jkn1121

I just want to make sure, D is True right?

ph3: Structure 2 ph7-8: Structure 2 ph11: Structure 1 A. T since the structure is hydrophic then structure 2 would be capactible B. F since it at ph 11, therefore sturcture is more basic to be capatible C. F, since the structure 2 has the charge to make it acidic

Because it has the charge and it goes with the lower ph since it more acidic

its a neutral structure and its in a chargeless stage but if the unpronated structure is with a pka of 9.2, i think it should be structure 2 instead, right?

Structure 1 is hydrophobic since its neutral and nonpolar can pass through the membrane easily I think the answer is A. F since the pka is very high it wont be soluble in a low ph B. T since the pka will be solube in a hgh ph C. F, ph too low D. T, ph similar close to ph 7,8

1. Benzylamine is a chemical used as a precursor in the synthesis of some drugs. The amino group in benzylamine has a pKa of 9.3. The other groups in this molecule can be considered not ionizable. Structure 1 The picture above shows the unprotonated form of the benzylamine. Draw the protonated state of this compound. Structure 2 * Suppose to be a NH3 Which structure (or structures) will predominated at the following pHs? (Structure 1 is the original picture (unprotonated) and structure 2 is the one you drew (protonated). pH 3: ______Structure 2____________________________ pH 7 – 8: _____Structure 1__________________________ pH 11: _______________Structure 1__________________ I dont undrstand how this can link into a T/F answerable Based on the structures above (your structure and the original unprotonated structure), which is true about benzylamine? (mark true or false) a. This molecule will be more soluble in water at low pH (for example at pH 3) _____________ b. This molecule will be more soluble in water at high pH (for example at pH 11) _____________ c. If ingested, this molecule is more likely to be absorbed (go through the cellular membrane) in the stomach, at a pH of ~ 1 _____________ d. If ingested, this molecule is more likely to be absorbed in the intestines, at pH ~ 7 or 8 _____________

If we have 3-Heptanone and 2-Heptanone, what value would we get for 3-Heptanone? Also is the C=O group in the 2-Heptanone structure is replaced by the C=S group or the C=Se group. What dipol moment values would you expect for the resulting structure. I'm not sure how to answer this

Actually that question was kind of off-topic

An enzymatic reaction takes place in 10 mL of a solution that has a total citrate concentration of 120 mM and an initial pH of 7.00. During the reaction 0.2 millimoles of acid are produced. a) Using the data in the table below, calculate the final pH of the solution. B) What would be the final pH in the absence of the citrate buffer? Assume that the other components of the solution have no significant buffering capacity and that the solution is initially at pH 7.00. c) What would be the final pH if the solution contained 120 mM of phosphate buffer instead of citrate buffer? Acid Ka (M) pKa ___________________________________ Citric Acid 7.41 x 10-4 3.13 (pK1) Citrate- 1.74 x 10-5 4.76 (pK2) Citrate2- 3.98 x 10-6 5.40 (pK3) H3PO4 7.08 x 10-2 2.15 (pK1) H2PO4- 1.51 x 10-7 6.82 (pK2) HPO42- 4.17 x 10-13 12.38 (pK3) I think using the pH= pka+ log (base/acid) I first take 10m x 120mM= 1200moles, then using 1/2(pka1+pka2) to determine which pka to use ill use pk1 and pk2 and then i get 3.945. Subtract then plug 1200mol for base, 3.945 for pka, and base as .2mM. final ph 7.72?

oh crap i take He g/mol instead of H2, so the answer is 302.21g

I was wondering if the Stronger the Acid, the lower the pka then in a case if a pka=10, then in would most likely be soluble in 0.1M HCl rather than a 0.1M NaOH ?

If P and T cancels out, so R is given and V is 24.8 dm^3, I calculate for n PV= nRT V= nR n= V/R n= 24.8/.0826 = 302.21mol * 4g/mol = 1208g

What do you mean?

For A, I said 9-10 since a buffer doesnt change the pH much and when I drew the titration curve the pH range around that area, I guess. I was wondering for D, the relationship between the pH and pka is that when the pH is lower than the pka, the concentration increases since comparing the conc. of NH3, at pH 9, the conc of NH3 is 0.080, then when the pH was 10, the conc. was .028

Ok, so if you are figuring out the volume, M= mol/L ------> L=mol/M L= 0.052/5M NaOH = 0.0104 or is it 0.052/0.1M = 0.52

Are you trying to find the molarity of OH?

so you subtract the difference to get 0.052 where did the 0.028 come from i only see 0.020

so the final answer is 0.028 of 5M NaOH needed to be added

So you have 2 equations: 2.512 = (0.1 - [g-NH3+])/[g-NH3+] [g-NH2] = (0.1-[g-NH3+] ________________________________ [g-NH2] = (0.1-[g-NH3+] = (0.1- .080) = 0.02 2.512= 0.1- ( 0.080/[g-NH3]) 2.512/0.02 = 125.6

I think I get up to the point where its : 2.512 = 0.1 -[(g-NH2+])/[g-NH3+]) then you cross multiply to get 2.512*[g-NH3+] =0.1- NH2, I dont understand [g-NH3+] = 0.1/3.512 part

Standard Stage: Vol= 24.8 dm^3 Pressure: 1 atm I'm not sure where to start or formula to use, not sure if the formula PV=nRT?

Ok, I see what I did... I added the negative sign, so once I get 2.512= 0.1 -[(g-NH2+])/[g-NH3+]) would you then divide? Is the 0.080 part of theNH3+ conc.?

0.4= log ((0.1 -[g-NH3+])/[g-NH3+]) Anti Log .398 = 0.1 [(g-NH3+])/[g-0.080+])

So are you saying that the pH range can be from 1-11 since the concentrations are equal to each other?

Would Part C besince 5M*1L= 5moles 10= 9.6 log(5/HA) , so then .4= log (5/HA) then 10^ -0.4 = 5/HA, to 2.51 (HA)=5 then A =2?