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[Rctkr Ieckadl Kelere IcorrdeuWhen 22.0 mL of & 6.50*10 ' magnesium acetate solution ! combined with 22.0 mLof & [.77*10 ' M sodium hydroxide solu...

Question

[Rctkr Ieckadl Kelere IcorrdeuWhen 22.0 mL of & 6.50*10 ' magnesium acetate solution ! combined with 22.0 mLof & [.77*10 ' M sodium hydroxide solution does & prec pllaic furmn? (yes (r n0)For lhete conditions Ihe Reaction Quolieni Q,is equalSubmit AnowerRotry Entiro Oroup0 more Groun ntrampta femalningApru=

[Rctkr Ieckadl Kelere Ico rrdeu When 22.0 mL of & 6.50*10 ' magnesium acetate solution ! combined with 22.0 mLof & [.77*10 ' M sodium hydroxide solution does & prec pllaic furmn? (yes (r n0) For lhete conditions Ihe Reaction Quolieni Q,is equal Submit Anower Rotry Entiro Oroup 0 more Groun ntrampta femalning Apru=



Answers

Two solutions of sodium acetate are prepared, one with a concentration of 0.1 M and the other with a concentration of 0.01 M. Calculate the pH values when the following concentrations of HCl have been added to each of these solutions: $0.0025 \mathrm{M}, 0.005 \mathrm{M}, 0.01 \mathrm{M}$ and $0.05 \mathrm{M}$

So here we are, given an example uh information about specific reactions. So we're given information that we have a concentration of silver of 0.200 Mueller's. And were given information that we have a concentration of nitric acid. Only the H-plus matters of 0.10 Mourners. And were asked how much of our contract basis we have to add in order for this reaction to occur. So in order for uh in order for precipitation just to occur, we have to have our Q sp slightly greater than R. K. Sp or technical technicality. We need our Q. S. P. Uh just for equilibrium, it would be equivalent to R K S. P. So the Q S P is also similar to the KsB and that would be the concentration of our silver kaine and our A city and nine. So this is equivalent to 2.3 times 10 to the -3. And since we already have this concentration, okay, we can find that the concentration of acetate Would have to be equal to 0.0 115 Molars. However, with the presence of nitric acid, it's not initially going to be easy to get to that concentration. Since Almost all of this really all of the acetate and nine he's going to react to form the weak acid form. This is a nearly complete reaction and the K. Is very very large. So this would be 10 to the 5th. The reverse. Uh this would be one over the K. A. So this has a very very large equilibrium constant. So basically, until all of your age pluses consumes, it's really not possible for The reaction to occur. So you have to first add 0.10 molars of your H Plus and once you're H plus is consumed, we're going to assume temporarily that the reverse reaction isn't going to lead to that much of a significant difference in the amount of acid. And then we can essentially have to add on top of the you need the required concentration Of 0.0115 molars. So this in total requires 0.11 two molars. And this is moles per liter. And for sodium acetate, we have sodium and to uh two carbons, three hydrogen is and to oxygen's the molar masses 82 g promote. Okay. Mhm. Mhm. And as a result, we find that we need a mass of about 9.2 g in this case, and this is our final answer.

During an acid catalyzed Esther Hydra Laissus. Then the rate constant off the reaction we can calculate given the following formula. What we have is K is equal to 2.303 over the change in temperature multiplied by the log off G infinity, take away Be one B t. Take away. Be one where everyone is the first titra value. The infinity is the final value. Ah, Bt is through the teacher value after tea interval change in temperature is delta T to the consistency of K will prove the reaction is fast. Order on the mean value of K gives the rate constant. Where K is 1.84 times 10 to the minus three seconds to the minus one That is our rate constant.

This is another one of my dreaded voiceovers when I was having problems with my audio so you can see in the bottom of reference. And the problem is as follows. We're asked the calculate the mass of sodium ass state that must be added to use air givens 500.0 mL of 0.200 Mueller acetic acid and I don't know why I'm writing that. Usually I like to write a Che si the ca we look up and it is 1.8 times 10 to the minus five and were asked. So a peek a of that is 4.745 and we are asked to make a 5.0 buffered solution ph of 5.0 buffer. So how many moles of sodium ass state do we have to add to create that buffer system that's buffered at 5.0 Mueller? So we're going to find our morality from Henderson Hasselbach, convert that to moles and then convert that to Mass not too hard. So here's the equation. PH equals the PK, plus the lot of concentration to be, or the log of the concentration of aid. I was wondering, Why did I put log in each of those which wouldn't be incorrect but may look unfamiliar to you. Let's start solving. We have 5.0 given us our Ph. We said our PK was 4.745 That will be attitude. The log of X. That's our concentration of our base. And we were given the concentration of the acid a 0.200 So there we are, adding, Excuse me, excuse me. Subtracting 4.745 from each side gives it the log of X. And then it's the log of 0.200 which equals log of X minus a negative point running out of room to write. You can't write very close to my screen edge on this. There we go. And now we can combine the 0.255 and the negative 0.699 We're gonna take that 10 to the negative 0.444 and the answer is zero point 360 Mueller and a C two h 302 That's my morality. So at malls, we're at morality Now. We're gonna go to moles. Remember, Miller? It is multiple leader. So we're gonna multiply that by how many milliliters we had. We'll have to convert that to leaders. So we have 0.5 00 leaders, and then we're gonna multiply that by the molar mass of a city gas it, which is 82.4 g per mole. And the answer is, I'm crossing off my units. 15 g of the sodium ass tape are required.


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