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Question

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Answers

Consider a solution prepared by mixing the following:
$$
\begin{array}{l}{50.0 \mathrm{mL} \text { of } 0.100 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}} \\ {100.0 \mathrm{mL} \text { of } 0.0500 \mathrm{M} \mathrm{KOH}} \\ {200.0 \mathrm{mL} \text { of } 0.0750 \mathrm{M} \mathrm{HCl}} \\ {50.0 \mathrm{mL} \text { of } 0.150 \mathrm{M} \mathrm{NaCN}}\end{array}
$$
Determine the volume of 0.100$M$ HNO $_{3}$ that must be added to this mixture to achieve a final pH value of $7.21 .$

Question number 57 as a week base. Strong acid penetration problem where you are given eight different volumes during the Thai Trish in process and asked to calculate the pH at each of those eight volumes. To do this, we need to recognize that when we start with a weak base in solution, pH is determined using a K B week based calculation. Then, once the Thai tray shin begins and strong acid has been added, some of the strong acid will convert the week base into a weak acid, and we will have a buffer solution. So pre equivalents all pH calculations will be buffer solution calculations using the Henderson Hassle Baulch equation At the equivalence point, all of the week base will have become weak acid, and we will perform a K A or weak acid pH calculation. Then post equivalents. We will have excess strong acid, and we will determine the pH simply based upon the excess strong acid concentration or the excess hydro knee. Um, concentration. So let's get started. The first volume of strong acid added is zero mill leaders. If we have added 0 mL of strong acid, then all we have in solution is the weak base to carry out the pH of a weak base solution. We simply need to first solve for the hydroxide concentration the hydroxide concentration, as you recall for a week base is simply equal to the square root of K B 5.2 times 10 to the negative four multiplied by the concentration of the week base 40.1 Moeller, this is our hydroxide concentration. If we divide this into K W, we will get the hydro knee. Um, concentration. If we take the negative log of the hydro knee, um, concentration, we will get our ph of 11.86 Next, we have added 10 million liters of the strong acid to our weak base because we started with 20 mL at 200.1 Moeller of the week base and we're adding 0.1 molar hcl than we have a 20 million leader equivalents point. So if we are at 10 million, leaders were halfway to the equivalents point. And as you recall the pH at the equivalence point equals p. K. Pekka is simply k be divided into K w. This will give us our K a value. We take the negative log of that, we'll get our PK and will recognize that at half equivalents pH equals P K. If you didn't recognize this, you could still use the Henderson Hasselbach equation and you will get the same answer. PH will be equal to P. K. Plus the log of the moles of base. The moles of base will be equal to the moles of base we start with which is 20 mL or two leaders at 20.1 Mueller minus the moles of strong acid added which will be the 10 mill leaders, or 100.1 leaders multiplied by its morality of 0.1. This will be the moles of strong base left in solution after 10 mL. I'm sorry. Weak base left in solution after 10 mL of strong acid has been added. We divide that by the moles of weak acid formed. The moles of weak acid formed will be equal to the moles of a strong acid added which is our 10 mL one leaders at 10.1 Moeller. This also gives us our pH equal to Pekka at 10.72 The next calculation, see, is similar to the previous. We still have a buffer solution. We're still pre equivalents at 15 mL, where equivalence is 20 million liters. So we still used to Henderson Hassle Baulch equation P H is equal to P K, plus the log of moles of base, which will be the moles of weak base. We start with again 20 mL at 200.1 Moeller, minus the moles of strong acid added, which could be calculated by taking the volume of strong acid added 15 mL, or 150.15 leaders multiplied by the molar ity. We then divide that by the moles of weak acid formed, which will be equal to the moles of strong acid, added the same calculation as we did appear. This then gives us a pH of 10.24 Because we are adding strong acid, we should see an increase. I'm sorry, a decrease in Ph throughout these titrate Asian calculations, which we're currently seeing next. For Part D, we've added 19 million liters. We are still pre equivalents, so we are going to still use the Henderson Hassle Baulch equation because we still have a buffer solution. But this time, instead of the 15 mL added, we've added 19 million liters or 190.19 leaders at 0.1 Mueller. So we'll also change that down here in the denominator when we calculate the moles of weak acid formed. Other than that, the calculation is the same, and we get 9.44 as our pH still decreasing as we are expecting it to. For Part E, we are at 19.95 mL, so still pre equivalents still a buffer solution still using the Henderson Hasselbach equation. This time, however, we're using our 19.95 mL instead of the 19, or the 15 19.95 mL is 0.1995 leaders. This will be the decrease in the moles of weak base and the increase in the moles of weak acid, and we get a pH of 8.12 Now we're a 20 mL at 20 millimeters. All of the week base has been converted into weak acid. If all of the moles of weak base have now become molds weak acid, then we have a weak acid, or K. A calculation. To calculate pH. We need to know the hydro knee um concentration that is being produced from the weak acid. Ah, hydro knee. Um, concentration will be equal to the square root of the K a value which we don't have. But we do have the K B value. We can calculate the K a value by dividing the KB value into K W. So k b divided indicate W gives us k A. So the hydro knee, um, concentration is the square root of K A multiplied by the concentration of our weak acid. We started with 0.1 Moeller weak base. All the weak base has become weak acid, but we doubled the volume. So the concentration of the weak acid is half the concentration of the week base we started with because we doubled the volume. This will give us our hydro knee, um, concentration once we take the square root. Then if we take the negative log of this hydro knee, um, concentration, we will get our pH at 6.1, still decreasing, then going to part G. We are post equivalents. If we are post equivalents, we have excess strong acid because there's the strong acid. Is our tie trahant To calculate the pH, we simply need to figure out the excess strong acid concentration, which we can determine by first calculating the excess strong acid moles. These air the total moles of strong acid we have added at 25 mL or 2005 leaders 0.1 Moeller. We'll subtract off the moles of the strong acid that reacted, which will be equal to the moles of weak base. We started with 20 million liters 200.1 Moeller and then we divide by our new volume 20 million liters. Plus the 25 million liters is 45 million liters, or 450.4005 leaders. This then gives us our excess hydro knee, um, concentration. We take the negative log that to get our Ph 3.90 for the last one. It's the same calculation as we just did. But now we will be using 25 mL 0.25 leaders instead of 20.5 mL, or 0.2005 leaders and it and the new volume, of course, is going to be the 20 plus the 25 or the 250.5 total, and we get a pH of 1.95

My name is now. We will answer the question, calculated the H off solutions that result of flown each mix here. He didn't it to mind for this. Make sure using Henderson or I see so we will use the easier missile, which is hundreds soul hassle. The first, the mixture in the question. Hey, said 50 million off open toe 15 polarity formic acid with 75 milliliters off point. Once real clarity sodium for mate. So to answer these, make sure we ride. The Henderson has said Bella question, which is B H equals B ke a off the formic as it belongs. Log the concentration off solely in for me, divided by the concentration for me. Cast here is a concentration most TV's. A concentration off solution calculates a wiki A, which is negative logs a k a. Off the formic as it which he is 1.77 times. By then, they get to four. Here the concentration off sodium for me in a mixer. Initially, we use 75 many off concentration on 750.13 and the mixing was forming, which she essentially, it's a concentration is 0.1 fine times going toe 5 50 Millie. So most of them there it's a concentration in solution. His initial concentration times about initial volume divided by the total volume off Make sure which is 75 lost 50. Also it every consideration of its chief for me formic acid in the picture is its initial volume times one initial concentration dividing boys toe Tell Vanni so buy some statement. By these value, we can calculate the B H A directly, which is three point a six for the second. The mixture, which is a moaning and the union could dry. Here we use the B o edge. It's a beauty is equal B k B last low concentration off ammonia NH Queer eye in a make sure divided by the concentration off money and make he took a queen bee. Could be it will be negative law. The baby off ammonia, which is 1.8 times by 10 negative five And the concentration off ammonium chloride in solution is its initial concentration, which is oh goingto juan times by it's the initial phone you divided by the total volume in the total volume. Here is a volume off ammonium chloride last ammonium off money, which is 250 plus 100 doing t five. Also the concentration off ammonia. He's all goingto one. It's a concentration 125 and divided by the total volume 251 to find. Now we can calculate the B 00 H, which we found its equal. 5.6. Then we can calculate the be edge, which is 14 because we work on room temperature. Negative. The B 00 H was eats 5.86 When it's equal e point mine full. Thank you.

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.

Question 56 is a weak acid week, a strong base titrate Asian for weak acid, strong based, high trey shins. We need to carry out a weak acid calculation at the equivalence point. I'm sorry at the initial point Buffer solution calculations, using the Henderson hassle Baulch equation pre equivalents a week base calculation at the equivalence point and then an excess strong base calculation post equivalents. So let's get started for the first one. We have zero mill leaders of strong base that has been added. If we have 0 mL of strong base that has been added, then all we have is the weak acid solution that is present. If all we have is the weak acid solution will carry out a weak acid pH calculation where the pH could be determined by taking the negative log of the hydro knee, um, concentration and the hydro knee. Um, concentration is simply equal to the square root of the K, a value which was given to us at 1.54 times 10 to the negative five multiplied by the concentration of the weak acid, which was given to us at 0.1 Moeller. So this right here gives us our hydroxide concentration are hydro knee, um, concentration whips. This right here gives us our hydro knee. Um, concentration. If we take the negative log of that, we'll get our P H and R. P. H is 2.91 for all the other calculations. We need to know where the equivalence point is. If we have 20 mL, appoint one Moeller butin OIC acid and we're titrate ing it with 0.1 Moeller sodium hydroxide because the concentrations air the same and the stock geometry is one toe one than the volumes will be the same. So 20 mL of sodium hydroxide is what is required to neutralize 20 mL of 200.1. Moeller butin OIC acid. So 10 million leaders is halfway to the equivalence point. Many of you might recognize that the pH for a weak acid, strong based hydration at the half equivalents point is equal to P. K. If you didn't recognize that, you would still know that free equivalence is a buffer solution and requires the Henderson hassle Baulch equation. So that's what I've done here is pH will be equal to P. K A. The negative log of the k A value plus the log of the moles. A base over the moles of acid. This is the Henderson Hassle Baulch equation. How will we calculate the moles of base? The moles of weak base formed will be equal to the moles of strong base added. The most of strong base added is simply equal to the volume of strong base 10 million leaders, or 100.1 leaders multiplied by the molar ity at 0.1 Moeller. We then divide this by the moles of weak acid that's still left in solution that will be equal to the volume of weak acid we started with, which is 20 mL or two leaders multiplied by the molar ity at 20.1 Moeller. Those moles will decrease by the moles of strong base, added the moles of strong base, added again will be equal to what we had up here in the numerator. We then carry out this calculation to get a ph of 4.81 which is because this ends up going to zero simply equal to P. K. Now for the next 1. 15 million leaders were still pre equivalents, so we still have a buffer solution, and we will still use the Henderson Hassle Baulch equation. It'll take on a format very similar to what we did in Part B, where pH is going to be equal to P. K, plus the log of the moles of base over the moles of acid. The moles of acid that we will have left will be equal to the moles of acid we start with, which is the full 20 mil leaders, or 200.2 leaders multiplied by the concentration of 0.1 Moeller minus the molds that reacted, which will correspond to the molds of strong base, added 15 mL, or 150.15 leaders multiplied by its concentration of 0.1 Moeller that will also be equal to the moles of weak base formed. Essentially, the moles of strong base added, is equal to the moles of week based formed that will give us a pH of 5.29 And as we expect as we are adding the strong base, we should see with each subsequent calculation an increase in pH, which we are seeing. This is one way to verify your answers. Now we go to Part D. Part D. Is 19 mL again, still pre equivalents. So it'll be a calculation very similar to what we had before. The only difference is now the volume of strong base added, which in this case is going to be equal to the 0.19 leaders or the 19 million liters and then up here. We will also have the volume of strong base added, which will correspond to when multiplied by the molar ity. The moles of week based formed, and we get a pH of 6.9 for the last one. We are just before the equivalents point at 19.95 mL, so we still have a buffer solution. We will still use the Henderson Hassle Baulch equation. So again, P h equals P K plus the log of in our denominator, it'll be the moles of weak acid that has not yet reacted, which is the molds of weak acid. We start with minus the moles of weak acid that reacted, which is equal to the moles of strong base added, namely 19.95 mL, or 0.1995 leaders. This, then, is also equal to the molds of week based formed, namely the 19.95 mL, or 0.1995 leaders multiplied by the molar ity. This then gives us a pH of 7.41 and we're still seeing an increase in Ph as we should now, once we reach the equivalence 0.20 mL, What we have done is converted all of the weak acid into weak base. So all we have in solution is weak base we can solve for the hydroxide concentration by doing a week based calculation. A week based calculation is the hydroxide concentration will be equal to the square root of K B. We don't have K B, but we do have a A. If we divide k into kw, this is our cabe. So hydroxide is equal to the square root of K B multiplied by the concentration of the week base will. How will we get the concentration of the week base? Every mole of weak acid has now become weak base. So we'll take the volume times the molar ity of the weak acid we started with to get the moles of weak base we now have and then divide it by the new volume. The new volume is 40 mL, or four leaders so hydroxide will be equal to the square root of K B multiplied by the concentration of the week base, and we get 5.70 times 10 to the negative six. The pH will then be equal to the negative log of the hydro knee, um, concentration and the hydro knee. Um, concentration will be equal to the hydroxide concentration divided into K. W. And we get a pH of 8.76 Now for Part G. We're just past the equivalence point a 20.5 million liters. So we have excess strong base in solution. The excess strong basin solution could be used to calculate the excess hydroxide concentration so pH will be equal to the negative log of the hydro knee. Um, concentration. Where the hydro knee, um concentration is the hydroxide concentration divided into K W. So pH equals negative log of the hydro knee. Um, concentration, which is hydroxide divided, indicate W. How do we get the hydroxide concentration? The hydroxide concentration will be equal to the total amount of strong base sodium hydroxide we've added, minus the moles of strong base that reacted which will be equal to the moles of weak acid. We started with divided by the new volume. So all of this down here is concentration of hydroxide. Divide that into kw to get hydro knee. Um, take the negative log to get our ph and we get 10.10 for the last one will do a calculation very similar to what we just did. But in this case, we are a 25 million liters. So we will have the total volume multiplied by the concentration of the strong base added, minus the moles that reacted, which will be the moles of weak acid. We started with multiplied by its concentration, divided by the new volume which is now 45 mL or 45 leaders and we get a ph of 12.5.


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