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Day 2 (pH Profile) Sequentially prepare the following mixtures in a glass cuvet:5aSb 2.006a6b7a7b 8a8b9a9bmL 0.10 M phosphate, pH 6.0 mL 0.10 M phosphate, pH 6.5 mL...

Question

Day 2 (pH Profile) Sequentially prepare the following mixtures in a glass cuvet:5aSb 2.006a6b7a7b 8a8b9a9bmL 0.10 M phosphate, pH 6.0 mL 0.10 M phosphate, pH 6.5 mL 0.10 M phosphate, pH 7.0 mL 0.10 Mphosphate, pH 7.5 mL 0.10 M phosphate, pH 8.0 mL deionized water mL chymotrypsin stock (keep on ice) mL 0.0010 MHC2.002.002.002.002.002.002.002.002.000.500.500.500.500.500.500.500.500.500.500.400.400.400.400.400.400.400.400.400.40For each run, follow procedure for Day Steps 2-6 except use Biospec-180

Day 2 (pH Profile) Sequentially prepare the following mixtures in a glass cuvet: 5a Sb 2.00 6a 6b 7a 7b 8a 8b 9a 9b mL 0.10 M phosphate, pH 6.0 mL 0.10 M phosphate, pH 6.5 mL 0.10 M phosphate, pH 7.0 mL 0.10 Mphosphate, pH 7.5 mL 0.10 M phosphate, pH 8.0 mL deionized water mL chymotrypsin stock (keep on ice) mL 0.0010 MHC 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.40 0.40 0.40 0.40 0.40 0.40 0.40 0.40 0.40 0.40 For each run, follow procedure for Day Steps 2-6 except use Biospec-1800 in kinetic mode to determine rates_ See the following page for parameter settings (CHY) The "b" runs serve as controls for the non-enzymatic hydrolysis of NPA, and are necessary because the rate of aqueous hydrolysis of NPA is strongly pH dependent:



Answers

Tris (hydroxymethyl)aminomethane, commonly called TRIS or Trizma, is often used as a buffer in biochemical studies. Its buffering range is $\mathrm{pH} 7$ to $9,$ and $K_{\mathrm{b}}$ is $1.19 \times 10^{-6}$ for the aqueous reaction
$$
\left(\mathrm{HOCH}_{2}\right)_{3} \mathrm{CNH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left(\mathrm{HOCH}_{2}\right)_{3} \mathrm{CNH}_{3}^{+}+\mathrm{OH}^{-}
$$
a. What is the optimal pH for TRIS buffers?
b. Calculate the ratio $[T R I S] /\left[T R I S H^{+}\right]$ at $p H=7.00$ and at
p H=9.00
c. A buffer is prepared by diluting 50.0 $\mathrm{g}$ TRIS base and 65.0 $\mathrm{g}$ TRIS hydrochloride (written as TRISHCl) to a total volume of 2.0 $\mathrm{L}$ What is the pH of this buffer? What is the pH after 0.50 $\mathrm{mL}$ of 12 $\mathrm{MHCl}$ is added to a 200.0 -mL portion of the buffer?

First off for part A the optimal but the optimal pH for the truce buffer Will the Pete will be the pH that equals the PK of tress. So if we know the KB Weaken divided into K W to get K A take, the negative log of that and 8.8 is the optimal pH of the tress buffer. If the P H is equal to seven, then the ratio can be calculated by using the Henderson Hassle Baulch equation. With target, pH equals P K A plus the log of the ratio. Rearranging it, we get a ratio of 8.4 times 10. The negative, too, at Ph. Nine will do the same calculation, and we'll get a ratio that's equal to 8.32 Then for Part C. If we had 50 grams of tress with the molar mass of 1 21.14 this will be the moles of tress. We divide that by the moles of Tris Hydrochloride, which is 65 g divided by its smaller mass, and we get one is our ratio. So the log of 10 and P h equals P K. In the case of the addition of HCL toe on Lee, 1/10 of the total volume, 200 mL instead of two leaders, then the moles of tress present in 200 mL is 1/10. What we had in two leaders and the moles of Tris hydrochloride is 1/10 what we had in two leaders. We then subtract off the moles of strong base, added. I'm sorry. Subtract off the moles of strong acid added from the weak base and add on the moles of strong acid added to the weak acid because every mole of strong acid added will convert a mole of weak base into weak acid. This then gives us a pH of 7.95

The given problem are solved in four parts for but a we need to find the pH off the solution. We can find the pH through the formula with just P H is equal to peek a plus. Log off salt divided by the asset which is given over here, substituting the values we have the final pH of the solution, which turns out to be 4.2 for but be let us assume that the volume off buffer solution is one liter. Then the number off mourns off Ben's weak acid. It will be 0.1 moles, given that 20% off the asset by most converted into Ben's Oni on upon adding a base, that means 0.20 moles off asset is decreased on 0.20 moles off Bendel meet increases. We can find the number off moves by subtracting 0.1 bye 0.0 toe, which turns out to be 0.8 moles for the road. We can find the pH off the solution, which turns out to be 4.37 For part C, let us assume that the volume off buffer solution is one leader again, we can find the number off moles off benzo neat, using the same procedure as earlier, and the number off more stones are Toby 0.1 more, given that 20% off the Contra Kate, based by moves converted into Ben's weak acid. Upon adding an acid solution, we can calculate the number off moles by adding zero point one more plus 0 +102 mole, which is 0.12 moles. We can further find the pH of the solution. The stones Out Toby 4.2 for But D The justification? No, the answer. In part B and party, I don't agree with each other because in part B, the effect off based upon acidic buffer is absorbed. Vera's for the part. See, the effect off acid on an acidic buffer is a blow, so the result nph for the two cases are different.

This question is asking us to determine how many grams of sodium di hydrogen, phosphate and di sodium hydrogen phosphate we need to prepare one leader of phosphate cover at Ph. Seven. So I have written the molecular weight of the two species involved, and, um as pH increases, uh, sodium Hydrogen phosphate is deep protein ated into di sodium hydrogen phosphate, and I have just written the desired pH and the P K a, um, of these species for the sake of ease. So the first thing we need to do is determine how much of the acid and the conjugate base are present in the solution. The question has told us that the total concentration of phosphate is, uh, 0.1 Mueller, which is basically the some of the acid and the conjugate base. Um, so we will just arbitrarily define what these are. So I will say the acid is X and the that makes the conjugate base 0.1 minus X. We will use the Henderson Hasselbach equation, which is a very useful equation and chemistry, too. Um, figure out these concentrations, so yeah, we already know the pH. Oh, I wrote the Ph wrong earlier. That is my bad. PH should be, uh, seven point, though, So we know the pH. It is seven. Uh, we know the PKK, which is 7.2, and we are trying to find the ratio, and we have previously defined are contacted base as 0.1 minus X and our acid Aztecs. The first step in rearranging this equation is subtracting 7.2 from seven, which is negative 0.2. And the inverse of log is taking 10 to the power of what we had on our other side, which is, I believe, 0.631 and we set that equal to zero point one minus X over X and 631 times X equals 0.1 minus x. So we're just gonna group, um, the terms with X on one side, giving us 1.631 X equals 0.1. And then when we divide by 1.631 on both sides, we get X equals zero point on 06 to Mueller, which is our concentration of acid as we defined X earlier on. So our concentration of the conjugate base as we have previously defined is 0.1 minus X, which we now know is 0.62 which equals 0.38 Mueller. So just to write that in terms that make it easier to think about our phosphate buffer, uh, this is our acid equals H A equals 0.6 to Mueller and our conjugate piece. Oops. Let's try that again. Our contact of based di sodium hydrogen phosphate aged one. Yeah. Star contacted Base. Awesome. Now that we know the concentrations, we can get into the math. So as we know, polarity is moles divided by volume and leaders. So when we have 0.6 to Mueller, we really have 0.62 moles. Her leader and these are calculations for our acid. So we're gonna do some study geometry. Um, if you remember previously ah, we were given the molecular weight in the question and the molecular weight of sodium di hydrogen phosphate is 100 and 38 g per remote. So, um, this is basically how you do start geometry. You try to have the units equivalent across, and, um, we know that one mole is 138 g. I hope you know how to do strike geometry and we basically multiply everything along the top, multiply everything along the bottom and then divide the tube. So once we do that, we end up with 8.56 g of sodium dire hydrogen phosphate. And basically what this times H 20 means is that it's a hydrate. Now we do the same thing for our conjugate base. Mhm 0.38 moles per liter and then one mole was 142 g as per the molecular weight. And then when we follow the same procedure, we end up with 5.34 g of di sodium hydrogen phosphate hydrate. So this is our answer right here. I wasn't the prettiest, but yep, that's basically it. And thank you for joining me for this video. I hope my process made sense


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