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The pH of a solution containing 5 x 10 6 M selenous acid (HzSeO3") is adjusted to pH 7.8....

Question

The pH of a solution containing 5 x 10 6 M selenous acid (HzSeO3") is adjusted to pH 7.8.

The pH of a solution containing 5 x 10 6 M selenous acid (HzSeO3") is adjusted to pH 7.8.



Answers

Determine the $\mathrm{pH}$ of a solution with the given hydronium concentration. Classify the solution as an acid base or almost neutral. (a) $6.8 \times 10^{-8}$ (b) $1.2 \times 10^{-7}$

Here we have a true or false question, and we're asking, is it true or false that at 10 to be a minus eight mm solution for official as Ph. Eight. So your first response might be to go straight to the pH formula, and you put in your values and you find great ph of eight. But we have a problem here. This is incredibly dilute, and it is an asset. And no matter how much you dilute our acid, we're not going to get a basic solution. False and dilute acid cannot give a solution. So what? What's happening? Why is the equation not working? It's because it is diluted in water. So the mhm although we're working with acid and it's extremely dilute, um, it's being outnumbered by the effects of water. So water is associating into H plus and and two minus. And these these hydrocarbons here massively outnumber, massively outnumber the hydrogen ions from the acid, so the the acid might as well it does not exist here. Basically, we're just looking at Ph. Uh, seven because we're looking at the page of water at this point. Um, but if you really want to factor in the impact of water, you will end up getting a value of 6.986 point nine. Oops, 98 Ferrar ph bits effectively neutral at this point. And this value here is if you take into account the solution of hydrochloric acid and also take into a Campbell water and comparative.

Each yours classified as a strong acid, which undergoes 100% ionization front, said Lowry. Equation here each pr h 20 30 plus B R minus. So if the HBR 6.9 times 10 to the negative eight smaller 1 to 1 more racial 6.9 times 10 to the negative eight Moeller ph is defined its negative log. Major Israel, plus actually going negative log 6.9 times 10 to the negative eights, but should be equal suit 7.16

So the first thing we'll be doing here is calculating the ratio of conjugal based acid in the starting solution. Using the Henderson Hasselbach equation. So we have. PH is equal to p k a. Add log a minus over h pain. We have no point. We have eight point, not 8.9 is equal to 7.4 ad love a minus over H Bay. So we have no K minus over H A equal to 9.6. So then a minus over h A is equal to the anti log minus 9.6. We got a value of four, so just run to the next page. So the solution has 100 million equivalents of the compound, so 18 million equivalents are in the form of the contract base. 20 million equivalents are in the form of the acid, so the ratio is four. So HCL is a strong acid and associates completely so if we have third email if not point of one point, not more. That HCL to study many equivalents of H plus is added to the solution. So 30 Miller equivalent titrate starting the equivalent of the contra bass. So the ratio becomes one. Solving the half of that equation for Ph. Again, What we get is pH. It's equal to P K a vlog gay minus over h A where we take pH was equal to 7.4 add log. So Ben one has been determined for a ratio the value of 7.4 ph.

Everyone we're going to Dio has a multi step problem here and we will be given a k A for Ben's OIC acid. We will be given the concentration of Ben's OIC acid it 0.10 Let me verify that 0.10 Moeller and we will be given the k A for Ben's OIC acid, which is six 6.5 times 10 to the minus fifth. So this is the information that we know and the Benz OIC acid. I'm gonna cheat a little bit and I'm just going to refer to it as hh A because it's just too long for me to write down over and over again. So h a it could be considered ah mahn a product weak acid so it will donate. Ah, hydrogen ended an eye on. Okay, so for this problem, we're going to work with the following information Let me switch colors here. We know that k A for acid, the equilibrium constant is going to be equal to and let me write down in a h a. And that's going to be an a quiz solution and we can consider this plus water if we want to, But I'm not going to worry about it right now because water is a liquid equals h plus or yields H plus plus our an ion. And both of these are also a quis. So our equilibrium expression will be reactive or products h plus and these represent concentrations over h A. Okay, So these three values and I wonder if I can highlight here these three values are the values that I need to get to put into our leak. Will Librium? Um, expression. Okay, so we're going to use an ice. I see a. And remember that ice gonna move this down a little bit? The eye is for our initial concentration, and our initial concentration of H A is 0.10 Moeller. And our initial concentrations of the hydrogen ion and the an iron are both zero make lines here. Try to make lions. We can see that. Wow, Look how straight those are. Okay, now, my change here, I'm gonna have a quantity x and X because I'm going to have the same number or the same quantity of hydrogen ions and an ions. So those b plus X and this is going to diminish by some quantity X some of my final concentration My equilibrium concentration will be x for the hydrogen ion X for the IA an eye on and 0.10 minus X for acid. I'm gonna write this down on our next page Just the equilibrium concentration. And let me move this down a bit. H A. The concentration was 0.10 minus X, and both my H plus and my anons were both X. Okay, Now we could make. Since our K A is so low 6.5 times 10 to the minus fifth, that means my ex is going to be insignificant. And if we want to do this and figure out a big quadratic equation, we could probably prove this. But rarely will it even make a rounding difference. So I'm going to be able to make the assumption that 0.10 minus X is pretty much the same thing as 0.10 So let me make that a little bit more legible. 0.10 Now we're going to put that into our K expression, which I will write in purple here again, Que a equals my age plus my a minus and by ups H A. So let's substitute our values in here. Que is 6.5 times 10 to the minus fifth equals And remember each of these air X. So I get X squared over the concentration of H A lips, which is where Sumi is 0.10 Do the math here. Don't forget you're gonna multiply each side by this quantity and then take the square root of that when we solve for X. If we do this correctly, I got 0.0 zero 255 And that's my concentration. Remember to find the pH, which is what we're ultimately looking for. We take the negative log of our concentration, which is the negative log of 0.255 And that should actually have to Well, I won't run yet. 255 And my Ph will be 2.59 Now if I look, I had two significant figures present in the, um, mathematical operation I did in this step. So I can have since I had two significant figures, I can have two decimals. That's the problem.


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