Biochem Question

[cypress]

From this equation:

 

2.512*[g-NH3+] + [g-NH3+] = 0.1 because

 

2.512 of [g-NH3+] plus 1 of [g-NH3+] is 3.512 of [g-NH3+]

 

so

 

2.512*[g-NH3+] + [g-NH3+] = 0.1 ----> 3.512*[g-NH3+] = 0.1

 

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

 

Try again..... you start with 0.080 and end with 0.028 so you have to add what amount?

[jkn1121]

so you subtract the difference to get 0.052

 

where did the 0.028 come from i only see 0.020

Edited August 31, 2010 by jkn1121

[cypress]

Yes, you are not finished with the question. To reduce the concentration of the NH3+ form from 0.08 m to 0.028 in 1L you need to add 0.08-0.028 m of OH-

if you have 5M NaOH how much hydroxide solution is needed to make this amount of OH- ?

[jkn1121]

Are you trying to find the molarity of OH?

 

 

[cypress]

so you subtract the difference to get 0.052

 

where did the 0.028 come from i only see 0.020

 

I think you calculated it from 0.1/3.512 = 0.028 which is the concentration of the NH3 form at pH 10.0

 

Are you trying to find the molarity of OH?

 

Not the molarity, the volume of 5M hydroxide that you must add to raise the pH from 9.0 to 10.0. The volume that would contain 0.052 m of OH- which is the amount of OH- required to reduce the concentration of g-NH3+ from 0.08 to 0.028 right?

[jkn1121]

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

[cypress]

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

 

 

The first equation is correct (since you are adding 5M NAOH) and the answer is 0.0104 or 10.4 ml including significant digits its 10 ml. good.

 

What did you come up with for a) ?

[jkn1121]

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

[cypress]

I'm not too fond of your answer for a) but if you are ok with it so be it.

 

You described the relationship correctly but why not just use the same equations we have been using to find the pH when 99% of the glycine is in the NH3 form?

[jkn1121]

What do you mean?

[jkn1121]

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 ?

Edited August 31, 2010 by jkn1121

[cypress]

Your question does not figure into the answer to d) Use the equations and process used in the previous questions to answer what pH would be when 99% of glycine is in its NH3+ form. Lay out the formula and I'll check your work.

[jkn1121]

Actually that question was kind of off-topic