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Determine the $\mathrm{pH}$ of $(\mathrm{a})$ a $0.40 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}$ solution, (b) a solution that is $0.40 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}$ and $0.20 M \mathrm{CH}_{3} \mathrm{COONa}$

$$\mathbf{p} \mathbf{H} \approx \mathbf{4} . \mathbf{4 4}$$

Chemistry 102

Chapter 16

Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria

Aqueous Equilibria

University of Central Florida

Drexel University

University of Maryland - University College

Lectures

00:41

In chemistry, an ion is an atom or molecule that has a non-zero net electric charge. The name was coined by John Dalton for ions in 1808, and later expanded to include molecules in 1834.

24:14

In chemistry, a buffer is a solution that resists changes in pH. Buffers are used to maintain a stable pH in a solution. Buffers are solutions of a weak acid and its conjugate base or a weak base and its conjugate acid, usually in the form of a salt of the conjugate base or acid. Buffers have the property that a small change in the amount of strong acid or strong base added to them results in a much larger change in pH. The resistance of a buffer solution to pH change is due to the fact that the process of adding acid or base to the solution is slow compared to the rate at which the pH changes. In addition to this buffering action, the inclusion of the conjugate base or acid also slows the process of pH change by the mechanism of the Henderson–Hasselbalch equation. Buffers are most commonly found in aqueous solutions.

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your two parts to this question will first start with the first part which is asking you for the pH of the initial solution of straight up acetic acid in solution. Here you're given the reaction ch three. You each just acetic acid, A quiz in equilibrium with it's contradict base and protein. You're given an initial concentration of sterile 0.40 Moller. Um and we're going to set this up like a nice table. So you have your initial concentrations. Um, in this case, we're going to label these 20 Whenever this reaches of state of equilibrium, you add something is added to this side of the reaction, which is the product. And then something is subtracted from the left side of the reaction on your final point of equilibrium. Concentrations are as follows Now in the text book, you have a table of equilibrium constants. In the case of acetic acid, R K, A value is equal to 1.8 times 10 to the negative fifth, and this is equal to our concentrations of products over the concentration, no reaction, so we can plug these values in and sort of simplifying. We get X squared over 0.40 minus X. Now you can solve this. You think quadratic formula quadratic equation, and you can still get your X value. Um, we know that we can use the small ex approximation whenever RK values are either less than around 10 to the fourth or greater than, um, for sorry, less than 10 to the negative fourth or greater than 10 to the fourth. In this case, RK value is less than 10 to the negative 4th 10 negative fifth So we can use our small x approximation. And we can kind of simplify this equation too. X squared over 0.4. Uh, whenever you saw for this whether you use the quadratic equation or you use the small ex approximation, you should get a value around X is equal to 2.7 times 10 to third, which is equal to your concentration of proton. Authenticate pH. You take negative log of this value, and when you plug that in, you get around 2.57 as your pH. So this is party. Now we're going to look at part B. So be, um, is an example of the common ion effect So you're adding to the same reaction that we initially had acetic acid use dissociating. And you these two ions, it's always got to rewrite your equations. Um, you have the same set up where you have your ice table, your initial concentrations, ours your point for Mueller and 0.2 Moeller because you're adding, um, sodium acid Tate to the system, which gives you more of the acetate ions dissolving into solution. And according to look at lease principle, this equation will move to the left as a result, and you'll have a change in ph so we can observe that phenomena using this equation. So I'm going to start out with zero protons. For the sake of simplification. We won this acid associates. We get some of the is moving right. You have, in addition to the product side of the reaction. And so you get new values of is your 0.4 minus x Moeller. Then on this side, you gets your 0.2 plus X. You just get acts. So again we're going to solve for or concentration of x our case people to 1.8 times send the negative fifth. That doesn't change and now our new equilibrium equation. It's going to be this. You deserve products over our reaction reactive. And then we can use the small ex approximation again. The situation to get your point to you X over is your appoint floor. Now we can sell for X. Our X is equal to 3.6 times 10 to the negative fifth. You know that our X is the concentration. Oh, our protons. And then whenever you saw for ph. Negative log of the concentration of protons, which when you saw this, you get 4.44 This is how you saw for the pH of an initial acidic solution and then the pH of a common ion effect.

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