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Determine the pH during the titration of 68.0 mL of 0.366 M hypochlorous acid (K, 3.Sx10*) by 0.366 M NaOH at the following points_ (Assume the titration is done at...

Question

Determine the pH during the titration of 68.0 mL of 0.366 M hypochlorous acid (K, 3.Sx10*) by 0.366 M NaOH at the following points_ (Assume the titration is done at 25 cC.)Before the addition of any NaOHAfter the addition of 16.0 mL of NaOHAt the half-equivalence point (the titration midpoint)d) At the equivalence pointAfter the addition of 102 mL of NaOH

Determine the pH during the titration of 68.0 mL of 0.366 M hypochlorous acid (K, 3.Sx10*) by 0.366 M NaOH at the following points_ (Assume the titration is done at 25 cC.) Before the addition of any NaOH After the addition of 16.0 mL of NaOH At the half-equivalence point (the titration midpoint) d) At the equivalence point After the addition of 102 mL of NaOH



Answers

Consider the titration of 40.0 mL of 0.250 M HF with 0.200 M NaOH. How many milliliters of base are required to reach the equivalence point? Calculate the pH at each of
the following points:
(a) After the addition of 10.0 mL of base
(b) Halfway to the equivalence point
(c) At the equivalence point
(d) After the addition of 80.0 mL of base

Yeah. To solve for the volume of sodium hydroxide needed. We'll start with the volume of the acid, convert the volume two mil leaders. I'm sorry the volume in mill leaders to leaders by dividing by 1000. Then multiply by the polarity to get molds of the acid. We then recognize the reaction is 1-1. So we get mold sodium hydroxide and then we'll divide by the polarity of sodium hydroxide to get peter sodium hydroxide And then multiply by 1000 to get ml sodium hydroxide. So 59 ml of sodium hydroxide is required to carry out this type tradition. This is sodium hydroxide. Then to solve for the ph at the equivalence point the hydroxide concentration because a base has been created will be equal to the square root of the KB value which is going to be the K. A. Value for acetic acid divided into K. W. Multiplied by the concentration of the acetate that is present at the equivalence point which will be the moles of acetate which is the molds of acetic acid. We start with which is its volume multiplied by its polarity Divided by the new volume which is the 42 million L plus what we just added 59 mil leaders converted to leaders And we get 347 times 10 to the -6. This being the hydroxide concentration. We can then solve for P. H. By taking the negative log of the hydro knee um concentration which is the hydroxide concentration divided into K. W. And we get 8.54. Next for part B. We'll do the same thing. We'll start with the volume of the asset that we have convert from mill leaders to leaders than leaders. To mull with more clarity here, we recognize that we need two moles of sodium hydroxide for every mole of sulfurous acid. Well then convert the Mosul of sodium hydroxide to leaders by dividing by its polarity and then from leaders to mill leaders. So this is what's required to reach the second equivalence point half of that is what's required to reach the first equivalence point. The ph of the first equivalence point where we have the unflattering species. H S 03 minus can be calculated as ph equals one half PK one plus PK two. This is typically the best way to solve for ph when we have an ample Terek species And we get 452. Now to solve for the ph at the second equivalence point will have the soul fight and ion. And we can solve for the hydroxide concentration by taking the square root of The KB one value which will be the K two divided into KW. Multiplied by the concentration of sulfite at the equivalence point. The moles of H203 that we start with will now be equal to the moles of sulfite that we have which will be the initial volume multiplied by the initial polarity. We then divided by the new volume which will be the volume. We started with Plus 132 million liters, And we get a hydroxide concentration of 45 times 10 to the negative five ph will then be the negative log of the hydro knee um concentration, which will be the hydroxide concentration that we just calculated, divided into K. W. Giving us a ph of 969.

In this problem. We have three titrate Asians for which we need to calculate the pH at the halfway point and at the equivalence point. The first one has a weak acid being tight, traded with a strong base. So at the halfway point, we have pH equal to P. K A. M P. K will simply be the negative log of the K value given to us at 6.4 times 10 the negative five. All weak acid strong based iterations have pH equal to Pekka at the halfway point. At the equivalents point, all of the weak acid has turned into a week base. So we need to calculate the concentration of the week base and used the K B expression to solve for the hydroxide concentration. Rearranging the KB expression we get hydroxide concentration is equal to K B, which is going to be K A divided into K W. One point of time. Send the negative 14 multiplied by the concentration of the week base. The concentration of the week based could be calculated by first calculating the moles of weak acid. We started with. We had 1.1 leaders 100 mL at 1000.1 Moeller. So this product gives us the moles of weak acid we started with which is now the moles of week based formed within divide by the new volume. We started with 100 mL. At the equivalents point, we added an additional 100 mL. So the volume is 0.2 leaders or 200 mL. We then get ah hydroxide concentration of 2.8 times 10 the negative six Mohler pH. Then will be the negative log of the hydro knee um concentration, which is the hydroxide concentration we just calculated divided into K W. We take the negative log of that ratio and we get a pH of 8.45 for the next one. We're starting with a weak base and adding a strong acid. Also the pH at the halfway point when we start with a weak base and add a strong acid is equal to P K A. So we need to solve for PK To do that, we'll take the negative log of the K value which was not given to us. But KB was given to us. So if we divide KB and decay w, we get k a we take the negative log of that ratio, we get our PK A, which is our pH at the halfway point 10.75 At the equivalence point all of the week base has turned into a weak acid, so we'll use the K expression Rearrange. It's all for the hydro knee um concentration, which will be the square root of K A, which will be K B, which was given to us divided into K W, multiplied by the concentration of the weak acid, which is going to be the moles of weak base. We started with because that is now the moles of weak acid formed, which is the volume 0.1 leaders multiplied by the concentration 0.1 divided by the new volume. We started with 100 mL because the concentration of the acid is double the concentration of the week base, then it requires half assed much to reach the equivalence point. So if we start with 100 mL of weak base, we'll need 50 mL to reach the equivalence point of the strong acid. So at the equivalence point, we have a total of 150 mL, or 1500.15 leaders. This then gives us a hydrogen concentration of 1.0 times 10 the negative six. We take the negative log of that to get our pH. The equivalence point, and it's 5.96 then for the last one at halfway we have a strong acid reacting with a strong base, so we simply need to calculate the hydro knee, um, ion moles that air in excess. It'll be the moles. We start with 100 million leaders 1000.1 leaders multiplied by the concentration, minus the moles of strong base added, which is 100 mL at 1000.2. Moeller concentration. I'm sorry. It is 100 mL at 1000.25 Mueller concentration. We know this because the concentration of the strong base added is half the concentration of the week. Asked that we started with so we need double the volume to reach the equivalence point or 200 mL. So the volume at half equivalence is 100 mL or 1000.1 leaders. We then divide by the new total volume. If we started with 100 mill leaders and we added 100 to reach the equivalence point, we now have 200 mL or point to leaders. The hydro knee, Um, concentration, then is 0.1 to 5, and the pH is gonna be the negative log of that or 50.9. Now, with the equivalence point, we have a strong acid being titrate ID with a where we have a strong yeah, acid being tight, traded with a strong base. So at the equivalence point, everything should be neutralized in the Ph would be seven.

Question 56 is a weak acid week, a strong base titrate Asian for weak acid, strong based, high trey shins. We need to carry out a weak acid calculation at the equivalence point. I'm sorry at the initial point Buffer solution calculations, using the Henderson hassle Baulch equation pre equivalents a week base calculation at the equivalence point and then an excess strong base calculation post equivalents. So let's get started for the first one. We have zero mill leaders of strong base that has been added. If we have 0 mL of strong base that has been added, then all we have is the weak acid solution that is present. If all we have is the weak acid solution will carry out a weak acid pH calculation where the pH could be determined by taking the negative log of the hydro knee, um, concentration and the hydro knee. Um, concentration is simply equal to the square root of the K, a value which was given to us at 1.54 times 10 to the negative five multiplied by the concentration of the weak acid, which was given to us at 0.1 Moeller. So this right here gives us our hydroxide concentration are hydro knee, um, concentration whips. This right here gives us our hydro knee. Um, concentration. If we take the negative log of that, we'll get our P H and R. P. H is 2.91 for all the other calculations. We need to know where the equivalence point is. If we have 20 mL, appoint one Moeller butin OIC acid and we're titrate ing it with 0.1 Moeller sodium hydroxide because the concentrations air the same and the stock geometry is one toe one than the volumes will be the same. So 20 mL of sodium hydroxide is what is required to neutralize 20 mL of 200.1. Moeller butin OIC acid. So 10 million leaders is halfway to the equivalence point. Many of you might recognize that the pH for a weak acid, strong based hydration at the half equivalents point is equal to P. K. If you didn't recognize that, you would still know that free equivalence is a buffer solution and requires the Henderson hassle Baulch equation. So that's what I've done here is pH will be equal to P. K A. The negative log of the k A value plus the log of the moles. A base over the moles of acid. This is the Henderson Hassle Baulch equation. How will we calculate the moles of base? The moles of weak base formed will be equal to the moles of strong base added. The most of strong base added is simply equal to the volume of strong base 10 million leaders, or 100.1 leaders multiplied by the molar ity at 0.1 Moeller. We then divide this by the moles of weak acid that's still left in solution that will be equal to the volume of weak acid we started with, which is 20 mL or two leaders multiplied by the molar ity at 20.1 Moeller. Those moles will decrease by the moles of strong base, added the moles of strong base, added again will be equal to what we had up here in the numerator. We then carry out this calculation to get a ph of 4.81 which is because this ends up going to zero simply equal to P. K. Now for the next 1. 15 million leaders were still pre equivalents, so we still have a buffer solution, and we will still use the Henderson Hassle Baulch equation. It'll take on a format very similar to what we did in Part B, where pH is going to be equal to P. K, plus the log of the moles of base over the moles of acid. The moles of acid that we will have left will be equal to the moles of acid we start with, which is the full 20 mil leaders, or 200.2 leaders multiplied by the concentration of 0.1 Moeller minus the molds that reacted, which will correspond to the molds of strong base, added 15 mL, or 150.15 leaders multiplied by its concentration of 0.1 Moeller that will also be equal to the moles of weak base formed. Essentially, the moles of strong base added, is equal to the moles of week based formed that will give us a pH of 5.29 And as we expect as we are adding the strong base, we should see with each subsequent calculation an increase in pH, which we are seeing. This is one way to verify your answers. Now we go to Part D. Part D. Is 19 mL again, still pre equivalents. So it'll be a calculation very similar to what we had before. The only difference is now the volume of strong base added, which in this case is going to be equal to the 0.19 leaders or the 19 million liters and then up here. We will also have the volume of strong base added, which will correspond to when multiplied by the molar ity. The moles of week based formed, and we get a pH of 6.9 for the last one. We are just before the equivalents point at 19.95 mL, so we still have a buffer solution. We will still use the Henderson Hassle Baulch equation. So again, P h equals P K plus the log of in our denominator, it'll be the moles of weak acid that has not yet reacted, which is the molds of weak acid. We start with minus the moles of weak acid that reacted, which is equal to the moles of strong base added, namely 19.95 mL, or 0.1995 leaders. This, then, is also equal to the molds of week based formed, namely the 19.95 mL, or 0.1995 leaders multiplied by the molar ity. This then gives us a pH of 7.41 and we're still seeing an increase in Ph as we should now, once we reach the equivalence 0.20 mL, What we have done is converted all of the weak acid into weak base. So all we have in solution is weak base we can solve for the hydroxide concentration by doing a week based calculation. A week based calculation is the hydroxide concentration will be equal to the square root of K B. We don't have K B, but we do have a A. If we divide k into kw, this is our cabe. So hydroxide is equal to the square root of K B multiplied by the concentration of the week base will. How will we get the concentration of the week base? Every mole of weak acid has now become weak base. So we'll take the volume times the molar ity of the weak acid we started with to get the moles of weak base we now have and then divide it by the new volume. The new volume is 40 mL, or four leaders so hydroxide will be equal to the square root of K B multiplied by the concentration of the week base, and we get 5.70 times 10 to the negative six. The pH will then be equal to the negative log of the hydro knee, um, concentration and the hydro knee. Um, concentration will be equal to the hydroxide concentration divided into K. W. And we get a pH of 8.76 Now for Part G. We're just past the equivalence point a 20.5 million liters. So we have excess strong base in solution. The excess strong basin solution could be used to calculate the excess hydroxide concentration so pH will be equal to the negative log of the hydro knee. Um, concentration. Where the hydro knee, um concentration is the hydroxide concentration divided into K W. So pH equals negative log of the hydro knee. Um, concentration, which is hydroxide divided, indicate W. How do we get the hydroxide concentration? The hydroxide concentration will be equal to the total amount of strong base sodium hydroxide we've added, minus the moles of strong base that reacted which will be equal to the moles of weak acid. We started with divided by the new volume. So all of this down here is concentration of hydroxide. Divide that into kw to get hydro knee. Um, take the negative log to get our ph and we get 10.10 for the last one will do a calculation very similar to what we just did. But in this case, we are a 25 million liters. So we will have the total volume multiplied by the concentration of the strong base added, minus the moles that reacted, which will be the moles of weak acid. We started with multiplied by its concentration, divided by the new volume which is now 45 mL or 45 leaders and we get a ph of 12.5.

So starting with a here 18 02 plus in a oh, age will titrate. So age you know, two equals 25 times 0.132 which is equals a 3.3 more. And then in a oh, age is equal to 10 times your 100.1 when 16 which is equal to 1.16 mol and then h and no. Two is an excess. So use a Henderson HASA back equation and Ph Will Ego 3.14 plus log of 1.1 6/2 0.14 which is equal to 2.87 And then for B, we have aged, you know, two equals 3.3 wall in a reach is equal to 2.32 and then we have excess eight, you know, to logical 0.98 more. So then I use the same Anderson house about equation and this is our answer


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