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4. Calculate the pH at each point for the titration 0f 25 mL of.2 M CH:COOH with M NaOH:a before any NaOH is addedb after 5 mL NaOH is addedc after 12.50 mL NaOH is...

Question

4. Calculate the pH at each point for the titration 0f 25 mL of.2 M CH:COOH with M NaOH:a before any NaOH is addedb after 5 mL NaOH is addedc after 12.50 mL NaOH is addedd.after 25 mL NaOH is addede: after 26mL NaOH is added

4. Calculate the pH at each point for the titration 0f 25 mL of.2 M CH:COOH with M NaOH: a before any NaOH is added b after 5 mL NaOH is added c after 12.50 mL NaOH is added d.after 25 mL NaOH is added e: after 26mL NaOH is added



Answers

Calculate the pH during the titration of 40.00 $\mathrm{mL}$ of 0.1000 $\mathrm{M} \mathrm{HCl}$ with 0.1000 $\mathrm{M}$ NaOH solution after the following additions of base:
(a) 0 mL (b) 25.00 mL (c) 39.00 mL (d) 39.90 mL
(e) 40.00 mL (f) 40.10 mL (g) 50.00 mL

Let's calculate the pH after 0.2 Mole of any O. H is added to one leader of the following solutions from exercise 25 for a going back to exercise 25 we have one leader of 0.100 Moeller each c three h 502 which is a weak acid. And to this we're gonna add some any Ohh! So H C three h 502 also which minus or produce c three h five 02 minus in each to own. And we can set up a nice table here. Initially, we have 0.1 Moeller and the polarity of the O. H is equal to the majority of the O. H. Minus. So 0.20 moles in one leader would be 10.20 Mueller just to be 0.20 Moeller zero minus 0.20 Mueller minus 0.20 Moeller and plus 0.20 Moeller. This will yield US 0.80 Moeller zero and went 0 to 0 Moeller To calculate the pH of the solution, we'll need Thio User Henderson Hasselbach Equation P H is equal to the peek. A peek a is 1.3 times 10 to the negative five. Plus, the log of the conjugate base is 0.20 Mueller over the acid. His 0.80 Mueller. We'll find that the peak A is a 4.89 and the second half is minus 0.60 Therefore, my peach, after a point 20 mole of any O. H. Has been added 2.1 Moeller one leader h three h c three h 50 to the pH works out to 4.29 for be. We have from exercise 29. We have one leader of 0.1 Moeller. Then a C three h 502 on C three h 50 to minus is a week Peace. So we have any Ohh. And, um this n a c three h 502 s, o n e, which is a strong base n a c three h 50 twos, Weak base. So here are pH depends on the strong base, as it will completely ionized and dominate. Ph. Therefore the majority of any Ohh physical too minus son. Any wages equal to the military of always minus 0.20 moles per one leader. This works at that 10.20 Mueller P O. H is equal to negative log, which minus which is equal to negative log 0.20 and we get a pH value of 1.70 P. H is equal to 14 minus 1.70 on the pH of our solution is equal to 12.30 for see we're going to add are zero to Mueller nuh into pure water. And so, um, we're going thio. So for the concentration of O. H minus, which is equal to any which, when 0 to 0 malls in one leader in this is equal that 10.20 Mueller P O. H. It's equal the negative log oh H minus, which is equal the point. Karim Negative log 0.20 and that's equal to 1.70 p. H. Is equal to 14, minus 1.70 on this is equal to P. H 2 12.30 So impure water, our Ph others will results at 12.30 and for D from Exercise 25. We have a mixture of 250.1 Moeller H C three h 502 and 0.1 Moeller of N A C three, h 502 or sodium being the spectator C three H 50 to minus and we're gonna have one leader of this mixture so set up our equation here. And to this, we're gonna add the 0.20 mole of nuh so h c three h 502 Add Oh h minus. Well, pretty C three, h five, 02 minus and H 20 and we'll separates table here on we have 0.1 Mueller 0.1 Moeller. And from part A. We saw for the morality of oh H minus nuh But H minus is equivalent to that strong basis 00.20 Mueller minus 0.20 Mueller for zero minus 00.20 Moeller 4.80 Mueller and plus 0.20 Mueller 4.1 to 0 Mueller. Now let's calculate the pH using your Henderson Hasselbach equation. Peek a negative log of 1.3 times 10 to the negative five plus log of similarity of the basis 0.1 to 0 Moeller. End of the conjugate acid here is 0.80 Mueller 4.89 plus 0.18 and we'll find the pH of the resulting mixture equal to 5.7 mhm.

If you're titrate 150 mL of 1500.1 Moeller strong acid h I with sodium hydroxide. And after adding 20 mL of 200.25 Mueller, we can calculate the hydro knee, um, concentration. That's left in solution by recognizing H. I completely dissociates. And the moles of hydro knee um, that is produced will be equal to the moles of H I present, which will be 150 mL, or 1500.15 leaders multiplied by its concentration. This will be the moles of strong acid moles of hydro knee. Um well, then subtract off The moles of strong base added 20 mL 200.2 zero Leaders multiplied by its concentration because every mole of sodium hydroxide we add will decrease the moles of strong acid present. And then we divide by the new volume 150 mL plus 20 mL and then take the negative log of this hydro knee, um, concentration to get our ph of point one. I'm sorry. 1.23 In order for the pH to be seven, we need to reach the equivalence point. So we need to figure out the volume of sodium hydroxide that is needed to reach the equivalence point. And because the reaction is one toe one, we can simply take the moles of theme the strong acid H I that we start with and divide by the polarity of sodium hydroxide, and we will get the volume of sodium hydroxide 60 mL.

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

You're part of the problem. The very first thing that we will need to do is you will have to find the initial moles of any Ohh. And how we do that is simple dimensional analysis going to take the value of any wage that we're adding and we're gonna multiply it by one leader is every 1000 milliliters multiplied by the moles of H. C O per every one liter of solution. And when we do this, we will get 0.58 moles. Now we need the volume of eight. CEO need it in our inner solution. So we're gonna take the moles we found before and want to play it by our 1 to 1 molar ratio of our reaction and multiply it for everyone. Leader of a solution. We have 0.75 moles of HCL and we will get 70 77.3 middle leaders And now we have to find the moles of HCL that we have consumed. So we're going to take five milliliters of H. C. O multiplied by everyone. Leader is 1000 with the leaders Times 0.7 rules for every one liter of solution and then again multiplied by our 1 to 1 ratio of our reaction. And now we need to find the sodium hydroxide that is remaining. So we take our initial molds that we found subtract the molds that we just found that is consumed in our reaction. And we will get our remaining bulls 0.542 molds. And now we have to find the polarity of sodium hydroxide. In this first part, she will take the molds that we just calculated, divided by our total volume 55 milliliters. Multiply it by 1000 for everyone. Leader to get this into leaders, and we will get our concentration or polarity 0.985 Moeller And now we're one step closer to finding our pH. Sir pH in our first part is 14 plus the log of our hydroxide concentration. And that's simple. Now we have all the information we need will take 14 plus the log of our calculation from before, and you will be able to get her pH, which is 12.993 now on department. You're gonna follow some of the same steps so we will find the moles of sodium hydroxide that's consumed. Who will take our 50 milliliters of HCL solution. Multiply it by everyone. Leader is 1000 milliliters times 0.75 Moles of H C E o bided for everyone leader and then again multiplied by our 1 to 1 ratio in our reaction and we will get 0.38 moles of an a O. H that is consumed. And now again, we have to find our house or sodium hydroxide that is remaining so you would take our initial moles. 0.580 Divide it. I'm sorry. Subtract it by the most we just found that were consumed and we will get 0.2 moles that is remaining in this part b of the reaction. But then again, we'll find the mole Larry of sodium hydroxide. We will take the moles we just previously found 0.2 Divide that for divide that for every 100 milliliters of our solution. And we have to multiply that by 1000 with the leaders for everyone leader, because remember Mole Aridjis and leaders and then we will get our mole Arat AEA's 0.2 to find the pH. We will take 14 plus the log of 0.2 and we will find that are ph for part B is 2.3 now on to part C. Again, we're gonna follow very, very similar procedure. We first need the moles of each seal that we add. So we will take 0.1 leader of HCL and multiply it by 0.1 mole of eight seal for everyone. Leader of solution we have. So you will get the molds that we add or 0.1 Then we find the moles of HCL that are consumed. And again we've already calculated this. This is 0.58 Now again, same step is the other parts we take the moles of HCL that is remaining. And how do we find that we find the initial 0.1 Subtract the most consumed and we will get the most remaining 0.4 And then we can calculate the molar ity of H CEO With this specific edition a volume we will take the mold. So we just found 0.4 Divide that by the volume of our solution you will get the polarity as 0.4 Now we can find the pH. All we have to do now is take the negative lock of our hard Droney. Um, concentrations will take the negative log of 0.4 And we seem to get 1.4. And this is how we find ph in three different scenarios volume.


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