# science( chem)

Find the pH of a solution prepared by disolving 1.00g of glycine amide hydrochloride (BH+) (mwt=110.54g/mol) plus 1.00g of glycine amide (B)(mwt=74.08) in 0.10L
PKa=8.04

I got pH= 8.21

Is that correct? This question actually has 4 extra parts but I wanted to make sure I got this part right before I went on to the next part. If this part is correct then I'll post the other parts and see if those are right.
thanks

I get 8.21 also.

okay

part b)
How many grams of glycine amide should be added to 1.00g of glycine amide hypochloride (BH+) to give 100ml of sol with a pH of 8.00?

For this I got 0.611g of (B)
I double checked my work but if you find this to be incorrect I'll post my work.

Thanks Dr.Bob

I forgot that this was refering to the sol in part a)

um..ignore this ..I got ahead of myself b/c I'm doing part c which refers to part a
sorry

I might as well post part c while I'm at it..

Part c)
What would the pH be if the solution in part a) were mixed with 5.00ml of 0.1M HCl ?

For this I got pH = 8.17
Is it fine?

And if anyone wants to look at my answer..
Part d) What would the pH if the solution in part c) were mixed with 10.00ml of 0.100M NaOH?

I'm not sure if what I did before is right since nobody said anything..asumming what I calculated before was correct...since I'm not really sure how to do this I'll post my work on this Q

since before I added in part c) 5.0e-4mol of H+ and now I'm adding 1.0e-3mol NaOH

[HB+]= (9.04e-3molHB+)+(5.0e-4mol H+)-(1.0e-3mol OH-)/( 0.115L)= 0.07426M HB+

[B]= (0.01349M B)-(5.0e-4mol H+)+(1.0e-3mol OH-)/(0.115L)= 0.1216M B

[H30+]=(9.12e-9)[(0.07426M HB+)/(0.1216M B)]=5.568e-9

pH= -log(5.568e-9)=8.25

It looks like it's correct since the pH went up a little since I added more NaOH than I did HCl but I'm not sure If that is correct.
I would appreciate someone checking my work for me..Thanks =)

I have 0.611 g also.

I have 8.17 also.

I didn't go through any of your work. My answer was 8.25. We need to look at common sense. The solution in A was 8.21 if we didn't err. Adding HCl to that should make it more acid and it does with 8.17. Adding NaOH should make it more basic and it does with 8.25. So the numbers pass the common sense test.

I was thinking that too but I made a error on the calculations before and got a similar answer but only had a 0.01 difference.

For part e) I have a problem with this Q and I'm confused

What would the pH be if the solution in part a) were mixed with 90.46ml of 0.1M NaOH? (this would be the exact quantity required to neutralize the glycine amide hydrochloride (BH+).

For this one I was thinking of the neutralization at the eq point eq where the eqzn to solve this would relate to the basic salt produced
BUT there is a second source of B from the buffer so...

I was thinking
that since the eqzn from the bk was
[OH-]=�ãKb*cB

that it would be
cB= (9.046e-3mol B from the acid)+(0.01349mol B from the conjugate base)/(0.09046L+ 0.1L)= 0.11832M B

Kw=Ka*Kb
Kb=1.096e-6

[OH-]=�ã(1.096e-6)(0.11833M B)=
pOH=-log(3.60e-4)=3.44
14-3.44= 10.56

somehow this looks completely wrong...help please...=)

that weird squiggly line above the a in the [OH-] eqzn is a radical but I guess you'd know that but I don't know if it is possible to make a radical sign...it (do you know ?)

I can't make the radical sign. I write, for square root 4, sqrt(4) = 2.
I don't have time right now to look at the final problem but I'll get back if no one else does.

Okay...

If that neutralizes the acid exactly, then you can calculate mols base formed from that neutralization, add the base you had at the beginning, divide mols by L to get M, then write the hydrolysis equation for the salt you have and solve it as you would for any other salt. You suggested that, more or less, when you said you were thinking of the equivalence point. That will get it I believe.

I did that but I'm still not sure about the equivalence pH which was 10.56 which just looks wrong but ...I would still appreciate it if you looked at what I did as this is the part I'm most unsure about out of all the parts to this question

this is extremely random but zimmerman reagent is very dangerous from what I just saw...My jeans have about 4 more holes and 1 is 1/2 an inch in diameter...scary....

When the jeans are washed the holes get bigger and small holes appear where there were no holes before.

I was thinking that the detergent reacted with the residue of the reagent but is that possible or it doesn't react with detergent?

but wait a sec...is that last question right or not? I really want to know if I calculated it correctly please..=)

I went through the calculation and obtained 10.55 and your method is ok. I used 0.0135 for 1/74.08 and not 0.0134. That made the molarity come out to 0.11838 which I rounded to 0.1184 and not 0.1183. I think that is the difference.

See response above. I got a little mixed up with all of the parts and going back and forth between them. But I think your method is ok for all of the parts.

XD

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