order to continue to the equivalence point, I'm gonna want to completely react pH in or two without adding any excess college so that only water and can or two are present in the solution. So in order to that, we have to have an equal number of moles, though agent or two and carriage. Since there is a 1 to 1 ratio between the two, it's the first. Let's calculate how many moles of agent or two we have. Polarity is mulls over leaders. So our number of moles of but you know to is equal to the polarity of agent or two, which is 0.390 times the number of leaders of agent or two inches zero points here 234 So we know we have point. Syria's urine hero 913 morals agent or two. That means we also need 0.913 Moles of college to completely react e it's you know, too. So now with this information we can solve for the volume, OK, Which we need to add using the same equation. Here we have, uh, leaders is equal to number of moles over the polarity. The number of leaders of K. O. H we need is 0.913 moles over 0.588 Moller. And this is equal to 0.155 leaders of college or, uh, 15.5 millimeters of college. So we have a volume of K wage needed to reach the equivalence point. And now let's find the pH at the equivalence point. All that exists in solution at the equipments Point is okay, you know, to and water. So let's look at the reaction between these two in solution. He used lions will separate from each other, and some of these. And so potassium is what's called expect today, Ryan. So we can rewrite this equation without potassium Now, like the open nice table. Our first step is to figure out how much, uh or what the concentration of and our two minuses in the initial step. We don't have to worry about water since the liquid and we don't use the liquids in our calculations. So you know that in order to reach the equivalence point were we reacted 0.0 years your 913 moles of H and R two with the same number of moles of carriage. Since there is a 1 to 1 ratio of each of these with and or two minus, we know that there will be the same amount of moles of theano to minus at the equivalence point. So we know we have your 0.913 moors of and our two minus at the equivalence point. No, in terms of volume, we started with 23.4 milliliters of a geno to and we added 15.5 milliliters of carriage. If we add those and convert to leaders, we get a total volume of 0.389 leaders. This thing gives us a polarity of your 0.235 Mueller, you know, two months So now we finally have our initial concentration been or two minus which can go in our eyes table here in our initial concentrations for the products are ceremony regional or changes than some unknown X. So we react X and O to minus. And, um, we form X of each of the approx 10 or equilibrium is just our initial plus our change. So por n o to minus 0.35 minus sex for each of the products just x You know, we know that the KP, uh, you know, to minus is equal to the concentration of agent or two times the concentration of O H minus over the concentration of and O to minus. And we also know that K W is equal to K a times K B. This is important because we can look up the K a value of, um, ano two minuses conjugate acid, which is agent or two. We can look that up in the textbook, and the value of that is 7.1 times 10 to the negative fourth. So we can use this number here to convert, um to K B, which we can then use in this equation here to sell for our values of X. So let's do that? Um, plugging in r k A. Into this equation. Here we get a K B uh, 1.41 times 10 to the negative. Love it. So 1.41 times 10 to the negative 11 is equal to or concentration of agent or two just X. You know, our table times are concentration of always dryness, which is also X just right. This is X squared, divided about our concentration of, um, you know, to minus which is your 0.235 96 in order to excelling for acts a little bit easier. Let's make the assumption that acts this so small that subtracting it from points here are 235 here is not gonna make a difference. We could make this assumption assumption as long as we check it at the end to make sure it was valid. So if we make that assumption, then we have 1.41 times 10 to the negative of a negative. 11th is equal to X squared over 0.235 Solving for X get that X is equal to 5.75 times 10 to the negative seven, and this tells us that our assumption is valid. Since 0.235 minus 5.75 times 10 to the negative some there's still point over 23523 significant figures. If our assumption was invalid, we would have to go back and, um, resold this problem for X using the quadratic equation. But since our assumption was violent, we can just move on. Now that we have our value of X, we also have our value of the concentration of hydroxide, uh, at our equivalents point. That's because our, um, equivalents that are equilibrium. We have six Moller O H minus, and now we finally have enough information to sell for a peach. P H is equal to 14 minus p o h or 14 minus the negative log of the concentration of which mints put me in our concentration of hydroxide, which is equal to X. We get a final pH 7.76 trading H two c are free with K o H. Years old to a equivalence plants because H two c 03 has two protons to lose. First, it will form HC or free minus, which will then further react with KO each to form CEO for you to minus. So let's look at these reactions one at a time. First, we start with 17.3 new leaders. 0.13 Moeller um h to see a free and, uh, your 0.5 a Moeller carriage. Just like in the last problem, we can multiply the polarity by the volume to get the number of moles. So if we moved to play 0.130 by 0.173 we get our number of moles of H 20 for you, which is, um, 0.0 2 to 5. Now we know we need 0.255 moles of ko age as well. Since we have a 1 to 1 ratio of K wage H two c 03 that reaction or tight rations. And just like in the last problem again, we can't. So for the volume of K wage needed, it's we just divide this by our polarity. Here is your 0.588 to get a volume uh, your 0.382 you're thistles also equal to 38.2 mil. Leaders No one's We've added this much k wage. We are our first equivalence point. So all we have in the solution is que HCR free and water. So we look at a reaction between the two HCR three minus is how it will exist in solution with water. All right, we have each to sear free and away triteness. Then from here we can set up our ice people. So first, how much h c 03 minus do we have initially? Well, we started our tight rations reaction with ah zero points. Here's your 2 to 5 knows uh, h two C or free which react with 0.2 to 5 moles of carriage. And since there is a 1 to 1 ratio, uh, each of these with h c 03 minus, we have the same amount of moles about agency or three minus after the reaction, once we get to the equivalence point. So you know we have points. Here's your 2 to 5 moles of H C 03 minutes, but now we need the concentration. So we divide this by our volume, which will be our original volume of age to see a free, which is 17.3 milliliters plus are added volume of carriage which we calculated to be 83.2 milliliters very 38 point to the Lakers. This gives us a concentration of corno for 05 Muller so we could put that here and we can ignore water again. Since it's a liquid and our initial concentrations of each product is zero muller, our change is then some amount. Ex uh, HCFC minus is reacting to create X each Do Seo free. And next which witness So a equilibrium. We have 0.405 minus next for a concentration of THC in three minutes and X for each other products now really quickly. Our k A values for each to see are free. Okay, One is 4.5 times 10 to the negative seven and K a two is 4.7 times 10 to the negative 11. Okay, a one corresponds to the loss of the first per ton. So Equation one and que two corresponds to the loss of the second Burton Or the second question here. No using K one, we can do the same process that we did here in the first problem. You using the value, uh, k A. That we have for each to see her free. We tend so for the K B H C 03 minus. And we get 2.22 times 10 to the negative. Now we know that the K B is equal to the concentration off the products. So the constellation of H 20 free at equilibrium multiplied by the constellation of Elektra minus that equilibrium, which is X squared, divided by the concentration Uh uh HCR three minus in equilibrium, which is 0.405 minus six. Again, we contrary making that assumption that X is so small it won't make a difference if it's here or not. And if we do that, we console for X, then we get a value, uh, tree times 10 to the negative fifth, which tells us that our assumption is valid. So no, we know are final construction of O H. Minus, which we can use to sell for a peach. P H is equal to 14 minus that I could have long of the concentration off r O H minus at equilibrium. So putting in X for our concentration of 40 to minus Here we got a pH uh, 9.48 So that's our final page for the court first equipments plan. And now we have to repeat the whole process to find me, um, volume of K, which required, and ph for the second equivalents point. So, as we know from before we start this second part of the tight rations with points here's or 2 to 5 moles of H C. 03 minus. Again, there is a 1 to 1 ratio between each CEO three minus and care, which so we're gonna need an equal number of moles of college to react completely with the HCL from minus to reach the equipments point. So, to your point years here, 2 to 5 knows, uh, carriage over our mole aren t of college, which is 0.458 and we get the same amount as we did last time. So this is the volume of K wage will need to add to the situation after reaching the first corpulence print. So you want a total volume of carriage that we need to add from the very beginning to get to the second equivalents point. It will be the addition between both of these and we get a total volume of 76.4. No leaders of carriage. So again, at this point, we're at the equivalents point, which means we're only have Casey 03 minus and H 20 present in the solution. We can set up another equation here. It's your three to minus from the Casey 03 months, plus the water. I'm a personally dissociates too. HCR free minus. And I would dryness both in the agrees, please. And again, we make a nice table. And for our initial, um, concentration of CEO three to minus we have reacted 0.0 2 to 5 moles of it's either free miners to form the same number of moles of Here are 32 months on. Our volume is going to be our, um, original 17.3 milliliters of each too seriously, plus our total addition of K o. H, which is 76 point formal leaders. This gives us a total concentration. Those your 0.0 24 speak and write that down here, ignoring water again because it's a liquid and our initial concentrations for the products is again. Zero bowler are changes again minus X on the left and plus backs on the right. So our equilibrium is 0.0 to 4. My suspects who are so free to minus and X for each of the rocks from our value of K A to for each to stay free, you can calculate KB horse ear free to minus to me 2.13 times 10 to the negative fourth. And we know this is equal to X squared from our products divine applying your 0.2 whore minus X from our recommence. If we make our small X assumption and we solve for X, we get a value of is your 0.0 2 to 6 which actually does not validate our sumption. So we have to do it the long way. Rearranging this equation, we get the expression X squared plus 2.13 times 10 to negative four times fix minus 4.81 times 10 to the negative seven equals zero. You're using the quadratic equation. We consult this to get X equals 5.49 times 10 turning to port any negative values you got 17 forex, you can just ignore since we know we need a positive concentration for, um, our products here and this year should say HCR three minus. All right now we consult for our Ph the same way we did over here when we get a final pH value about 10.7.

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.

Question. Number 58 includes two problems of a weak acid being titrate ID with a strong base for the first week. Acid. It's mano pro dick, and there's Onley one equivalents point. But for the second week, acid, it's die Pro Dick, and there are two equivalents. Points were asked to calculate the volume of sodium hydroxide required to reach each equivalence 0.1 in part A and two in Part B. Then we were asked to calculate the pH at each equivalence point for part A. We have 42.2 million liters of acetic acid at 0.5 to 0 Moeller. So if we were to take the 0.0 4 to 2 leaders, which corresponds to 42.2 mL, multiply it by the concentration. We will get moles of acetic acid. We're starting with one mole of acetic acid reacts with one mole of base sodium hydroxide because all acid based filtration reactions to reach a an equivalence point. Just one equivalents point is a one toe one stoke geometry. Once we have moles of base required, we can divide by the molar ity of the base Solution 0.372 Moeller and we will get the leaders of base required to neutralize this volume of acid, and we could multiply by 1000 to get milliliters 58.99 mL. Then we are asked to calculate the pH of the solution. At the equivalents point pH is always equal to the negative log of the hydro knee. Um, concentration. So how are we going to calculate the hydro knee? Um, concentration? Well, at the equivalence point, all of the acetic acid has turned into acetate, so we have the week consequent base of acetic acid. So we need to carry out a week base KB calculation in order to get the hydroxide concentration which we can then use to get the hydro knee. Um, concentration. So if the P H is equal to the negative log of the hydro knee, um, concentration and we are calculating the hydroxide concentration first, then we need to divide the hydroxide concentration in two K W 1.0 times 10 the negative 14 to get our pH. So P H is equal to negative log of the hydro knee. Um, concentration, which is the hydroxide concentration divided, indicate W. So How do we get the hydroxide concentration? Remember, if we have a weak base, the hydroxide concentration is equal to the square root of the K B value. We don't have the KP value for acetate, but we do have the K A value. The K B value will be the K A value 1.8 times 10 to negative five divided into K W. So the hydro knee, um concentration is the square root of K B multiplied by the week based concentration. Well, all of the weak acid acetic acid has turned into the week base. So if we take the volume times the polarity of acetic acid, we will get the moles of acetic acid we're starting with, which is now the moles of weak base acetate that we have. If we divide that by the new volume, then here. What's in parenthesis? Parentheses is the polarity of the week base at the equivalence point, the new volume being the 42.2 mL, or 0.422 leaders, plus the equivalents point volume 58.99 million liters or 0.5899 leaders. So this is the week based concentration. This is K B. If we multiply them together and take the square root this is hydroxide we defied hydroxide indicate w This gives us the hydro knee. Um, concentration. We take the negative log of the hydro knee, um, concentration and we get ph 8.54 now for the next one, we have a die protic acid sulfurous acid. The first equivalents point will correspond to just the removal of one hydrogen. So if we start with the volume of sulfurous acid in leaders, multiply it by the molar ity of the sulfurous acid. We will get moles of sulfurous acid. One mole sulfurous acid. If only one hydrogen is coming off to reach the first equivalents point corresponds to one mole of base sodium hydroxide. When we have mold, sodium hydroxide will divide by the molar iti to get leaders sodium hydroxide required to neutralize the first hydrogen of this volume of sulfurous acid. And then we can convert it to leaders by multiplying by 1000. So because we did a stoy geometry of one toe one. This is the volume that corresponds to just the first equivalents point to calculate the pH at that first equivalents point, we actually have a AMFA Terek species. The H s 03 minus that can behave as an acid and a base. So the pH is going to be equal to one half PK one plus p k two PK one for sulfurous acid is your values might differ. I'm using the values that I found 1.3 times 10 to the negative, too. And then, plus the peak A of the second hydrogen coming off of sulfurous acid 6.0 times, 10 to the negative eight. And we get a pH of 4.55 at the first equivalents point. This technique on Lee works when we have an AMFA Terek species such as create that, such as what is created when one hydrogen comes off of a die product acid. Next, we're going to calculate the second equivalents point. The second equivalents point will correspond to double the first equivalents point because the moles of sulfurous acid we start with will be equal to the moles of H. S 03 minus at the first equivalents point and then we would require the exact same amount of sodium hydroxide to convert that exact same number of now H s 03 minus into eso tu minus. So the second equivalents point would be reached at another 66.3 million liters or a total of 3100 and 32.6 million liters to solve For the pH. We're going to do something very similar to what we did in this part of the problem. PH will be equal to the negative log of the hydro knee. Um, concentration hydro knee. Um, concentration will be equal to kw, divided by hydroxide. So how will we calculate the hydroxide concentration? Well, at the second equivalents point, all we have is s 03 to minus, which is a week base to solve for the hydroxide concentration created by eso three to minus. We need to take the square root of K B for eso three to minus K B for eso three to minus is going to be a k A divided into K w. And then multiply that by the concentration of eso three to minus. The concentration will be equal to the moles of sulfurous acid. We started with however many moles of sulfurous acid we had the beginning. That's now how many moles of eso three to minus we have that can be calculated by taking the volume of sulfurous acid divided by are multiplied by the initial volume divide that now by the new volume. So this is moles as the three to minus. We have divide by the new volume the 42.2 million liters, or 0.4 to 2 liters, plus the volume. We just added 132.6 million liters, or 0.1 through +206 liters, and we get a pH of 8.33 As we expect, we should see an increase in Ph from the first equivalence point in comparison to the second equivalents point.


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