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A solution is made by mixing $5.00 \times 10^{2} \mathrm{mL}$ of $0.167 M \mathrm{NaOH}$ with $5.00 \times 10^{2} \mathrm{mL}$ of $0.100 \mathrm{M}$$\mathrm{CH}_{3} \mathrm{COOH} .$ Calculate the equilibrium concentrations of $\mathrm{H}^{+}, \mathrm{CH}_{3} \mathrm{COOH}, \mathrm{CH}_{3} \mathrm{COO}^{-}, \mathrm{OH}^{-},$ and $\overline{\mathrm{Na}}^{+}$

$$\begin{array}{l}{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=0.05 \mathrm{M}} \\{\left[\mathrm{OH}^{-}\right]=0.0335 \mathrm{M}} \\{\left[\mathrm{Na}^{+}\right]=0.0835 \mathrm{M}} \\{\left[\mathrm{H}^{+}\right]=2.99 \cdot 10^{-13} \mathrm{M}} \\{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]=8.30 \cdot 10^{-10} \mathrm{M}}\end{array}$$

Chemistry 102

Chapter 16

Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria

Aqueous Equilibria

University of Central Florida

Rice University

University of Maryland - University College

University of Kentucky

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.

09:55

A solution is made by mixi…

06:25

04:12

01:47

Consider a solution formed…

Okay, here we're told we mix 500 millimeters of 0.167, oh with 501 molar, acetic acid or ch 3 ch and right away. We have equal volumes and the n o h is a higher concentration than the acetic acid. So we're going to have an axis in a left over by the end of this, and then we're asked to find the concentration of every species in the solution, and i went ahead- include the peak of acetic as 4.756, because we'll need that by the end of It now i'm going to start with the easiest concentration, that's gonna, be in a plus and the reason why is gabe asiest, because it's not involved in any of the acid base reactions it simply dissolves and acts as a spectator ion. So all we have to do is consider the dilution here so using the tuition formula. 1 v, 1 equals c 2 v 2 point: we have 1 concentration and we're going to a more duliconcentration. That'S what we're trying to find. So we remove the v 2 over t 1 v 1 divide by v, 2 to get c 2 point all right, so we're gonna take or 0.167 molariwe have 500 millimeters initially, and then we added to another 500 miller for a total of 1000 miller in the Total volume or the total volume between the 2 editions- and we end up with a concentration of 0.0835 molar in a o, h, okay, so this straight ahead is the concentration for nate plus, because when we stick an h in solution, it splits apart and a plus N o h, minus and the n a plus, doesn't perform any further chemistry, so we already have the concentration of n a plus right here, just 0.835 check. We can grasp that 1 off now everything else is gonna be or complicated. Is the roll acid bases they're? Don'T all react with each other, and so we have to consider some more chemistry, but before we do that, let's go ahead and dilute or acetic acid, because we'll need that diluted to again we take our concentration adjusted by 500 over 1000 milliliters ammung up 0.500 molar. His heat of acid, i'm going to call it h, a that's what i didn't not this at first h, a is the weak acid form and a minus is the conjugate base weak acid conjugate base, just because i don't want to write that orse acid over and Over again, so the h here represents the label proton and the minus represents the charge on the acetate ion after we lose the proton. So now, let's look at the chemistry that's happening. We have acetic acid, acting h minus to make a minus plus h 2 o. Were i not steel because it's the solvent it's pure liquid, it's not going to play effect in the afterall chemistry. All we really care about is the loss. The neutralization, this, oh with this acidacitic acid, so this is 0.05 and this is 0.0835 and you can do this in moles. I just think it's easier and molarity at this stage. So this is our lamenting reactant welse. We have less of it than we do, or ammonia or hydroxide, rather so that's gonna, 000.0335 and 0.0500 right. So now we have the concentrations of 2 more of these species. We have the concentration of you'll, do the whole thing and concentration of a minus which must right. That means we just have to find the concentration of the acids and i'm gonna prefaces. These are prudent if you wanted to be a hinderer. Accurate you'd have to do a little bit more work, but whatever adjustments you'd make, these are just rounding, theirs and we'll kind see with how small, the hay and h, 0 plus or that they're not going to play a huge impact. Making adjustments for that. So i'm gonna go ahead and call this their prefatial concentrations right. So now, let's find it to the real possibility. That'S going to be the next easiest 1. So we know k w. A concentration is equal to the concentration of h, 3, o plus in the concentration of o, h, minus and k. W is always 1 times 10 to the negative 14 point. It'S a constant, and it's always going to be like that. So if you know the conservation 1, we can find the concentration together. So let's go ahead and do that divide this over and that would give us 2.99 times 10 to the negative 13 point and that's going to be a molar and that's going to be equal to r h, 30 plus ions. When we found the h, there was 1 that leaves just h a left over to find, and so there, the chemistry that's happening to a minus reaches water in a reversible reaction to make h a plus, h, minus, and this reaction is described by k b, and This is why i said this: if you want to be super, accurate you'd have to make an adjustment here bes some of his being made in this fraction, but you will see that this is going to be very small compared to what this value is. So we can mostly just ignore it. So when we have this reaction, we're gonna figure out how much of this we're making value of that is going to be just like any other equilibrium expression. We have where it's h, a which is the products over the reactant, which is the case, a minus where noun water here, just because it's upper liquid and doesn't factor in our equilibrium expression. We know we can find this value from the peeka included. And then we re have this value. We'Ve already found this value for a minus. We have a value for this, but that's a different. This is h, we're making from this expression where this is th, that's already writ in solution, so these are not the same. Despite looking the same, so we have to find this value here, an we have this expression i had to think. Usually we don't have the value, what i'm doing this talk, but we have this value here right. This will unpack this equilibrium right, as this goes up. It'S going to shift to the left and vice versa, so we have 314 values here. We can just go ahead and plug in what we have wait. First, we have to find so 4.756 is our equals 14 minus 4.756 or popes 14 minus b. This case 4.76. So peals 9.244, then we take 10 to the minus 9.244 and i'm getting that from p k b, equal to negative log k, and so i move the negative over and i take to the power of and that's how i get. This expression equals 5.70 times 10 to the negative 10 point all right. So now we have every value except a here: let's go ahead and plug stuff in we have 5.7 times 10 to the negative tenatowell, i'm moving the minus over times minus concentration 0.05 and then i'm divided by this. Oh concentration, that 0.0335 and all told we'll get 8.51 times 10 to the negative 10 molars is h, the plus that's what we're solving for there. You go right. So a is 8.51 times the negative 10 point. So very little h a been less h, 3, o plus and then h, minus minus and plenty of al so constitution in a plus concentration of a which minus concentration of a minus concentration of h, 3 plus concentration of h. A.

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