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The $K_{\mathrm{a}}$ for benzoic acid is $6.5 \times 10^{-5} .$ Calculate the $\mathrm{pH}$ of a $0.10 \mathrm{M}$ benzoic acid solution.

$$p H=-\log \left[H^{+}\right]=2 \cdot 6$$

Chemistry 102

Chapter 15

Acids and Bases

Liquids

University of Central Florida

Rice University

University of Toronto

Lectures

03:07

A liquid is a nearly incompressible fluid that conforms to the shape of its container but retains a (nearly) constant volume independent of pressure. As such, a liquid is one of the four fundamental states of matter (the others being solid, gas and plasma). A liquid is made up of tiny vibrating particles of matter, such as atoms, held together by intermolecular bonds. Water is, by far, the most common liquid on Earth. Like a gas, a liquid is able to flow and take the shape of a container. Most liquids resist compression, although others can be compressed. Unlike a gas, a liquid does not disperse to fill every space of a container, and maintains a fairly constant density. A distinctive property of the liquid state is surface tension, leading to wetting phenomena.

04:38

A liquid is a state of matter in which a substance changes its shape easily and takes the form of its container, and in which the substance retains a constant volume independent of pressure. As a result of this, a liquid does not maintain a definite shape, and its volume is variable. The characteristic properties of a liquid are surface tension, viscosity, and capillarity. The liquid state has a definite volume, but it also has a definite surface. The volume is uniform throughout the whole of the liquid. Solids have a fixed shape and a definite volume, but they do not have a definite surface. The volume of a solid does not vary, but the volume of a liquid may vary.

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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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