Calculate the pHpH of a solution formed by mixing65 mLmL of a solution that is0.35 MM in NaHCO3NaHCO3 with75 mLmL of a solution that is0.11 MM in Na2CO3Na2CO3....

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

So now we'LL work on Problem forty from Chapter seventeen. So this had this question asks us to calculate the pH of the solution that river result from a mixture of solutions. So let's go and work on part A here. So in this problem, they're giving us a mixture of solutions between H F and sodium. So hydrofluoric acid and sodium chloride. Now, because we're dealing with two different solutions of two different concentrations and two different volumes, we need to be very careful to calculate the moles of each compound that's being added. So Force a Jeff for moles of H F. We can calculate that by multiplying the concentration. Times Lawyer three point twenty five Moller Times. We need to be careful to use the concentration and leaders, so it tells us one hundred fifty milliliters, which would be is your point one five zero leaders. And if we do this multiplication, we get zero point zero three seven five moles of a Jeff. We can do the same thing for sodium fluoride, so the concentration were given a zero point three zero, and the volume is two hundred twenty five milliliters, which is zero point two two five leaders, and that gives us a zero point zero six seven five moles of sodium fluoride. Now we could calculate the concentrations here, since we know the final volume will be the addition of two hundred fifty and one hundred fifty. But since we're going to use the handles and Henderson House Aback equation, we don't need to. Since we know the moles of each constituent so we can do The pH is equal to P. K, which for h f will be three point four six plus the log of the base over the acid So zero point zero six seven five divided by zero point zero three seven five. And when we do this six there, when we do this calculation here we get, uh, three point four six plus to your point two six, which is equal to three point seven two. So that's the pH for pern, eh? So if we go move onto Part B, we're given a different situation where we have Ethel mean being mixed with Ethel ammonium chloride. So here we can calculate the moles of see to age five base off the concentration on volume, so we have zero point one zero kn Moller times zero point one seven five Because the volume was a hundred and seventy five milliliters. You multiply this and we get zero point zero one seven five moles. We can do the same thing for be confident acid. Zero point two zero Moller Times The volume, which is zero point two seven five leaders, is equal to zero point zero five zero five five zero moles. So since we know the malls, we can go ahead and use B. Anderson also my equation. So we're going to calculate p o O h here because we're dealing with a base. So we get the negative log of the K B, which is five point six times ten to the minus four, plus the log of acid zero point zero five five zero over bass zero point zero one seven five. And so that gives us three point two five wass zero point five zero, which is the word of three point seven five. Now, this is Pete. Oh, age. So we can easily find the pH by subtracting from fourteen. Fourteen minus three point seven five will give us ten point two five as our PH

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.

So in this video, we're going to be answering questions for you, too, from Chapter 17 which is. Calculate the pH of the solution that results from age mixture a 150 milliliters of 1500.25 Moeller hydrofluoric acid with 225 milliliters of 2250.30 lower sodium chloride. Um, so basically, when we mix the sodium employed with hydrochloric acid, we're diluting the concentration of hydrofluoric acid and the same for sodium fluoride. So we wanna use the equation we used for concentrations and dilutions, which is just see one V one equal See to be, too, where see one of you on correspond to the volume and polarity of the of the original mixture that you're diluting and see. Two is the concentration in the final solution and Envy. Shoe is the total volume of the final solution, so we have 20.25 Moeller Hydrochloric acid times 150 milliliters. Mercy when everyone, and not equals X are unknown is the final concentration. Um, times are total volume, which is 375 mil. Leaders and X equals 3750.1 Moller and we do the same thing for sodium fluoride and why is R unknown, which is the final concentration? And we saw for why. And we get 0.18 Moeller. So now we know we have significant concentrations of an acid and it's congregate base. Well, that's just a buffer solution so we can use the hundreds and hostile back equation. So we have P h equals P K A plus log of the concentration of base divided by the concentration of acid. That's 3.14 hydrofluoric acid plus log of 0.18 divided by a 0.1, and our Ph is 3.36 So in part B, we have 175 in the leaders of 0.1 Moeller C two h five image d'oh with 275 mil leaders of 2750.2 molars t to H five in H three steel so we can tell that we have an acid in the conjugal based here. Oh, many gun where you're gonna have to figure out what the concentration is after we've mixed beings. After we've done Ida a dilution. Um, so we you see one be on equal, see to beat you again and for our base. We have 0.1 or Moeller times 175 milliliters equals are unknown. The X The final concentration times the total volume, which is 450 milliliters. We end up with X equals 4500.39 Moeller. We do the same thing for the acid, and we end up with y equals 0.1 to Moller. So again we have significant concentrations of an acid and its content based. Let's use the Henderson Hostile Black equation and we have 10.8 plus log of 0.39 divided by 0.1 to which gives us a pH of 11.3.

Hello. So in this question we have been given we have to find out the ph of the following. So in the first part just give an sc that you two is equal to zero 15. And so the volume of volume of etc for the volume of 80-03 to go to heaven 50 Ml So now and he's here to two is given in the question That is zero 13. Um and the volume of any CHO two is given I think yeah seven 55 point 00 M. L. So now we have to convert the It's your two in the mixtures. So let's go ahead. So for the mixture it has to be converted into Like this 0.05 later into zero point 15 and upon zero point 0.25 peter. So it comes out to be around They were going 06. No no for the any CH. 0. 2 mixture. So here zero point 07 five and two 0.13 upon 0.125 L. So we'll get zero point zero seven eight. Yeah so now the K. Of As to its U. three it CHO two is given is no 1.8 into Tennessee. The power -4. We have all notice. So the PKK PK of it's you too as 3.74 as we have calculated. So the formula is calculating the ph of the first point part is the formula. It is speak plus log based upon. Is it? This is the formula race and you have to learn it. This is it he said so no after substituting the values in this we'll get be it. He goes to 3.86. I'm not showing the full multiplication. You can just have written all the aspects all these data and now you can substitute in this formula and we'll get Phs 3.86. No coming onto the B part again. The and that's three zero point the name, vol of NH three is 1 25.0. Ml So and it for See it has given zero Also it is given 0.10 M. And the volume of it will be 2 50. Amen through. N est rien mixture will Come out to be zero. We can just calculated, I have already told you the formula for calculation that is one actually different calculation. 0.125. L. leaders into little going in 10 her phone 0.375. And we will get mixture equals to 0.33 No, for the same picture off. NH four field it welcome 0.66 M. No, the baby of finished three, his 1.70 16- 10 days is about -5. So PKB he goes to minus log Gaby. So he will get developed for PKB. So it is equals 2 -6 one point 7 16-10 issue. The power minus five Which is equals two 4.75. No PK plus PKB is 14 through. The value for PK is 14 minus PKV, which comes out to be 14 -4.75, Which is 9.25 to now, substituting the data. And the formula art formula for calculating the ph So PH equals two 9.25 plus log 0.033 upon 0.066 comes out to be 8.95 and this is your answer. Thanks.